4.4.1 Symbol: ⁸O⁶
Explanation: The atomic number is 6, which indicates the presence of six protons in the nucleus and six electrons orbiting around it. The number of neutrons can be calculated by subtracting the atomic number from the mass number, giving us 8 neutrons (14 - 6).
4.4.2 Symbol: ²⁴Na¹¹
Explanation: Sodium has an atomic number of 11 and a mass number of 24. This means it contains 11 protons and 13 neutrons in its nucleus.
4.4.3 Symbol: ¹³⁰I⁵³
Explanation: Iodine has 53 protons (atomic number) and a mass number of 130. By subtracting the atomic number from the mass number, we find that it has 77 neutrons in its nucleus.
4.4.4 Symbol: ¹⁷O
Explanation: Oxygen has an atomic number of 8 and a mass number of 17. Subtracting the atomic number from the mass number gives us the number of neutrons, which is 9.
4.5.1 Atomic structure of helium: The atomic number of helium is 2, and its atomic weight is 4. It has two electrons in its first shell. The atomic structure is represented as:
He: 2
4.5.2 Atomic structure of lithium: The atomic number of lithium is 3, and its atomic weight is 6.94. The atomic structure is shown below:
Li: 3
2,1
4.6 Symbol; Element; Net Charge: The information is incomplete, so the answer cannot be provided.
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rank the following in order of decreasing δ and energy of light absorbed. a: [cr(en)3]3 b: [cr(cn)6]3− c: [crcl6]3−
The order of decreasing δ and energy of light absorbed for the compounds [Cr(en)3]3+, [CrCl6]3-, and [Cr(CN)6]3- is as follows: [Cr(en)3]3+ > [CrCl6]3- > [Cr(CN)6]3-.
In the given order, [Cr(en)3]3+ has the highest value of δ and absorbs light with the highest energy. This can be attributed to the presence of the ethylenediamine ligands (en), which are strong field ligands. The strong field ligands cause a larger splitting of the d-orbitals in the central chromium ion, resulting in a higher energy gap between the ground state and excited states. Therefore, [Cr(en)3]3+ exhibits a higher δ and absorbs light with higher energy.
On the other hand, [Cr(CN)6]3- has the lowest value of δ and absorbs light with the lowest energy. This is because cyanide ligands (CN) are weak field ligands, leading to a smaller splitting of the d-orbitals and a lower energy gap. As a result, [Cr(CN)6]3- has the lowest δ and absorbs light with lower energy compared to the other two compounds.
In between these, [CrCl6]3- falls in the middle with intermediate values of δ and energy of light absorbed. Chloride ligands (Cl) are moderately strong field ligands, causing a moderate splitting of the d-orbitals and an intermediate energy gap.
In summary, the order of the compounds with decreasing δ and energy of light absorbed is [Cr(en)3]3+ > [CrCl6]3- > [Cr(CN)6]3-. This order is determined by the strength of the ligands and the resulting splitting of the d-orbitals, which influences the energy gap and the energy of light absorbed by the compounds.
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How many grams of {ZnSO}_{4} are there in 223 grams of an aqueous solution that is 21.8 % by welght {ZnSO} . { g } {ZnSO}_{4}
Given the aqueous solution is 21.8% by weight of {ZnSO4}.We can use this information to find out how many grams of {ZnSO4} are there in 100 grams of the aqueous solution. We then use this value to find out how many grams of {ZnSO4} are there in 223 grams of the solution.
Using the formula:% By weight of ZnSO4 = (Weight of ZnSO4 / Weight of Aqueous Solution) x 10021.8 = (Weight of {ZnSO4} / 100) x 100Weight of {ZnSO4} in 100 g of Aqueous solution = 21.8 gNow, we can use the concept of ratios to find the weight of {ZnSO4} in 223 g of the solution.Weight of {ZnSO4} in 1 g of the solution = 21.8/100 gWeight of {ZnSO4} in 223 g of the solution = 223 x 21.8/100 g
Weight of {ZnSO4} in 223 g of the solution = 48.67 gTherefore, there are more than 100 grams of {ZnSO4} in 223 grams of the given aqueous solution. Specifically, there are 48.67 grams of {ZnSO4}.
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Classify each of the following complexes as either paramagnetic or diamagnetic:
[Zn(OH2)6]2+, [Cr(en)3]3+
Paramagnetic refers to the ability of a substance to become magnetized in the presence of an external magnetic field. When all of the electrons are paired, the substance is diamagnetic and does not show magnetic properties. Zn(OH2)6]2+ has no unpaired electrons. It is therefore diamagnetic and does not show magnetic properties. Cr(en)3]3+ has an unpaired electron. It is therefore paramagnetic and shows magnetic properties.
Paramagnetic and diamagnetic are the two categories of magnetic properties. In the presence of a magnetic field, diamagnetic substances exhibit a repulsive force, whereas paramagnetic substances exhibit an attractive force. The electrons in diamagnetic substances are all paired up in their respective orbitals, thus they are unaffected by a magnetic field. Whereas, paramagnetic substances have unpaired electrons that will orient themselves in the direction of the magnetic field and exhibit magnetic behavior.
[Zn(OH2)6]2+
Zinc(II) has a d10 electron configuration, with no unpaired electrons in the outermost shell. So, [Zn(OH2)6]2+ complex ion does not have any unpaired electrons and will not exhibit any magnetic behavior. Thus, it is a diamagnetic complex.
[Cr(en)3]3+
When a complex is formed with a transition metal such as chromium, the coordination compounds can exhibit paramagnetic behavior if they have at least one unpaired electron. In the outermost shell, Cr(III) has 3 d electrons, which could be either paired or unpaired. Chromium(III) complex ion [Cr(en)3]3+ has three chelating ethylenediamine (en) ligands, resulting in an octahedral coordination geometry. All of the electrons in chromium are paired except one, which is in the t2g orbital, and it has one unpaired electron in the e g orbital, which causes it to become paramagnetic.
In conclusion, [Zn(OH2)6]2+ complex ion has no unpaired electrons and will not exhibit any magnetic behavior. Thus, it is a diamagnetic complex. Whereas, [Cr(en)3]3+ complex ion is paramagnetic because it has one unpaired electron, which causes it to become paramagnetic.
