Answer:
Manganese (II) ion, Mn²⁺
Explanation:
Hello,
In this case, given the overall reaction:
[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]
Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺
Best regards.
A volumetric flask contains 25.0 mL of a 14% m/V sugar solution. If 2.5 mL of this solution is added to 22.5 mL of distilled water, what is the % m/V of the new solution.
Answer:
The new solution is 1.4% m/V
Explanation:
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:
[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]
We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
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. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.
Answer
Naphthalene is a non electrolyte
If the unknown compound is an electrolyte it gives 2 or more ions in solution
( NaCl >> Na+ + Cl- => 2 ions
Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)
the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )
For naphthalene
delta T = 1.86 x m
for a salt that gives 2 ions
delta T = 1.86 x m x 2
hence the lowering in freezion point of unkown is greater then napthalene
Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page?
The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6
Answer:
Compounds whose real bond angle are the same as ideal bond angle;
SF6, BF3, CH4, PCI5
Compounds whose real bond angles differ from ideal bond angles;
H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5
Explanation:
According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.
The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.
For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.
Chemistry question. Image attached.
Answer:
The answer to your question is given below
Explanation:
The balanced equation for the reaction is given below:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
1. Writing an expression for the equilibrium constant, K.
The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Thus, we can write the equilibrium constant, K for the reaction as follow:
CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)
K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²
2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/
Answer:
10.328 m
Explanation:
normal atmospheric pressure = 101325 Pa
density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3
pressure = pgh
where p = density
g = acceleration due to gravity = 9.81 m/s^2
h = height of column
imputing values, we have
101325 = 1000 x 9.81 x h
height of column h = 101325/9810 = 10.328 m
What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12
Answer:
pH = 3.39
Explanation:
The equilibrium in water of ascorbic acid (With its conjugate base) is:
H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)
Where the acidic dissociation constant is written as:
Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]
H₂O is not taken in the Ka expression because is a pure liquid.
As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:
[H₂C₆H₆O₆] = 2.5x10⁻³M - X
[HC₆H₆O₆⁻] = X
[H₃O⁺] = X
Replacing in the Ka expression:
7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]
1.975x10⁻⁷ - 7.9x10⁻⁵X = X²
0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷
Solving for X:
X = -0.00048566→ False solution, there is no negative concentrations
X = 0.00040666 → Right solution
As [H₃O⁺] = X, [H₃O⁺] = 0.00040666
pH is defined as -log [H₃O⁺];
pH = -log 0.00040666,
pH = 3.39Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.
Answer:
2ErF3 + 3Mg → 2Er + 3MgF2
Explanation:
Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.
What does the empirical formula tell you about the compound?
The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?
Can you see what type of functional group this compound could have?
Answer:
See explanation.
Explanation:
Hello,
In this case, we can show how the empirical formula is found by following the shown below procedure:
1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:
[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]
2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:
[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]
3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:
[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]
4. Compute the moles of oxygen by using its molar mass:
[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]
5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:
[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]
6. Search for the closest whole number (in this case multiply by 2):
[tex]C_3H_6O_2[/tex]
Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:
[tex]M=12*3+1*6+16*2=74g/mol[/tex]
Which is about three times in the molecular formula, for that reason, the actual formula is:
[tex]C_9H_{18}O_6[/tex]
It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.
Best regards.
Which of the following is NOT a type of crystal structure? A. None of these B. Metallic C. Ionic D. Macromolecular (giant covalent)
Answer:
A. None of these.
Explanation:
A crystal structure is an arrangement of atoms or ions in a repeating three-dimensional array.
B. is wrong. Metal atoms, such as gold, arrange themselves into a crystal structure.
C. is wrong. Ionic solids, such as sodium chloride, arrange themselves into a crystal structure.
D. is wrong. Macromolecules (network solids), such as diamond, arrange themselves into a crystal structure.
The correct answer is None of these.
What is a crystal structure?A crystal structure is a three-dimensional collection of atoms or ions that repeats itself.Metal atoms(gold), Ionic solids (sodium chloride), and Macromolecules(network solids) arrange themselves into a crystal structure.Learn more about crystal structure here:-
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You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3
Complete Question
The complete question is shown on the first uploaded image
Answer:
The initial concentration is [tex]C_f = 0.0022 \ M[/tex]
Explanation:
From the question we are told that
The volume of solution A is [tex]V_i = 10.0 mL[/tex]
The concentration of A is [tex]C_i = 0.0200 \ M[/tex]
The volume of solution B is [tex]V_B = 10.0mL[/tex]
The volume of water is [tex]V_{w } = 70.0 mL[/tex]
Generally the law of dilution is mathematically represented as
[tex]C_i * V_i = C_f * V_f[/tex]
Where [tex]C_f[/tex] is the concentration of the mixture
[tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as
[tex]V_f = 10 + 10 + 70[/tex]
[tex]V_f = 90mL[/tex]
So
[tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]
substituting values
[tex]C_f = \frac{0.0200 * 10 }{90}[/tex]
[tex]C_f = 0.0022 \ M[/tex]
Note the mixture obtained is [tex]KIO_3[/tex]
A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)
a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.
