Answer:
a. Oxidized: Fe(s)
Reduced: Pb(NO3)2
b.Oxidized: Cu(s)
Reduced: AgNO3
c. Oxidized: Al(s)
Reduced: AgNO3
Explanation:
In a redox reaction, one reactant is been oxidized whereas the other is reduced.. The reduced reactant is the one that is gaining electrons and the oxidized one is loosing electrons.
In the reactions:
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
The Fe is as reactant as Fe(s) (Oxidation state 0) and the product is +3 (Because NO3, nitrate ion, is always -1). That means Fe is oxidized. The Pb as reactant is +2 and as product 0 (Gaining 2 electrons). Pb(NO3)2 is reduced
b. 2AgNO3(aq)+Cu(s)→2Ag(s)+Cu(NO3)2(a)
AgNO3 is +1 and Ag(s) is 0. AgNO3 is reduced. Cu(s) is 0 as reactant and +2 as product. Cu(s) is been oxidized
c. 3AgNO3(aq)+Al(s)→3Ag(s)+Al(NO3)3(aq)
Here, in the same way, AgNO3 is +1 as reactant and 0 as product. AgNO3 is reduced. And Al(s) is 0 as reactant but + 3 as product. Al(s) is oxidized.
Not all bonds are "created equal". From the following molecules, which one contains the most easily broken carbon to carbon bond? Group of answer choices H3C—CH3 F2C=CF2 H2C=CH2 HCCH
Answer:
H3C—CH3
Explanation:
The strength of a bond is indicated by the value of its bond dissociation energy. Simply put, the bond dissociation energy is the energy required to break the bond.
Carbon forms single, double and triple bonds with itself. As a matter of fact, carbon atoms can link to each other indefinitely. This is known as catenation and has been attributed to the low bond energy of the carbon-carbon single bond.
The bond energy of the carbon-carbon single bond is about 90KJmol-1 while that of carbon-carbon double bond is about 174KJmok-1. The carbon-carbon triple bond has the highest bond dissociation energy of about 230KJmol-1.
Hence, it is easier to break carbon-carbon single bonds than double and triple bonds respectively, hence the answer.
According to the forces of attraction, the molecule which can be easily broken is CH₃-CH₃ as it has a single bond with low dissociation energy as compared to double or triple bonds.
Forces of attraction is a force by which atoms in a molecule combine. it is basically an attractive force in nature. It can act between an ion and an atom as well.It varies for different states of matter that is solids, liquids and gases.
The forces of attraction are maximum in solids as the molecules present in solid are tightly held while it is minimum in gases as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.
The physical properties such as melting point, boiling point, density are all dependent on forces of attraction which exists in the substances.Single bonds have least dissociation energy while triple bonds have the maximum dissociation energy.
Thus,the molecule which can be easily broken is CH₃-CH₃.
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A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is
Answer:
1. EF = PSCl₃; 2. MF = PSCl₃
Explanation:
1. Empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our first job is to calculate the molar ratio of P:S:Cl.
Assume 100 g of the compound.
(a) Calculate the mass of each element.
Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.
(b) Calculate the moles of each element
[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]
(c) Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3
(d) Write the empirical formula
EF = PSCl₃
The empirical formula for this compound is PSCl₃.
2. Molecular formula
(a) Calculate the ratio of the molecular and empirical formula masses
n = (169.4 u)/(169.40 u) = 1.000 ≈ 1
(b) Calculate the molecular formula
MF = (EF)ₙ = (EF)₁ = PSCl₃
The molecular formula for this compound is PSCl₃.
18. Sucralose contains which two functional groups: (2 points)
A) benzene
B) halogen
C) carboxyl
D) hydroxy!
Answer:
The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.
Explanation:
An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.
The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.