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The freezing point of water: A. is 500^{\circ} \mathrm{C} B. does not exist C. decreases with increasing pressure D. decreases with decreasing pressure
The freezing point of water decreases with decreasing pressure. Thus, option D is correct.
The freezing point of water decreases with decreasing pressure. This phenomenon is known as the "freezing point depression." When the pressure on water decreases, such as at high altitudes or in a vacuum, the freezing point of water is lower than the standard freezing point at atmospheric pressure (0 °C or 32 °F).
As pressure decreases, the molecules in the water have less force pushing them together, making it more difficult for them to arrange themselves into a solid crystal lattice. Therefore, the freezing point of water decreases. This is why water can remain in a liquid state at temperatures below 0 °C (32 °F) in high-altitude regions or under low-pressure conditions, such as in certain laboratory experiments.
It's worth noting that while decreasing pressure lowers the freezing point of water, increasing pressure generally has the opposite effect, raising the freezing point.
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Part IV. Preparation of 100 {~mL} 25 % Solution X Materials: Solution X, measuring cylinder, distilled water, and parafilm. Show calculation steps. (1) Calculate the volume of
In Part IV of the experiment, we are preparing a 100 mL 25% solution X using Solution X, a measuring cylinder, distilled water, and parafilm. The calculation steps for this preparation are as follows:
Calculation of the volume of Solution X:
We know that we need 25 mL of Solution X to make 100 mL of a 25% solution X. The volume of Solution X needed can be calculated using the following formula:
Volume of Solution X = (25 mL/100 mL) x 100 mL = 25 mL
Therefore, 25 mL of Solution X is needed to prepare 100 mL of a 25% solution X.
Calculation of the volume of distilled water:
To calculate the volume of distilled water needed, we can use the following formula:
Volume of distilled water = Total volume - Volume of Solution X
= 100 mL - 25 mL
= 75 mL
Therefore, 75 mL of distilled water is needed to prepare 100 mL of a 25% solution X.
Mixing of Solution X and distilled water:
Now that we have calculated the volume of Solution X and distilled water needed, we can mix them together to prepare the 25% solution X. We can use a measuring cylinder to measure 25 mL of Solution X and pour it into a clean, dry beaker. Next, we can measure 75 mL of distilled water using the same measuring cylinder and add it to the beaker containing Solution X. We can then thoroughly mix the contents of the beaker using a stirring rod to ensure that the Solution X is well dissolved in the distilled water.
Finally, we can use parafilm to cover the beaker and label it with the name of the solution, concentration, and date of preparation. This will help prevent contamination and ensure that the solution can be easily identified if needed.
Hence, by following the above-mentioned steps, we have successfully prepared 100 mL of a 25% solution X.
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6. A U-tube is fitted with a semi-peeable membrane and then filled. On the left side pure wate introduced, while the right side is given a 0.200 {M} aqueous solution of {KI} \
U-tube is a device made up of a glass or plastic tube in the shape of the letter U that is bent at its center at the same point. U-tube is often used in laboratories to compare densities or liquid levels in two vessels that are open to the air, with the purpose of determining the liquid level height difference between the two arms.
KI is a potassium iodide, which is an inorganic chemical compound. It is a salt with a crystalline structure that is white to colorless and occurs naturally in minerals and seawater. The purpose of adding this solution to the right side is to determine the concentration of the solution in the left side of the tube, which has pure water in it.
As a result, the iodide ions will move from the 0.200 M solution of KI to the left side of the U-tube, which has pure water. This will result in an increase in the concentration of KI in the left arm of the U-tube.
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a 5.000 g sample of an organic hydrocarbon is combusted and the products measured. in the reaction, 15.37 g of carbon dioxide and 7.186 g of water are produced. assuming the oxygen used for the combustion was in excess, determine the empirical formula of the hydrocarbon
The empirical formula of the hydrocarbon is CH2.
To determine the empirical formula of the hydrocarbon, we need to find the moles of carbon and hydrogen in the given amounts of carbon dioxide and water. Calculate the moles of carbon dioxide (CO2) and water (H2O) using their respective molar masses.
Moles of CO2 = 15.37 g / molar mass of CO2
Moles of H2O = 7.186 g / molar mass of H2O
Determine the ratio of moles of carbon to moles of hydrogen in the hydrocarbon. Since the empirical formula represents the simplest whole-number ratio of atoms in a compound, we divide the number of moles by the smallest value obtained.
In this case, the moles of carbon in the hydrocarbon are equal to the moles of carbon dioxide, and the moles of hydrogen are twice the moles of water.
Therefore, the empirical formula of the hydrocarbon is CH2.
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4. (3 pts) Thiophenol ({C}_{6} {H}_{5} {SH}) is a weak acid with a {pK}_{a} of 6.6 . Would you expect thiophenol to be more soluble in a 0.1
Thiophenol ({C6H5SH}) is a weak acid with a pKa of 6.6. Solubility is a measure of a substance's ability to dissolve in a solvent.
When the solute's molecules interact favorably with the solvent's molecules, solubility is maximized. As a result, the solubility of a substance is frequently influenced by the solvent's properties. As a result, the solubility of thiophenol in a 0.1M sodium hydroxide (NaOH) solution can be determined as follows. The answer is the first one. When thiophenol ({C6H5SH}) is added to the NaOH solution, it will deprotonate. The following equation depicts the deprotonation of thiophenol to form the thiophenol anion ({C6H5S-}): C6H5SH (aq) + NaOH (aq) → C6H5S- (aq) + H2O (l)This deprotonation reaction is favored because the Na+ ion interacts favorably with the C6H5S- ion, while the H2O molecule interacts poorly with the C6H5SH molecule. As a result, thiophenol is more soluble in a 0.1M NaOH solution than in water because the reaction drives the equilibrium to the right and the thiophenol ion's solubility is greater in the basic solution than in water.
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which quantity must change to have a transmutation? a) oxidation number b) atomic mass c) atomic number d) electrical charge
The correct option is c) Atomic number.