Answer and Explanation:
The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.
If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.
Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol
Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol
moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-
According to this, the following are the answers to the sentences:
a. The number of moles of CH₃COOH will remain the same ⇒ FALSE
The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions
b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE
Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)
c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE
The equilibrium concentration of OH⁻ is increased
d. The pH will decrease⇒ FALSE
pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:
[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]
pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15
Thus, the pH will increase.
At 25.0°C the Henry's Law constant for methane CH4 gas in water is ×1.410−3/Matm.
Calculate the mass in grams of CH4 gas that can be dissolved in 75.mL of water at 25.0°C and a CH4 partial pressure of 0.68atm. Round your answer to 2 significant digits.
Answer:
1.1 × 10⁻³ g
Explanation:
Step 1: Given data
Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm
Volume of water (=volume of solution): 75 mL
Partial pressure of methane (P): 0.68 atm
Step 2: Calculate the concentration of methane in water (C)
We will use Henry's law.
[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]
Step 3: Calculate the moles of methane in 75 mL of water
[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]
Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane
The molar mass of methane is 16.04 g/mol.
[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings
Answer:
[tex]V=27992L=28.00m^3[/tex]
Explanation:
Hello,
In this case, the combustion of methane is shown below:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]
Best regards.
Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence
Considering the definition of bond and the different type of bonds, valence is not one of the types of bonds.
What is a chemical bondA chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.
In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
Ionic bondAn ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.
Metallic bondMetallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.
SummaryIn summary, valence is not one of the types of bonds. The types of bonds are covalent, ionic and metallic.
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When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
How many valence electrons are in the electron dot structures for the elements in group 3A(13)?
Answer:
here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.
hope it helps...
The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
What are Groups in the Periodic Table?The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.
Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.
There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.
The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.
Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
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How does the carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere?
Answer:
The calculator add the CO2 released from the use of electricity, released from driving and the CO2 from the waste that we disposed.
Explanation:
The carbon dioxide, CO2 is what the human body does not need, therefore, we breathe it out, hence taking in oxygen(respiration process). The plants need oxygen for the production of their own food.
The carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere through the consideration of several factors such as the kind of food that we eat.
Therefore, if we are to use the carbon calculator to determine the amount of CO2 that each individual releases into the atmosphere we will have:
The amount of CO2 that each individual releases into the atmosphere =( CO2 released from the use of electricity) + (CO2 released from driving) + (the CO2 from the waste that we disposed).
If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?
Answer:
the density if vinegar will also be needed
Explanation:
Because this is an experiment of volumetric analysis
Identify the correctly written chemical reaction
A. Reactant + Reactant = Product
B. Reactant + Reactant → Product + Product
C. Reactant + Product → Reactant + Product
D. Product + Product Reactant + Reactant
Answer:
B. Reactant + Reactant -> Product + Product
Explanation:
Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction
Products are substances that are produced as a result of the reaction
Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving. In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards
And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.
To keep things simple:
Products can't be at the beginning of a reaction since they weren't formed yet.
Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products
As such, only one possible chemical reaction would follow that reasoning:
Reactant + Reactant -> Product + Product
Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.
What is a chemical equation?A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.
In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.
Hence, option B is correct.
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List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest
Answer:
CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH
Explanation:
We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.
The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.
Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?
boiling
freezing
chromatography
sorting
Answer:
a. boiling
Explanation:
An electrode has a negative electrode potential. Which statement is correct regarding the potential energy of an electron at this electrode?
A. An electron at this electrode has the same potential energy as it has at a standard hydrogen electrode.
B. An electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode.
C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.
Answer:
C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.
Explanation:
The standard hydrogen electrode (SHE) is used to measure the electrode potential of substances. The standard hydrogen electrode is arbitrarily assigned an electrode potential of zero. Recall that electrode potentials are always measured as reduction potentials in electrochemical systems.
For an electrode that has a negative electrode potential, electrons at this electrode have a higher potential energy compared to electrons at the standard hydrogen electrode. Electrons flow from this electrode to the hydrogen electrode.
On the other hand, a positive electrode potential implies that an electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode. Hence electrons will flow from the standard hydrogen electrode to this electrode.
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]
If you combine 24.2 g of a solute that has a molar mass of 24.2 g/mol with 100.0 g of a solvent, what is the molality of the resulting solution
Answer: 10 moles/kg.