To determine the concentration of chloride ion (Cl-) in a 100 mL sample of ground water, a chemist adds a large enough volume of AgNO3 solution to precipitate all Cl- as AgCl. The mass of the resulting precipitate is 93.9 mg. What is the chloride ion concentration in milligrams of chloride per liter of ground water
Answer:
[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]
Explanation:
Hello,
In this case, we can represent the chemical reaction as:
[tex]Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)[/tex]
In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:
[tex]m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-[/tex]
Finally, for the given volume of water in liters (0.100L), we compute the required concentration:
[tex][Cl^-]=\frac{23.2mgCl^-}{0.100L}\\[/tex]
[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]
Best regards.
The concentration of chloride ions in the groundwater sample is 230 mg/L.
Calculation of chloride ion concentration:Based on the given information,
The mass of the resulting precipitate, that is, AgCl is 93.9 mg or 0.0939 g. The molar mass of AgCl is 143.2 g/mol.Now the number of moles of AgCl precipitate can be calculated as,
n = Given mass/Molar mass
Now putting the values we get,
[tex]n = \frac{0.939 g}{143.32 g/mol} \\n = 6.5 * 10^{-4}[/tex]
Thus, 6.5 × 10⁻⁴ moles of AgCl comprises 6.5 × 10⁻⁴ chloride ions. Therefore, 6.5 × 10⁻⁴ of chloride ions are present in the sample of 100 ml.
Now the molar mass of chloride ion is 35.453 g/mol, the mass of chloride ion will be,
Mass = Mole × Molar mass
Mass = 6.5 × 10⁻⁴ moles × 35.453 g/mol
Mass = 0.0230 g or 23 mg
The volume of the groundwater sample is 100 ml or 0.1000 L.
Now the concentration of the chloride ions in the sample given is,
C = 23 mg/0.1000 L
C = 230 mg/L
Thus, the concentration of chloride ions in the groundwater sample is 230 grams per liter.
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What is the percent composition of sulfur in SO2?
OA
32.1 g
(32.1 g + 16.0 g + 16.0 g)
x 100%
32.1 g
O B.
x 100%
(16.0 g + 16.0 g)
O c.
(32.1 g + 16.0 g + 16.0 g)
X 100%
32.1 g
(32.1 g)(6.02 x 102)
O D.
100 g
Answer:
5 .04 percent.
Explanation:
it is also known as sulfur dioxide. so it's 5.0.4 percent.
Your question is incomplete. Read below to find the content.
0.4 % is the percent composition of sulfur in SO2.
How is SO2 prepared from sulfur?In the laboratory, sulfur dioxide is prepared by the reaction of metallic sulfite or a metallic bisulfite with dilute acid. For example, a reaction between the dilute sulphuric acid and sodium sulfite will result in the formation of SO2.
What do you mean by percentage composition?The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. Here, the quantity is measured in terms of grams of the elements present.
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What is/are the major organic product(s) of the following reaction, Question 2 options: A) CH3CH2CCH Br B) CH3CCCH3 Br C) CH2CH2 HCCH Br D) HCCCH2CH2Br E) HCCBr
Answer:
CH3CH2C≡CH
Explanation:
The particular reaction under study is known as the alkykation of acetylide ions. An acetylide ion can be alkykated using a suitable alkyl halide. The overall scheme of the reaction is;
CH≡C^- + RX -----> RC≡CH + X^-
This reaction is most effective when primary alkyl halides are used. It involves SN2 substitution of a halide in the alkyl halide by an acetylide ion. Secondary, tertiary or even bulky primary substrates are known to yield alkenes and alkynes owing to elimination by E2 mechanism.
Suppose you have a container filled with air at 212 oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N2 and 21% O2 for simplification. Calculate the mass of air and moles of O2 in the container.
Answer:
[tex]m_{air}=0.947g[/tex]
[tex]n_{O_2} =0.00686molO_2[/tex]
Explanation:
Hello,
In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:
[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]
In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:
[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]
Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:
[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]
Best regards.
This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False
Answer:
The given statement is false.
Explanation:
However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.So that the given is incorrect.