Transmutation is the conversion of one chemical element or isotope into another. The quantity that must change for a transmutation is atomic number. Transmutation can be described as the conversion of one chemical element into another. It can also be described as a change in the atomic nucleus that results in the conversion of one element into another. In order for a transmutation to occur, the number of protons in the nucleus of the atom must change. This means that the atomic number must change. The atomic number is the number of protons in the nucleus of an atom. If the number of protons changes, then the element itself will change. For example, the transmutation of uranium into lead is a well-known example of this process. Uranium has an atomic number of 92, while lead has an atomic number of 82. In order for this transmutation to occur, the number of protons in the nucleus of the atom must change from 92 to 82.
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According to the Michaelis-Menten equation, what is the ratio of V/Vmax when [S] = 15Km? Express the ratio as a decimal rounded to 2 decimal places_ b. (5 pts) According to the Michaelis-Menten equation, if the ratio of V/Vmax is 0.30,what is the value of [SJKu ? Express the ratio as a decimal rounded to 2 decimal places. c-d. (10 pts) An experiment is performed in which the enzyme acetylcholinesterase converts two different substrate molecules_ A and B, to product. The table below shows kinetic data for the enzyme operating on and The first two columns show velocity data at different concentrations of A; the last two columns show velocity data at different concentrations of B_ Note that the bottom row shows the calculated Vmax for A and for B. [A] (uM) V (uM/sec) [B] (uM) V (uWsec) 19 5 33 20 70 12 66 45 135 18 86 65 175 28 110 90 215 48 139 100 228 110 175 130 265 180 190 440 405 220 195 700 443 Vmax 220 Vmax 530 By inspecting the table (no math needed), determine the Km of the enzyme for substrate A in terms of UM: Enter the value of Ku (without unit) in question 8 on the online answer form By inspecting the table (no math needed); determine the Kv of the enzyme for substrate B in terms of UM Enter the value of K (without unit) in question 9 on the online answer form: Assume that for the enzyme, the Kn values of the substrate indicate the binding affinities of the substrates for the active site. Which substrate, or B, has higher_binding affinity for the active site? Select the correct answer from the options in question 10 on the online answer form
a. The ratio of V/Vmax when [S] = 15Km according to the Michaelis-Menten equation cannot be determined without additional information.
b. If the ratio of V/Vmax is 0.30 according to the Michaelis-Menten equation, the value of [S] cannot be determined without additional information.
c. By inspecting the table, the Km of the enzyme for substrate A in terms of μM cannot be determined.
The Michaelis-Menten equation describes the relationship between the substrate concentration ([S]), the maximum reaction velocity (Vmax), and the Michaelis constant (Km) in enzyme kinetics.
However, the ratio of V/Vmax when [S] = 15Km cannot be determined without knowing the specific values of Vmax and Km or having additional data points.
b. Similarly, if the ratio of V/Vmax is given as 0.30, the value of [S] cannot be determined without additional information. The Michaelis-Menten equation relates the ratio V/Vmax to the substrate concentration [S], Vmax, and Km.
Without knowing any of these values, it is not possible to determine the specific concentration of [S].
c. By inspecting the table, we can gather information about the velocities at different concentrations of substrates A and B.
However, the Km of the enzyme for substrate A in terms of μM cannot be determined solely by inspecting the table.
The Km value represents the substrate concentration at which the reaction velocity is half of Vmax. In the given table, the Km value is not directly provided.
The Michaelis-Menten equation is a fundamental concept in enzyme kinetics, describing the relationship between substrate concentration and enzyme activity.
The equation provides insights into the catalytic efficiency and substrate binding affinity of enzymes.
To determine specific values such as V/Vmax, [S], Km, or substrate binding affinity, precise experimental measurements or additional data points are required.
Understanding these parameters helps in studying enzyme kinetics, optimizing enzyme reactions, and designing effective enzyme inhibitors or activators.
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pleas assign peaks for the 13C NMR of trans cinnamic acid
Trans-cinnamic acid is an organic compound with the formula C6H5CH=CHCO2H. The 13C NMR spectrum of trans-cinnamic acid will have the following peaks assigned: The phenyl ring exhibits a total of five distinct peaks in the 13C NMR spectrum.
Chemical shift (ppm)Carbon atoms160.13C=O129.5α-carbon (next to carbonyl group)128.
0β-carbon (double bond carbon)131.2, 129.3, 128.5, 126.8, 126.0
Phenyl ring (five carbons)132.1, 129.6, 129.5, 129.2, 128.6
For trans-cinnamic acid, the number of carbon environments is five, as it has a carbonyl group (C=O) and a phenyl ring. In the 13C NMR spectrum, the carbonyl group is usually the highest peak and the chemical shift is the lowest. The chemical shift for α-carbon is greater than that of the β-carbon because the α-carbon is closer to the carbonyl group.
The chemical shift values for the β-carbon are higher than those for the α-carbon because they are further away from the electron-withdrawing carbonyl group.In the phenyl ring, all five carbon atoms have different chemical shift values. Carbon 2 (C2) has the highest chemical shift, whereas carbon 6 (C6) has the lowest chemical shift.
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from n=3 to n=6 J (energy) s−1 (frequency) m (wavelength) radiation is emitted radiation is absorbed (b) from n=9 to n=3 J (energy) s−1 (frequency) m (wavelength) radiation is emitted radiation is absorbed (c) from n=7 to n=4 ] (energy) s−1 (frequency) m (wavelength)
From the question;
1) The frequency is 2.75 * 10^14 Hz
2) The frequency is 3.25 * 10^16 Hz
3) The frequency is 1.4 * 10^14 Hz
What is the energy levels?The energy levels can be obtained from the Rydberg formula.
