Answer:
Gases are at 1 bar pressure.
Explanation:
Option a is incorrect . Standard Pressure in chemistry is always taken in atm and not in bar. The standard condition for electro chemical cell measurements are
Gases must be at 1 atm pressure.
Liquids must be in pure state
Solids must also be in pure state
Solutes are at 1.0 M concentration
And the cell voltages are positive.
In the context of electrochemical cells, the statement 'The cell voltage is always positive' is incorrect as cell voltage can be negative in an electrolytic cell.
Explanation:The statement that is not true of the standard conditions for electrochemical cell measurements is that 'The cell voltage is always positive.' While the cell voltage can be positive in a galvanic/voltaic cell as it produces an electrical current, it can be negative in an electrolytic cell, where electrical energy is being consumed to bring about a chemical change.
The other statements about gases at 1 bar pressure, liquids and solids being in a pure state, and solutes at a 1.0 M concentration are indeed part of the standard conditions for measuring the properties of an electrochemical cell.
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Would you expect the nonbonding electron-pair domain in NH3 to be greater or less in size than for the corresponding one in PH3?
* greater
* lesser
Answer:
Less
Explanation:
The bond angle in NH3 is greater than the bond angle in PH3.
According to the valence shell electron pair repulsion theory, the shape of molecules depends on the number of electron pairs that surround the central atom. The electronegativity of the central atom determines the bond angle of the molecule. The greater the electronegativitty of the central atom, the greater the bond angle.
Nitrogen is more electronegative than Phosphorus hence, the bond angle in NH3 is greater than the bond angle in PH3.
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What is the energy of a photon (in units of h) that has a wavelength of 300 nm?
(c = 3.00 x 108 m/s)
1 x 10^17 h
1 x 10^15 h
1 x 10^6 h
1 x 10^-2 h
Answer:
[tex]Energy=1\times 10^{15}h\ J[/tex]
Explanation:
Considering:-
[tex]E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Planck's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
[tex]\lambda[/tex] is the wavelength of the light
Given, Wavelength = 300 nm
Also, 1 m = [tex]10^{-9}[/tex] nm
So,
Wavelength = [tex]300\times 10^{-9}[/tex] m
Applying in the equation as:-
[tex]E=\frac {h\times c}{\lambda}[/tex]
[tex]E=\frac{h\times 3\times 10^8}{300\times 10^{-9}}\ J[/tex]
[tex]Energy=1\times 10^{15}h\ J[/tex]
Complete and balance the following nuclear equations by including the missing particle. You may want to reference (Pages 906 - 909) Section 21.2 while completing this problem. Part A 25298Cf+105B → 310n+?
Answer:
The missing particle is [tex]_{103}^{259}\textrm{Lr}[/tex].
Explanation:
In a nuclear reaction, the total mass and total atomic number remains the same.
For the given fission reaction:
[tex]^{252}_{98}\textrm{Cf}+^{10}_{5}\textrm{B}\rightarrow ^A_Z\textrm{X}+3^1_0\textrm{n}[/tex]
To calculate A:
Total mass on reactant side = total mass on product side
252 +10 = A+ 3
A = 259
To calculate Z:
Total atomic number on reactant side = total atomic number on product side
98 +5 = Z + 0
Z = 103
Hence, the isotopic symbol of unknown element is [tex]_{103}^{259}\textrm{Lr}[/tex]
(a) [tex]\[ \frac{252}{98}\text{Cf} + \frac{10}{5}\text{B} \rightarrow 3 \frac{1}{0}\text{n} + \frac{259}{103}\text{Lr} \][/tex]
(b) [tex]\[ \frac{2}{1}\text{H} + \frac{3}{2}\text{He} \rightarrow \frac{4}{2}\text{He} + \frac{1}{1}\text{H} \][/tex]
(c) [tex]\[ \frac{1}{1}\text{H} + \frac{11}{5}\text{B} \rightarrow 3 \frac{4}{2}\text{He} \][/tex]
To complete and balance the nuclear equations, we need to ensure that both the atomic numbers (proton numbers) and mass numbers (nucleon numbers) are conserved on both sides of each equation.