A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation:
Calculate the value of ΔG∘rxnΔGrxn∘ for the following reaction at 296 K. Ka = 2.9 × 10–8 and assume Ka does not change significantly with temperature. $$HClO(aq)+H2O(l) HClO−(aq)+H3O+(aq)
Answer:
[tex]\Delta G_{rxn}=42.7\frac{kJ}{mol}[/tex]
Explanation:
In this case, for the dissociation of hypochlorous acid, we know that the acid dissociation constant (Ka) is 2.9x10⁻⁸, which is related with the Gibbs free energy as shown below:
[tex]\Delta G_{rxn}=-RTln(K)[/tex]
But in this case K is just Ka, therefore, at 296 K, it turns out:
[tex]\Delta G_{rxn}=-8.314\frac{J}{mol*K}*296K*ln(2.9x10^{-8})\\\\\Delta G_{rxn}=42.7\frac{kJ}{mol}[/tex]
Such result, means that the reaction is nonspontaneous at the given temperature, it means it is not favorable (not easily occurring).
Best regards.
Draw the Lewis structure of acetaldehyde (CH₃CHO) and then choose the appropriate pair of molecular geometries of the two central atoms.
a) tetrahedral/trigonal pyramidal.
b) trigonal planar / linear.
c) tetrahedral / trigonal planar.
d) trigonal/tetrahedral.
e) planar /trigonal planar.
Given that,
Draw the Lewis structure of acetaldehyde (CH₃CHO).
We know that,
The Lewis structure shows the number of electrons around an atom.
According to structure,
We need to find the molecular geometries of the two central atoms
Using molecular geometries
For first central atom,
Number of bond pair = 2
Here, double bond to O count as single bond
The number of lone pair is zero.
The geometry is Trigonal planar.
For second central atom,
Number of bond pair = 4
The number of lone pair is zero.
The geometry is tetrahedral
Hence, The molecular geometries of the two central atoms are trigonal planar and tetrahedral.
(d) is correct option.
The central carbon atoms in acetaldehyde have a tetrahedral geometry and a trigonal planar geometry respectively.
Acetaldehyde has two central carbon atoms. The Lewis structure of acetaldehyde shows the arrangement of electrons around the atoms in the compound. The lone pairs are shown as dots while the bond pairs are represented using a single dash.
The first central carbon atom in acetaldehyde has a tetrahedral geometry while the second central carbon atom in acetaldehyde has a trigonal planar geometry.
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1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-lb woman. Show all your calculations. (1lb = 453.6 g)
2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations
Answer:
1. 0.138g of valium would be lethel in the woman
2. 125mg/min is the drip of the patient
Explanation:
1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.
A person that weighs 200lb requires:
200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)
90.72kg × (1.52mg / kg) =
137.9mg ≡
0.138g of valium would be lethel in the woman2. The IV contains 1.5g = 1500mg/mL.
If the patient is receiving 5.0mL/h, its rate in mg/h is:
5.0mL/h × (1500mg/mL) = 7500mg/h
Now as 1h = 60min:
7500mg/h × (1h / 60min) =
125mg/min is the drip of the patientFill in the blanks
1. the ____ atoms of the ____ groups and the hydrogen atoms attached to the ____ atoms form ____.
(answer choices; nitrogen, carbon, a-helices or b-pleated sheets, carbonyl, globular of firbrous structures, oxygen, amino, random coil, thiol)
2. In the a helix, hydrogen bonds form between the ____ oxygen atom and the ____ hydrogen atom of a peptide bond in _____. In the B-pleasted sheet, hydrogen bonds form between the ____ oxygen atom and the ____ hydrogen atom of a peptide bond in ____.
(answer choices; parallell sections of a long polypeptide chain, the next amino acid unit, carboxyl, the next turn of the helical chain, thiol, amino, carbonyl)
Answer:
1. oxygen; carbonyl; nitrogen; α-helices or β-pleated sheets
2. carbonyl; amino; the next turn of the helical chain; carbonyl; amino; parallel sections of a long polypeptide chain.