We know that;
1/λ = RH(1/n1^2 - 1/n2^2)
1/λ = 1.097 * 10^7 (1/3^2 - 1/6^2)
λ = 1.09 * 10^-6 m
E = hc/λ
E = 6.6 * 10^-34 * 3 * 10^8/ 1.09 * 10^-6
= 1.82 * 10^-19 J
E = hf
f = E/h
f = 1.82 * 10^-19 J/ 6.6 * 10^-34
f = 2.75 * 10^14 Hz
2)
1/λ = 1.097 * 10^7 (1/3^2 - 1/9^2)
λ = 9.2 * 10^-9 m
E = hc/λ
E = 6.6 * 10^-34 * 3 * 10^8/ 9.2 * 10^-9
E = 2.15 * 10^-17 J
E = hf
f = 2.15 * 10^-17 J/ 6.6 * 10^-34
f = 3.25 * 10^16 Hz
3)
1/λ = 1.097 * 10^7 (1/4^2 - 1/7^2)
λ = 2.2 * 10^-6 m
E = 6.6 * 10^-34 * 3 * 10^8/2.2 * 10^-6
= 9 * 10^-20 J
f = 9 * 10^-20 J/6.6 * 10^-34
f = 1.4 * 10^14 Hz
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Use equation(s) to show how you would synthesize the
following and use curved arrows
to outline the mechanism of each.
a) Bromocyclopentane from an alkene
b) 2-butanol from an alkene
a) Bromocyclopentane can be synthesized from an alkene through a radical bromination mechanism, involving the addition of bromine radicals (Br·) to the alkene.
b) 2-Butanol can be synthesized from an alkene through acid-catalyzed hydration, where the alkene undergoes addition of water (H₂O) and subsequent proton transfer reactions.
a) To synthesize bromocyclopentane from an alkene, the reaction can be carried out using a radical bromination mechanism. The overall reaction equation is as follows:
Alkene + Br₂ → Bromocyclopentane
The mechanism involves three steps: initiation, propagation, and termination.
Initiation: The bromine molecule (Br₂) is homolytically cleaved by UV light or heat, forming two bromine radicals (Br·).
Br₂ → 2Br·
Propagation:
A bromine radical (Br·) abstracts a hydrogen atom from the alkene, generating an alkyl radical.
Br· + Alkene → Alkyl Radical
The alkyl radical reacts with a bromine molecule (Br₂), resulting in the formation of a bromoalkane and a new bromine radical.
Alkyl Radical + Br₂ → Bromoalkane + Br·
Termination: The bromine radicals (Br·) can undergo various termination reactions, such as recombination or reaction with impurities or solvent molecules, to form stable products and stop the radical chain reaction.
Overall, these steps outline the mechanism of the radical bromination reaction that converts an alkene into bromocyclopentane.
b) To synthesize 2-butanol from an alkene, the reaction can be carried out using acid-catalyzed hydration. The overall reaction equation is as follows:
Alkene + H₂O + H⁺ → 2-Butanol
The mechanism involves the addition of water to the alkene under acidic conditions, leading to the formation of an intermediate carbocation, followed by nucleophilic attack and subsequent proton transfer.
Protonation of the alkene:
The alkene reacts with the acid catalyst (H⁺), resulting in the protonation of the double bond.
Alkene + H⁺ → Carbocation
Nucleophilic attack by water:
Water (H₂O) acts as a nucleophile and attacks the carbocation, leading to the formation of an oxonium ion.
Carbocation + H₂O → Oxonium Ion
Proton transfer:
A proton is transferred from the oxonium ion to a water molecule, resulting in the formation of 2-butanol and regeneration of the acid catalyst.
Oxonium Ion + H₂O → 2-Butanol + H⁺
This mechanism demonstrates how an alkene can be converted to 2-butanol through acid-catalyzed hydration, involving the addition of water and subsequent proton transfer reactions.
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What did we say ATP is similar to?.
Answer:
ADP is similar to a drained battery, while ATP is like to a charged battery. With the addition of water to the substrate, ATP can be hydrolyzed into ADP, releasing energy.
Explanation:
calculate Ph of a buffer solution when 25 mL of 2M sodium acetate is mixed with 5mL of 1M acetic acid and diluted to 100mL of pure water . Assume pka ~ 5 for acetic acid. What would be Ph if the solution were further diluted with pure water to 1L.
The pH of the buffer solution, when initially diluted to 100 mL, is 6. If further diluted with pure water to 1 L, the pH remains 6 as the concentration of [A-] and [HA] does not change.
calculate the pH of the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Volume of sodium acetate (NaAc) = 25 mL
Concentration of sodium acetate (NaAc) = 2 M
Volume of acetic acid (HAc) = 5 mL
Concentration of acetic acid (HAc) = 1 M
Volume of final solution = 100 mL
1: Calculate the moles of NaAc and HAc used:
Moles of NaAc = concentration * volume
Moles of NaAc = 2 M * 0.025 L (since 25 mL = 0.025 L)
Moles of NaAc = 0.05 mol
Moles of HAc = concentration * volume
Moles of HAc = 1 M * 0.005 L (since 5 mL = 0.005 L)
Moles of HAc = 0.005 mol
2: Calculate the total moles of acetate ions ([A-]) and acetic acid ([HA]) in the solution:
Total moles of acetate ions ([A-]) = moles of NaAc
Total moles of acetate ions ([A-]) = 0.05 mol
Total moles of acetic acid ([HA]) = moles of HAc
Total moles of acetic acid ([HA]) = 0.005 mol
3: Calculate the concentration of acetate ions ([A-]) and acetic acid ([HA]) in the solution:
Concentration of acetate ions ([A-]) = moles of acetate ions / volume of final solution
Concentration of acetate ions ([A-]) = 0.05 mol / 0.1 L (since 100 mL = 0.1 L)
Concentration of acetate ions ([A-]) = 0.5 M
Concentration of acetic acid ([HA]) = moles of acetic acid / volume of final solution
Concentration of acetic acid ([HA]) = 0.005 mol / 0.1 L
Concentration of acetic acid ([HA]) = 0.05 M
4: Calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 5 + log(0.5/0.05)
pH = 5 + log(10)
pH = 5 + 1
pH = 6
The pH of the buffer solution, when initially diluted to 100 mL, is 6.
If the solution is further diluted with pure water to 1 L, the pH of the buffer will remain the same since the concentration of [A-] and [HA] will not change.
Therefore, the pH will still be 6.
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which of the following should have the lowest boiling point? group of answer choices n2 na2s nh3 hf
Answer:
Nitrogen gas (N2) should have the lowest boiling point among the given options. This is because N2 is a nonpolar molecule with weak London dispersion forces between its molecules, which results in a relatively low boiling point. Sodium sulfide (Na2S) is an ionic compound, so it has a very high boiling point due to strong electrostatic forces between its ions. Ammonia (NH3) and hydrogen fluoride (HF) are polar molecules that can form hydrogen bonds between their molecules, which results in higher boiling points than N2.