(a) [tex]\( \frac{252}{98}\text{Cf} + \frac{10}{5}\text{B} \rightarrow 3 \frac{1}{0}\text{n} + ? \)[/tex]
First, determine the total atomic and mass numbers on the left side:
- Cf: Atomic number = 98, Mass number = 252
- B: Atomic number = 5, Mass number = 10
Sum of atomic numbers on the left side: 98 + 5 = 103
Sum of mass numbers on the left side: 252 + 10 = 262
On the right side, we have 3 neutrons:
- Each neutron: Atomic number = 0, Mass number = 1
- Total for neutrons: Atomic number = 0, Mass number = 3
So, the missing particle must balance the equation:
- Atomic number: 103 - 0 = 103
- Mass number: 262 - 3 = 259
Thus, the missing particle is [tex]\( \frac{259}{103}\text{Lr} \)[/tex] (Lawrencium).
The complete equation is:
[tex]\[ \frac{252}{98}\text{Cf} + \frac{10}{5}\text{B} \rightarrow 3 \frac{1}{0}\text{n} + \frac{259}{103}\text{Lr} \][/tex]
(b) [tex]\( \frac{2}{1}\text{H} + \frac{3}{2}\text{He} \rightarrow \frac{4}{2}\text{He} + ? \)[/tex]
First, determine the total atomic and mass numbers on the left side:
- H: Atomic number = 1, Mass number = 2
- He: Atomic number = 2, Mass number = 3
Sum of atomic numbers on the left side: 1 + 2 = 3
Sum of mass numbers on the left side: 2 + 3 = 5
On the right side, we have one helium-4 nucleus:
- He: Atomic number = 2, Mass number = 4
So, the missing particle must balance the equation:
- Atomic number: 3 - 2 = 1
- Mass number: 5 - 4 = 1
Thus, the missing particle is [tex]\( \frac{1}{1}\text{H} \)[/tex] (a proton).
The complete equation is:
[tex]\[ \frac{2}{1}\text{H} + \frac{3}{2}\text{He} \rightarrow \frac{4}{2}\text{He} + \frac{1}{1}\text{H} \][/tex]
(c) [tex]\( \frac{1}{1}\text{H} + \frac{11}{5}\text{B} \rightarrow 3? \)[/tex]
First, determine the total atomic and mass numbers on the left side:
- H: Atomic number = 1, Mass number = 1
- B: Atomic number = 5, Mass number = 11
Sum of atomic numbers on the left side: 1 + 5 = 6
Sum of mass numbers on the left side: 1 + 11 = 12
Let ? be [tex]\( \frac{4}{2}\text{He} \) (alpha particle)[/tex]. Since we need 3 of these particles:
- Each He: Atomic number = 2, Mass number = 4
- Total for 3 He: Atomic number = 6, Mass number = 12
So, the 3 particles are 3 [tex]\( \frac{4}{2}\text{He} \)[/tex] (3 alpha particles).
The complete equation is:
[tex]\[ \frac{1}{1}\text{H} + \frac{11}{5}\text{B} \rightarrow 3 \frac{4}{2}\text{He} \][/tex]
Complete Question:
Complete and balance the following nuclear equations by supplying the missing particle:
(a) 252/98Cf + 10/5B -> 3 1/0n + ?
(b) 2/1H + 3/2He -> 4/2He + ?
(c) 1/1H + 11/5B -> 3?
Which oxide(s) of manganese have a mass percent of the metal that is greater than 50%?MnO MnO2 Mn2O3All of these210
Answer:
All of these oxides of manganese have a mass percent of metal greater than 50%
Explanation:
Step 1: Data given
Molar mass of Mn = 54.94 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate mass % of Mn in MnO
Molar mass of MnO = 54.94 + 16 = 70.94 g/mol
% Mn = (54.94/70.94)*100 % = 77.45 %
Step 3: Calculate % of Mn in MnO2
Molar mass of MnO2 = 54.94 + 2*16 = 86.94 g/mol
% Mn = (54.94/86.94) * 100% = 63.19 %
Step 4: Calculate % of Mn in Mn2O3
Molar mass of Mn2O3 = 2*54.94 + 3*16 = 157.88 g/mol
% Mn = ((2*54.94)/157.88)*100 % = 69.6 %
All of these oxides of manganese have a mass percent of metal greater than 50%
The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer with a pH of approximately 3?