Explanation:
The secondary structure of proteins is of two major conformations, the α helix and β conformations which are very stable.
1. The oxygen atoms of the carbonyl groups and the hydrogen atoms attached to the nitrogen atoms form α-helices or β-pleated sheets.
2. In the α-helix, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in the next turn of the helical chain. In the β-pleated sheet, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in parallel sections of a long polypeptide chain.
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm L / mole K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.
Answer:
1. 0.98 atm
Explanation:
The following data were obtained from the question:
Mass of unknown liquid (m) = 0.3175 g
Temperature (T) = 99 °C
Pressure (P) = 748.2 mmHg
Volume (V) = 145.0 mL
Gas constant (R) = 0.08206 atm.L/Kmol
1. Determination of the pressure in atm.
760 mmHg = 1 atm
Therefore,
748.2 mmHg = 748.2/760 = 0.98 atm
Therefore, the pressure in atm is 0.98 atm.
Consider a triangle ABC like the one below. Suppose that C=83°, a = 43, and b = 44. Solve the triangle.
Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
If there is more than one solution, use the button labeled "or".
Answer:
Explanation:
In a triangle
a / sin A = b / sinB = c / sinC
Putting the values
43 / sin A = 44 / sinB
sinA / sinB = 43 / 44 = 1 / 1.023
A + B = 180 - 83 = 97
sinA / sin ( 97 - A ) = 1 / 1.023
sin 97 cos A - cos 97 sin A = 1.023 sin A
= .9925 cos A + .122 sin A = 1.023 sin A
.9925 cos A = .901 sin A
squaring
.985 cos²A = .8118 sin²A
.985 - .985 sin²A = .8118 sin²A
.985 = 1.7968 sin²A
sinA = .74
A = 47.73
B = 49.27
c / sin C = b / sin B
c = b sinC / sinB
= 44 x sin 83 / sin 49.27
= 44 x .9925 / .7578
= 57.62
1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G
Answer:
The reaction given in the question is:
2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)
The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:
ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.
The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.
Now the values of ΔH° and ΔS° are,
ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)
ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)
ΔH°rxn = -1277.56 kJ/mole
ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)
ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15
ΔS°rxn = 313.11 J/mole/K
Now the formula for calculating ΔG°rxn is,
ΔG°rxn = ΔH°rxn - TΔS°rxn
ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole
ΔG°rxn = -1370.86 kJ/mol
A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
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If the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%, how can the cell obtain more sugar? {Hint: Sugar is polar and does not pass through the cell membrane easily.}
Select one:
a. Sugar can undergo facilitated diffusion through a channel protein, and no energy is required.
b. The cell must use active transport to pump the sugar in. Energy is required.
c. Sugar can diffuse straight through the phospholipid bilayer.
d. The cell CANNOT obtain more sugar. It is doomed.
The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.
FACILITATED DIFFUSION:
Facilitated diffusion is the process whereby molecules move across a cell through the help of carrier/channel proteins. Facilitated diffusion is a type of passive transport and hence, does not require energy to occur. This is because movement occurs down a concentration gradient (high to low). According to this question, the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%. This means that the sugar can travel down a concentration gradient across the cell membrane. However, because sugar is a polar molecule and does not pass through the cell membrane easily, a carrier proteins is needed to aid its movement. Therefore, the cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.Learn more at: https://brainly.com/question/18122054?referrer=searchResults
The NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?
Answer:
The peaks are registered from tetramethyl silane (TMS)
Explanation:
Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.
Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.
I hope this helped.
Draw the structure of 1,4-hexanediamine.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.
View Available Hint(s)
Answer:
1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.
Explanation:
1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.
The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.
Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].
1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.
The structure of 1,4-hexanediamine is shown below.
Which best describes the total mass of a sample of water when it condenses
from a liquid to a gas?
A. The mass is less because the water molecules get closer together
and take up more space.
B. The mass is the same because the decrease in energy equals the
increase in the number of molecules.
C. The mass is the same because water molecules are not created or
destroyed during a phase change.