Explanation:
Nitrogen gas (N2) should have the lowest boiling point among the given options. This is because N2 is a nonpolar molecule with weak London dispersion forces between its molecules, which results in a relatively low boiling point. Sodium sulfide (Na2S) is an ionic compound, so it has a very high boiling point due to strong electrostatic forces between its ions. Ammonia (NH3) and hydrogen fluoride (HF) are polar molecules that can form hydrogen bonds between their molecules, which results in higher boiling points than N2.
1. Compound X has a solubility in toluene of 20mg per 100ml at 20C, and a solubility of 0.60 g per 100ml at 75C. You are given 0.52 g sample of compound X that is contaminated with 35mg of compound Y. Use this infoation to answer questions A&B. A. If compound Y is completely soluble in toluene at all temperatures, describe the steps to purify X to obtain the maximum % recovery. Calculate the % recovery. (4 pts) B. If compound X has a solubility in toluene of 20mg per 100ml at 20C, and a solubility of 0.60 g per 100ml at 75C, what can be done to purify compound Y by recrystallization?
Recrystallization allows for the purification of compounds based on differences in solubility between the desired compound and impurities. By choosing an appropriate solvent system, compound Y can be selectively recrystallized, resulting in a purer sample.
A. To purify compound X and obtain the maximum % recovery, you can follow these steps:
1. Determine the solubility of compound Y in toluene at the given temperatures (20°C and 75°C). Since it is stated that compound Y is completely soluble in toluene at all temperatures, its solubility is not a limiting factor.
2. Dissolve the 0.52 g sample of compound X, contaminated with 35 mg of compound Y, in the minimum amount of toluene required to fully dissolve compound X at the higher temperature (75°C). This ensures that both compound X and Y are in the solution.
3. Slowly cool the solution to room temperature (20°C). As the temperature decreases, compound X's solubility in toluene decreases, resulting in the crystallization of compound X. Compound Y, being completely soluble, remains in the solution.
4. Filter the solution to separate the solid crystals of compound X from the liquid solution containing compound Y.
5. Wash the solid crystals of compound X with a cold solvent (such as cold toluene) to remove any impurities or residual compound Y.
6. Allow the washed solid crystals of compound X to dry, either by air-drying or under vacuum, to remove any remaining solvent.
7. Weigh the purified compound X obtained from the solid crystals. Calculate the % recovery using the formula:
% recovery = (mass of purified compound X / initial mass of compound X) * 100
B. To purify compound Y by recrystallization, you need to consider its solubility characteristics. Since compound Y is completely soluble in toluene at all temperatures, recrystallization using toluene alone may not be effective.
However, you can explore recrystallization using a different solvent system that has a selective solubility for compound Y. The general steps for recrystallization are as follows:
1. Choose a suitable solvent or solvent mixture that exhibits a temperature-dependent solubility behavior for compound Y. The solvent should have a low solubility for compound Y at low temperatures and a higher solubility at elevated temperatures.
2. Dissolve the impure sample of compound Y in the minimum amount of hot solvent required to fully dissolve it. If necessary, you can use gentle heating to aid dissolution.
3. Filter the hot solution to remove any insoluble impurities or undissolved material.
4. Cool the filtered solution slowly to room temperature or lower temperatures, allowing compound Y to crystallize out. The slower the cooling rate, the larger and purer the crystals obtained.
5. Collect the crystals of compound Y by filtration and wash them with a cold portion of the recrystallization solvent to remove any remaining impurities.
6. Dry the purified crystals of compound Y, either by air-drying or under vacuum, to remove any residual solvent.
Recrystallization allows for the purification of compounds based on differences in solubility between the desired compound and impurities. By choosing an appropriate solvent system, compound Y can be selectively recrystallized, resulting in a purer sample.
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For the Gluep prepared with 2 Tbsp of borax, indicate the
similarities and differences between this gluep and the first
sample.
For the Gluep prepared with 2 Tbsp of borax, some similarities and differences between this gluep and the first sample are given below.
Similarities:Both the glueps contain the same ingredients such as Elmer’s glue, water, and food coloring. Both the glueps are non-toxic and safe for children to play with. Both the glueps are polymers and behave in a similar way to other polymer substances.
Differences:The first sample of gluep is more fluidic and easy to pour compared to the gluep prepared with 2 Tbsp of borax. The second gluep is more viscous and behaves like a solid when force is applied. The first sample of gluep is more transparent and clearer compared to the gluep prepared with 2 Tbsp of borax. The second gluep is more opaque and thicker. The first sample of gluep can be peeled off from the surface, while the gluep prepared with 2 Tbsp of borax behaves like a solid and cannot be peeled off.
Gluep is a simple and fun experiment that is easy to prepare with only a few common household ingredients. It is an example of a polymer that behaves as both a solid and a liquid. Elmer's glue contains a polymer called polyvinyl acetate (PVA) which is responsible for the glue's adhesive properties. When borax is added to the glue, the PVA molecules cross-link to form a network of chains, making the glue thicker and more elastic.
In conclusion, both the glueps have similarities and differences, with the first sample being more transparent and easier to pour while the gluep prepared with 2 Tbsp of borax being more viscous and behaving like a solid. Both glueps are polymers and non-toxic, making them safe for children to play with.
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a two step reaction mechanism is proposed for a gas phase reaction, as represented above. which of the follwoing correctly identifies both the chemical equation for step 1 and the rate law for the overall reaction? Step 1: (slow) Step 2: NO3(g) + CO(g) + NO2(g) + CO2(g) (fast) Overall: NO2(g) + CO(g) + NO(g) + CO2(g) A two-step reaction mechanism is proposed for a gas-phase reaction, as represented above. Which of the following correctly identifies both the chemical equation for step 1 and the rate law for the overall reaction? A ) The chemical equation for step 1 is 2 NO2(g) → NO(g) + NO3(g), and the rate law is rate = (NO,]. ( B ) The chemical equation for step 1 is NO2(g) + CO(g) + NO(g) + CO2(g), and the rate law is rate = [NO2][CO). c) The chemical equation for step 1 is NO3(g) + 2 CO(g) + NO(g) + 2 CO2(g), and the rate law is rate = [NO2][CO). D) The chemical equation for step 1 is NO3(g) + NO2(g) + 2 CO(g) + NO(g) + NO2(g) + 2 CO2(g), and the rate law is rate = [NO3][NO2][CO]".