HC3H5O3
CH3NH3+
Ka
8.3 x 10-4
2.3 x 10-11
A mixture of 100. mL of 0.1 M NaC3H5O3 and 100. mL of NaOH
A mixture of 100. mL of 0.1 M CH3NH3Cl and 100. mL of CH3NH2
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
A mixture of 100. mL of 0.1 M CH3NH3Cl and 50. mL of NaOH
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
[tex]pH=pKa+log(\frac{salt}{acid} )[/tex]
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
The branch of science that deals with the chemicals and the bond are called chemistry. there are two types of solution out of which one is acidic and the other is basic
The correct answer is C.
What is a buffer solution?The solution in which the ph remains constant is called buffer solution.The equation used to solve the question is as follows:-
[tex]pH = pKa +log\frac{salt}{acid}[/tex]
Thus the pH of the buffer solution will be near to the pKa of the acid used in making the buffer solution because the salt is used in the question with respect to the acid.
The pKa value of [tex]HC_3H_5O_3[/tex]acid is closer to the required pH = 4 than [tex]CH_3NH_3^+[/tex]acid. [tex]pKa = -log [Ka]HC_3H_5O_3[/tex] , the pKa is 3.1
[tex]CH_3NH_3^+[/tex]the pKa is 10.64
The total pKb value is
= 14-10.6
= 3.36
Hence, the correct answer is C which is A mixture of 100 mL of 0.1 M HC3H5O3 and 50. mL of NaOH
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A 2.50 g sample of powdered zinc is added to 100.0 mL of a 2.00 M aqueous solution of hydrobromic acid in a calorimeter. The total heat capacity of the calorimeter and solution is 448 J/K. The observed increase in temperature is 21.1 K at a constant pressure of one bar. Calculate the standard enthalpy of reaction using these data. Zn ( s ) + 2 HBr ( aq ) ⟶ ZnBr 2 ( aq ) + H 2 ( g )
Answer: The standard enthalpy of the reaction is -248.78 kJ/mol
Explanation:
The heat released by the reaction is absorbed by the calorimeter and the solution.
The chemical equation used to calculate the heat released follows:
[tex]q=c\times \Delta T[/tex]
where,
c = heat capacity of calorimeter = 448 J/K
[tex]\Delta T[/tex] = change in temperature = 21.1 K
Putting values in above equation, we get:
[tex]q=448J/K\times 21.1K=9452.8J[/tex]
Sign convention of heat:
When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of zinc = 2.50 g
Molar mass of zinc = 65.4 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of zinc}=\frac{2.50g}{65.4g/mol}=0.038mol[/tex]
To calculate the standard enthalpy of the reaction, we use the equation:
[tex]\Delta H^o_{rxn}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat released = -9452.8 J
n = number of moles = 0.038 moles
[tex]\Delta H^o_{rxn}[/tex] = standard enthalpy of the reaction
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=\frac{-9452.8J}{0.038mol}=-248757.9J/mol=-248.78kJ/mol[/tex]
Conversion factor used: 1 kJ = 1000 J
Hence, the standard enthalpy of the reaction is -248.78 kJ/mol
Draw all of the monochlorination products that you might obtain from the free-radical chlorination of ethylcyclopropane.You do not have to consider stereochemistry.Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner.Separate multiple products using the + sign from the dropdown menu.
Answer:
-) (2-chloroethyl)cyclopropane
-) (1-chloroethyl)cyclopropane
-) 1-chloro-2-ethylcyclopropane
-) 1-chloro-1-ethylcyclopropane
Explanation:
The monochlorination are products in which Cl is added only 1 time. Now, in the ethylcyclopropane we will have 5 carbons, so we will have 5 options for monochlorination (see figure 1). But, for carbons 2 and 3 we will have equivalent structures, therefore at the end, we will have only 4 possible structures. In carbon 5, carbon 4, carbon 3 and carbon 1 (see figure 2).