Explanation:
Given, Mass of solute = 24.2 g
Molar mass of solute = 24.2 g/mol
[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]
Mass of solvent = 100.0g = 0.1 kg [1 g=0.001 kg]
[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]
Hence, the molality of the resulting solution is 10 moles/kg.
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.
Answer:
The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]
Explanation:
Let assume that the gas is O2 gas
O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.
Under the same conditions;
the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.
From Graham's Law of Diffusion;
Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.
i.e
[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]
[tex]R = \dfrac{k}{d}[/tex] where K = constant
If we compare the rate o diffusion of two gases;
[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]
Since the density of a gas d is proportional to its relative molecular mass M. Then;
[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]
Rate is the reciprocal of time ; i.e
[tex]R = \dfrac{1}{t}[/tex]
Thus; replacing the value of R into the above previous equation;we have:
[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]
We can equally say:
[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]
[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]
[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]
[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]
[tex]M_2 = 32 \times 1.62076418[/tex]
[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]
Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng
Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are 4 atoms per unit cell.
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
143pm is the radius of an Al atomThe radius of the atom of Al in the FCC structure has been 143 pm.
The FCC lattice has been contributed with atoms at the edge of the cubic structure.
The FCC has consisted of 4 atoms in a lattice.
The mass of the unit cell of Al can be calculated as:[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole
4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles
The mass of 1 mole Al has been 26.98 g/mol.
The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g
The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.
The volume of the Al cell can be calculated as:Density = [tex]\rm \dfrac{mass}{volume}[/tex]
Volume = Density × Mass
The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g
The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]
The volume of the cube has been given as:Volume = [tex]\rm edge\;length^3[/tex]
6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]
Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm
Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm
Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
In an FCC lattice structure, the radius of the atom can be given by:Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
1 m = [tex]\rm 10^1^2[/tex] pm
1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.
The radius of the atom of Al in the FCC structure has been 143 pm.
For more information about the FCC structure, refer to the link:
https://brainly.com/question/14934549
A constant volume and mass of helium gas at 77°C is heated so that the pressure of the gas doubles. What is the new temperature of the gas in Celsius degrees?
Answer:
427°C .
Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial temperature (T1) = 77°C
Initial pressure (P1) = P
Final pressure (P2) = 2P
Final temperature (T2) =?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T (°C) + 273
Initial temperature (T1) = 77°C
Initial temperature (T1) = 77°C+ 273 = 350K
Step 3:
Determination of the new temperature. The new temperature can be obtained as follow:
P1/T1 = P2/T2
P/350 = 2P/T2
Cross multiply
P x T2 = 350 x 2P
Divide both side by P
T2 = (350 x 2P ) / P
T2 = 700K
Step 4:
Conversion of Kelvin temperature to celsius temperature.
This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 700K
T(°C) = 700 – 273
T(°C) = 427°C
Therefore, the new temperature of the gas is 427°C
Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1, select all the products (in formulae and general chemical terms) that could result from homolysis. In Part 2, select the reaction conditions that are most likely to promote homolysis.
Part 1. Choose all that may occur as possible products of a homolysis reaction.
Choose one or more:_______.
a. hydride ion
b. R3CO
c. Br2
d. H
e. a carbocation
f. H3C
g. H3CO-
h. hydrogen ion
i. a carbon free radical
Part 2. Choose the conditions under which homolysis is likely to occur.
Choose one or more:_______.
a. strong base
b. ultraviolet irradiation
c. high temperature
d. strong acid
e. infrared irradiation
f. low temperature
Answer:
1) R₃CO , H, H₃C, a carbon free radical
2) high temperature, ultraviolet irradiation
Explanation:
1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :
R₃CO
H
H₃C
a carbon free radical
2) Homolysis require high temperature, ultraviolet irradiation.
Write the following isotope in nuclide notation: oxygen-14
Answer:
[tex]14\\8[/tex]O
Explanation:
The top number always represents the mass number.
The bottom number always represents the atomic number.
The element always goes after the numbers.
If charge is present, that comes after the element.
Therapeutic drugs generally need to have some hydrophobic and hydrophilic components to be able to effectively reach their target organs and tissues given there are aqueous and nonaqueous parts of the body. The degree to which a compound is hydrophobic and hydrophilic can be determined by measuring its relative solubility in water and octanol, C8H17OH, and water. To do this, a sample of the compound is added to a mixture of water and octanol and mixed well. Water and octanol are immiscible so after the mixture settles, the concentration of the compound in water and the concentration of the compound in octanol is measured. The ratio of the concentrations is called the partition ratio:
The question is incomplete as some part is missing:
concentration in octanol Partition Ratio = concentration in water
a) What are the intermolecular forces of attraction between octanol molecules? Explain.
b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.
c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.
d) Would nonane (figure 2) be more soluble in water or octanol? Explain.
e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.
f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.
Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C
Answer:
1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.
2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).
3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.
4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.
5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.