The chemical equation for step 1 is 2 NO2(g) → NO(g) + NO3(g), and the rate law is rate = [NO2].
In the proposed two-step reaction mechanism, step 1 is the slow step, while step 2 is the fast step. In step 1, the chemical equation is 2 NO2(g) → NO(g) + NO3(g). This equation suggests that two molecules of NO2 react to form one molecule of NO and one molecule of NO3. Since step 1 is the slow step, it determines the overall rate of the reaction.
The rate law for the overall reaction is determined by the rate-determining step, which is step 1 in this case. The rate law is an expression that relates the rate of the reaction to the concentrations of the reactants. The rate law for the overall reaction can be written as rate = k[NO2], where k is the rate constant and [NO2] represents the concentration of NO2. This rate law indicates that the rate of the reaction is directly proportional to the concentration of NO2.
In summary, the chemical equation for step 1 is 2 NO2(g) → NO(g) + NO3(g), and the rate law for the overall reaction is rate = [NO2].
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When myoglobin is in contact with air (at sea level), how many parts per million of carbon monoxide ( μmol CO per mol of air) are required to tie up 5% of the myoglobin? The partial pressure of oxygen required to half-saturate myoglobin at 25∘C is 3.7kPa. The partial pressure of carbon monoxide required to half-saturate myoglobin in the absence of oxygen is 0.009kPa. Air is 21% oxygen and 79% nitrogen.
When myoglobin is in contact with air (at sea level), 8.4 μmol CO per mol of air is required to tie up 5% of the myoglobin.
How to solve this?We know that air contains 21% oxygen and 79% nitrogen, so the partial pressure of oxygen is given by;Partial pressure of oxygen = 21/100 x 101.3 kPa= 21.213 kPa.
The partial pressure of carbon monoxide required to half-saturate myoglobin is 0.009 kPa. This means that if the partial pressure of CO is 0.009 kPa, half of the myoglobin will have carbon monoxide (CO) bound to it.
Now let's calculate the partial pressure of oxygen needed to saturate myoglobin;The partial pressure of oxygen required to half-saturate myoglobin at 25∘C is 3.7 kPa.
Therefore, the partial pressure of oxygen required to saturate myoglobin completely is given by;Partial pressure of oxygen (P02) required to saturate myoglobin completely = 3.7 x 2 = 7.4 kPa.
Now we can calculate the amount of CO required to tie up 5% of myoglobin using the Hill equation.
The Hill equation is given by;θ=[P02]^n / ([P02]^n + [P50]^n), where;θ = fractional saturation[P02] = partial pressure of oxygen at 50% saturationn = Hill coefficient, and[P50] = partial pressure of oxygen required for 50% saturation.
Here, n = 1 because myoglobin binds oxygen cooperatively and P50 = 3.7 kPa.θ=0.5[7.4]^1 / ([7.4]^1 + [3.7]^1)θ=0.5[7.4] / ([7.4] + [3.7])θ=0.5[7.4] / 11.1θ= 0.249.
The fractional saturation of myoglobin is 0.249 when the partial pressure of oxygen is 3.7 kPa.
To calculate the partial pressure of CO required to tie up 5% of the myoglobin, we will use the same Hill equation, but this time we will substitute P02 with Pco because we want to find the partial pressure of CO required for 5% saturation.θ=[Pco]^n / ([Pco]^n + [P50]^n)Here, n = 1 because myoglobin binds CO cooperatively and P50 = 0.009 kPa.θ=0.05[7.4]^1 / ([Pco]^1 + [0.009]^1)θ= 0.37 / ([Pco] + 0.009)
We are looking for [Pco] such that θ=0.05 and [Pco] is in μmol CO per mol of air. This means that;θ=0.05= [CO bound to myoglobin] / [myoglobin].
Since we want to tie up 5% of the myoglobin, we can assume that all the CO is bound to the myoglobin. So;[CO bound to myoglobin] = 0.05 x [myoglobin]
Now, the number of moles of myoglobin in a given volume can be calculated using the ideal gas law;PV = nRT, where;P = pressureV = volume of the gasR = ideal gas constant T = temperature n = number of moles and n = PV/RT
We can assume that the volume of air is 1 mol since we are looking for the concentration of CO in μmol CO per mol of air. Also, the temperature is 25°C = 298K and R = 8.31 J/mol.K, so;n = 101.3 kPa x 1 mol / (8.31 J/mol.K x 298K)n = 40.7 mol. So the number of moles of myoglobin is;n = PV/RT = (7.4 kPa x 1 mol) / (8.31 J/mol.K x 298K) = 0.0029 mol
Now we can find the total number of μmol of myoglobin;Total μmol of myoglobin = 0.0029 mol x 6.02 x 1023 molecules/mol x 150 g/mol = 2.62 x 1019 μmol
Now we can calculate the number of μmol of CO required to tie up 5% of myoglobin;[CO bound to myoglobin] = 0.05 x [myoglobin]0.05 x 2.62 x 1019 μmol = 1.31 x 1018 μmol CO
We can now calculate the concentration of CO in μmol CO per mol of air;θ=0.05 = [1.31 x 1018 μmol CO] / [μmol CO per mol of air x 2.62 x 1019 μmol]μmol CO per mol of air = [1.31 x 1018 μmol CO] / [0.05 x 2.62 x 1019 μmol] = 8.4 μmol CO per mol of air.
Therefore, when myoglobin is in contact with air (at sea level), 8.4 μmol CO per mol of air is required to tie up 5% of the myoglobin.
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A chemist prepares a solution of mercury(I) chloride Hg2Cl2 by
measuring out 0.00000283μmol of mercury(I) chloride into a 200.mL
volumetric flask and filling the flask to the mark with water.
Calcula
The given information is as follows: Amount of mercury(I) chloride = 0.00000283 μmolVolume of the volumetric flask = 200 mLWe have to calculate the concentration of the solution, which is measured in molarity (M).Molarity is the number of moles of solute present in one litre (1 L) of the solution.