D. The mass is greater after water condenses because the mass of
the molecules increases.
Answer:
Its C I hopefully help you
Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.
Answer:
2NaOH(s) → Na₂O(s) + H₂O(g)
Hope that helps.
The blending of one s atomic orbital and three p atomic orbitals produces ________.
A three sp3
B four sp3
C three sp
D four sp2
E four sp
Answer:
B. four sp3
Hope that helps.
We have that for the Question "The blending of one s atomic orbital and three p atomic orbitals produces?"
Answer:
Option B = four [tex]sp^3[/tex]
Explanation:
When 1 s orbital blends with 3 p orbitals, they form a tetrahedrical shaped figure with each being a [tex]SP^3[/tex] orbital.. A total of 4 orbitals
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Natural atom of the same element may have the same _________?
A)proton
B)neutron
C)electron
D)All
Answer:B
Explanation:
Answer: i think it is c
Explanation: i checked my textbook.
In this experiment, you will analyze your sample by TLC by spotting pure benzophenone and your product and eluting with 5:1 hexanes/ethyl acetate. Choose the statement that BEST describes what you should observe for a successful experiment.
A) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value.
B) The benzophenone in the standard and reaction samples should travel the least and will have the same Rf value. No unreacted benzhydrol should be observed at a higher Rf value.
C) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. Unreacted benzhydrol should be observed at a lower Rf value.
D) The benzophenone in the standard and reaction samples should travel the farthest and will have the different Rf values. No unreacted benzhydrol should be observed at a lower Rf value.
Answer:
The correct answer is statement A.
Explanation:
In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.
For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.
The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.
What is TLC ?In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.
The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.
TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.
Thus, option A is correct.
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if your acetic acid buret was still wet inside with deionized water when you filled it with acetic acid?
Answer:
The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.
Balance this equation: __ UO2(s) + __ HF(ℓ) → __ UF4(s) + __ H2O(ℓ) Though you would not normally do so, enter the coefficient of "1" if needed. UO2(s) HF(ℓ) UF4(s) H2O(ℓ)
Answer:
The balanced equation is given below:
UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)
The coefficients are: 1, 4, 1, 2
Explanation:
UO2(s) + HF(l) —› UF4(s) + H2O(l)
The above equation can be balance as follow:
There are 2 atoms of O on the left side and 1 atom on the right side. It can be balanced by writing 2 in front of H2O as shown below:
UO2(s) + HF(l) —› UF4(s) + 2H2O(l)
There are 4 atoms of F on the right side and 1 atom on the left side. It can be balanced by writing 4 in front of HF as shown below:
UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)
Now, the equation is balanced.
The coefficients are: 1, 4, 1, 2.
Which statement describes global winds?
They flow from the same direction.
They travel over short distances.
They generate land breezes.
They blow away from the poles to the equator.
Idk the answer
Answer:
They blow away from poles to the equator.
Explanation:
Hello,
In this case, we must take into account that global wind systems are formed by the constant increase in the temperature of the Earth’s surface. Thus, they drive the oceans’ surface currents. In such a way, we can say wind is the basic movement of air from an area of higher pressure to an area of lower pressure, for that reason they blow away from the poles to the equator.
Best regards.
The statement that describes the global winds is they travel over short distances.
What is winds?
Wind is a pattern or type of the movement of the natural air or any other composition of gases over to the relative position of the planet's surface.
Global winds are those winds which can travel in a straight path and originated due to global convention currents. Global winds always move from west to east direction and travels short distances only.
Hence, option (2) is correct.
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If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?
The Average atomic mass of phosphorus is 29.9.
What is Average atomic mass ?The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.
Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.
An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.
The versions of an element with different neutrons have different masses and are called isotopes.
The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,
Average atomic mass of P = ∑(Isotope mass) (its abundance)
∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)
Abundance of isotope = % of the isotope / 100.
∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.
Hence , The Average atomic mass of phosphorus is 29.9.
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