Therefore, molarity (M) can be calculated using the formula as follows: Molarity (M) = Number of moles of solute/ Volume of solution (in litres)Given, the volume of solution is 200 mL, which is equal to 0.2 L. The number of moles of solute can be calculated as follows: Number of moles of
Hg2Cl2 = mass of Hg2Cl2/Molar mass of Hg2Cl2Molar mass of Hg2Cl2 = Atomic mass of mercury (Hg) × 2 + Atomic mass of Chlorine (Cl) × 2 = (200.59 g/mol × 2) + (35.45 g/mol × 2) = 401.18 g/mol + 70.90 g/mol = 472.08 g/mol Mass of Hg2Cl2 = 0.00000283 μmol × 472.08 g/mol = 0.001336 g = 1.336 mg Now, the number of moles of Hg2Cl2 = 1.336 mg/ 472.08 g/mol = 0.00000282 moles Therefore, the molarity (M) of the solution is: Molarity (M) = 0.00000282 moles/ 0.2 L = 0.0000141 M. Hence, the concentration of mercury(I) chloride Hg2Cl2 in the solution is 0.0000141 M.
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For a hypothetical reaction of A --> B occurring in the cell, the ΔG is +3 kJ/mol and the ΔGo' is -2 kJ/mol for a reaction occurring at 25oC.
What is the ratio of [A]/[B] found in the cell?
Possible answers are:
0.13
2.01
5
7.5
The ratio of [A]/[B] found in the cell is 2.01. Option B is correct.
Given that the ΔG for a hypothetical reaction of A = B occurring in the cell is +3 kJ/mol and the ΔGo' is -2 kJ/mol for a reaction occurring at 25oC.
We are to find the ratio of [A]/[B] found in the cell.
To calculate the ratio of [A]/[B] found in the cell, we will make use of the Gibbs free energy equation that is given as follows:
ΔG = ΔGo' + RT ln([B]/[A])
whereΔG = Gibbs free energy of the reaction
ΔGo' = Standard Gibbs free energy of the reaction
R = Ideal gas constant = 8.314 J/mol
K = 0.008314 kJ/mol K
T = temperature in Kelvin
= 298 K [A] and [B] are the concentrations of the reactants A and product B, respectively.
The ratio of [A]/[B] can be obtained by rearranging the Gibbs free energy equation as follows:
ln([B]/[A]) = (ΔG - ΔGo') / RT[B]/[A]
= e^[ΔG - ΔGo') / RT]
Substitute the given values into the above equation as follows:
[B]/[A] = e⁵ / (0.008314 × 298)] = 2.01
Therefore, Option B is correct.
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Balance equation and solve 3 and 4 in the image
Answer: Step By Step explanation:
Explanation: Assume the coefficients of compound and molecule to be
a, b, c and d respectively. Then solve it by the algebraic method of balancing equation used in the following attachment.
A chemist, Dr. V.A. Pore, wishes to detect an impurity in a certain compound that she is making. There is a test that has sensitivity of 0.9 and specificity of 0.95. That is, the test is positive for an impurity when an impurity is present 90% of the time, and the test is negative for an impurity when no impurity is present 95% of the time. About 15\% of Dr. Pore's compounds contain an impurity. 9. A compound is selected at random from Dr. Pore's output. The test indicates that an impurity is present. What is the conditional probability that the selected compound actually has an impurity? 1 10. Another compound is selected at random from Dr. Pore's output. The test indicates that an impurity is not present. What is the conditional probability that the selected compound is actually free of an impurity? 11. Two processes of a company produce rolls of materials. The rolls of Process I are 3% defective, and the rolls of Process II are 1% defective. Process I produces 60% of the company's output, Process II 40%. A roll is selected at random from the total output. Given that this roll is defective, what is the conditional probability that it is from Process I?
The correct options are:1.
Conditional probability that the selected compound actually has an impurity is 0.74.2.
Conditional probability that the selected compound is actually free of an impurity is 0.0185.3.
Conditional probability that the selected roll is from Process I given that it is defective is 0.64.
Here, we need to find out the probability that a selected compound has an impurity given that the test indicates an impurity is present.
P(A) = probability that a compound has impurity = 0.15
P(B) = probability that the test indicates an impurity is present
= 0.15 x 0.9 + 0.85 x 0.05
= 0.14 + 0.0425
= 0.1825P
(B|A) = probability that the test indicates an impurity is present given that the compound has impurity = 0.9
Therefore, by Bayes' Theorem,
P(A|B) = P(B|A) * P(A) / P(B)
= 0.9 * 0.15 / 0.1825
= 0.7370
≈ 0.74
Conditional probability that the selected compound actually has an impurity is 0.74.10.
Here, we need to find out the probability that a selected compound is actually free of an impurity given that the test indicates an impurity is not present.
P(A) = probability that a compound has impurity = 0.15
P(B) = probability that the test indicates an impurity is not present = 0.85 x 0.95 + 0.15 x 0.1 = 0.8075
P(B|A) = probability that the test indicates an impurity is not present given that the compound has impurity
= 0.1
Therefore, by Bayes' Theorem,
P(A|B) = P(B|A) * P(A) / P(B)
= 0.1 * 0.15 / 0.8075
= 0.0185
Conditional probability that the selected compound is actually free of an impurity is 0.0185.11.
Here, we need to find out the probability that the selected roll is from Process I given that it is defective.
Let A denote the event that a roll is from Process I and B denote the event that a roll is defective.
Then, we need to find out P(A|B).
P(A) = probability that a roll is from Process I = 0.6
P(B|A) = probability that a roll is defective given that it is from Process I = 0.03
P(B|A') = probability that a roll is defective given that it is from Process II = 0.01
P(A'|B) = probability that a roll is from Process II given that it is defective
Therefore, by Bayes' Theorem,
P(A|B) = P(B|A) * P(A) / [P(B|A) * P(A) + P(B|A') * P(A')]
= 0.03 * 0.6 / (0.03 * 0.6 + 0.01 * 0.4)
= 0.6429
≈ 0.64
Conditional probability that the selected roll is from Process I given that it is defective is 0.64.
Hence, the correct options are:1.
Conditional probability that the selected compound actually has an impurity is 0.74.2.
Conditional probability that the selected compound is actually free of an impurity is 0.0185.3.
Conditional probability that the selected roll is from Process I given that it is defective is 0.64.
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please help
1. How many significant figures do the following numbers have? a. 0.00345 b. 9.8 × 10^{-23} c. 340 d. 456.00 e. 3009
The significant figures in the given numbers are as follows:
a. 0.00345 : 3
b. 9.8 × 10^-23 : 2
c. 340: 2
d. 456.00: 5
e. 3009: 4
Significant figures are the digits in a number that carries meaning in terms of the accuracy or precision of the measurement. In a number, all the digits that are not zeros are significant, whereas trailing zeros are only significant if there is a decimal in the number. There are different rules for determining significant figures depending on the type of number.
Here are the rules for each type of number:
Rule for Non-zero numbers: All the non-zero digits are significant.Rule for leading zeros: All the leading zeros are not significant. Rule for Trailing zeros: The trailing zeros are significant only if there is a decimal in the number.Rule for exact numbers: The exact numbers have an infinite number of significant digits.Learn more about "significant figures" :
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it
is asking for the mass propane. the total number of atoms inside
the container is 6.880x10^26.
To find the mass of propane (C3H8) from the total number of atoms in a container, we need to use Avogadro's number and the molar mass of propane.
Avogadro's number is 6.022 x 10^23, which is the number of particles in one mole of a substance. The molar mass of propane is 44.1 g/mol, which means one mole of propane has a mass of 44.1 grams. We can use these values to find the mass of propane in the container, as shown below.
First, we need to find the number of moles of propane in the container. We can do this by dividing the total number of atoms by Avogadro's number:
6.880 x 10^26 atoms / 6.022 x 10^23 atoms/mol = 114.2 mol
Next, we can use the molar mass of propane to convert moles to grams:
114.2 mol x 44.1 g/mol = 5044 g
Therefore, the mass of propane in the container is 5044 grams.
Explanation:
Given,
Total number of atoms inside the container = 6.880 x 10^26
We are supposed to find the mass of propane (C3H8).
Now we will calculate the number of moles of propane present in the container. To calculate the number of moles, we use the Avogadro number.
Avogadro's number = 6.022 x 10²³ atoms/mole
Number of moles = Total number of atoms/ Avogadro's number
= 6.880 x 10²⁶ atoms / 6.022 x 10²³ atoms/mole
= 114.2 moles
Now, we will calculate the mass of propane using the molar mass of propane.
Molar mass of propane (C3H8) = 3 × Atomic mass of carbon + 8 × Atomic mass of hydrogen
= 3 × 12.01 u + 8 × 1.008 u
= 36.03 u + 8.064 u
= 44.094 u
Therefore, the molar mass of propane is 44.094 g/mol.
Mass of propane = Number of moles × Molar mass
= 114.2 moles × 44.094 g/mol
= 5044 g
The mass of propane inside the container is 5044 grams. The above explanation involves finding the number of moles using Avogadro's number and finding the mass of propane using its molar mass.
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Concept 2.1
Explain the following tes: matter, element, compound.
What are the four most abundant essential elements in
organisms?
What are four additional important elements in organisms?
W
Matter: Anything that has mass and takes up space is called matter.Element: A chemical substance consisting of atoms of the same number of protons in the nucleus.
For example, oxygen has eight protons in the nucleus, making it an element with an atomic number of 8.Compound: A substance formed when two or more chemical elements are chemically bonded together. Water, for example, is a compound that contains two hydrogen atoms and one oxygen atom (H2O). The four most abundant essential elements in organisms are carbon, hydrogen, nitrogen, and oxygen. Four additional important elements in organisms are calcium, phosphorus, potassium, and sulfur.
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the results of a separation using two-dimension gel electrophoresis are shown here.
The results of the separation using two-dimensional gel electrophoresis reveal the distribution and abundance of proteins in a sample.
Two-dimensional gel electrophoresis is a powerful technique used to separate complex mixtures of proteins based on their isoelectric point (pI) and molecular weight. The first dimension of this technique involves isoelectric focusing (IEF), where proteins are separated based on their charge. A pH gradient is established across the gel, and when an electric field is applied, proteins migrate towards the pH region where their net charge is zero, resulting in their separation according to their pI.
In the second dimension, the proteins from the first dimension gel are placed on top of a polyacrylamide gel, which is then subjected to sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE). In SDS-PAGE, proteins are separated based on their molecular weight. The proteins from the first dimension gel are now distributed along a single axis according to their pI and separated further by size during electrophoresis.
The resulting gel displays a complex pattern of spots, each representing a specific protein in the sample. By comparing the protein patterns obtained from different samples or conditions, researchers can identify changes in protein expression, post-translational modifications, or protein interactions. These results can provide insights into cellular processes, disease mechanisms, and biomarker discovery.
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Lithium, Sodium, and Calcium are all considered to be cations because they tend to when forming chemical bonds. gain protons lose electrons share protons share electrons gain electrons lose protons
Lithium, Sodium, and Calcium are all metals found in Group 1 and Group 2 of the periodic table, respectively. When these elements form chemical bonds, they tend to achieve a stable electron configuration by (b) losing electrons from their outermost energy levels.
This process results in the formation of positively charged ions known as cations.
By losing electrons, lithium, sodium, and calcium attain a lower energy state and a more stable electronic configuration, resembling the nearest noble gas configuration.
These cations then have a positive charge that attracts them to negatively charged species, such as anions, in ionic bonding.
Therefore, the correct answer is (b) lose electrons.
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Complete question :
Lithium, Sodium, and Calcium are all considered to be cations because they tend to when forming chemical bonds.
(a) gain protons
(b) lose electrons
(c) share protons
(d) share electrons
(e) gain electrons
(f) lose protons
Perform the following conversion:
83 grams = _________ megagrams
(Do not use scientific notation.)
The given value is 83 grams. So, 83 grams is equal to 0.000083 megagrams.
Converting grams to megagrams we get,1 megagram = 1,000,000 grams
So, 1 gram = 1/1,000,000 megagrams
Converting 83 grams to megagrams:
83 grams = 83/1,000,000 megagrams = 0.000083 megagrams
We can convert from grams to megagrams using the following formula:
1 megagram = 1,000,000 grams
Hence, 1 gram = 1/1,000,000 megagrams
To convert 83 grams to megagrams, we can use this formula and substitute the given value of 83 grams.
83 grams = 83/1,000,000 megagrams= 0.000083 megagrams
Therefore, 83 grams is equal to 0.000083 megagrams.
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