Final answer:
Monochlorination of ethylcyclopropane can produce several isomers, based on the position where the chlorine atom is added - either on the cyclopropane ring or the ethyl group attached to it.
Explanation:
The subject of the question is the monochlorination products from the free-radical chlorination of ethylcyclopropane. During the free-radical chlorination, chlorine radicals can react with the different types of hydrogens in ethylcyclopropane - those on the cyclopropane ring as well as those on the ethyl group. Chlorination at the ring will yield two isomers due to two different types of hydrogens (geminal and vicinal), while chlorination on the ethyl group could occur at either one of the methylene hydrogens or the methyl group hydrogens. When considering the ethyl group, monochlorination would give a 1-chloroethylcyclopropane or a 2-chloroethylcyclopropane, depending on which hydrogen is substituted. With regard to the cyclopropane ring, substitution could lead to two different chlorocyclopropanes due to the different positions of hydrogen on the ring.
Name the ester that is formed when pentanoic acid reacts with isopropanol.a.isopropyl pentanoateb.isopropyl pentanec.pentanoate isopropyld.isopropyl pentanoic acid
Answer: A
Explanation:
Please check the attachment for the answer.
Answer:
a. Isopropyl pentanoate.
Explanation:
Hello,
In this case, such esterification is:
[tex]CH_3CH_2CH_2CH_2COOH+OHCH(CH_3)_2\rightarrow CH_3CH_2CH_2CH_2COOCH(CH_3)_2+H_2O[/tex]
Wherein the first reactant is the pentanoic acid, the second one the isopropanol and the first product is the result, which is a. isopropyl pentanoate. This is because the isopropyl get separated from the hydroxile and bonds with the carboxile group of the pentanoic acid by also releasing water as the byproduct.
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An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of the acidic solution to completely neutralize all of the acid?
The volume of KOH must be added to 500 ml of the acidic solution to neutralize all of the acids is equal to 968 ml.
What is a neutralization reaction?A neutralization reaction can be described as a chemical reaction in which an acid and base react with each other to produce salt and water.
The neutralization reaction between HCl and NaOH:
HCl + KOH → KCl + H₂O
The concentration of HCl solution = 0.100M
The volume of the solution = 500 ml = 0.5 L
The number of moles of HCl = 0.100 × 0.5 = 0.05 mol
The neutralization reaction between H₂SO₄ and KOH:
H₂SO₄ + KOH → K₂SO₄ + 2H₂O
The concentration of H₂SO₄ solution = 0.100M
The volume of the acidic solution = 500 ml = 0.5 L
The number of moles of H₂SO₄ = 0.210 × 0.5 = 0.105 mol
Total number of moles needed to neutralize = 0.05 + 0.105 = 0.155 mol
The concentration of KOH solution = 0.150 M
The volume of the KOH solution = M/n
V = 0.150/0.155
V = 0.968 L
V = 968 ml
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Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . What is the theoretical yield of sodium chloride formed from the reaction of of hydrochloric acid and of sodium hydroxide? Round your answer to significant figures.
The theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide is 5.3 g
How to obtain the theoretical yield of sodium chloride?
We shall begin by determining the limiting reactant for the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide. This is shown below:
HCl + NaOH -> NaCl + H₂O
Molar mass of HCl = 36.5 g/molMass of HCl from the balanced equation = 1 × 36.5 = 36.5 g Molar mass of NaOH = 40 g/molMass of NaOH from the balanced equation = 1 × 40 = 40 gFrom the balanced equation above,
36.5 g of HCl reacted with 40 g of NaOH
Therefore,
3.3 g of HCl will react with = (3.3 × 40) / 36.5 = 3.6 g of NaOH
Now, we can see that only 3.6 g of NaOH reacted with 3.3 g of HCl
Thus, the limiting reactant is HCl
Finally, we shall calculate the theoretical yield of sodium chloride. Details below:
HCl + NaOH -> NaCl + H₂O
Molar mass of HCl = 36.5 g/molMass of HCl from the balanced equation = 1 × 36.5 = 36.5 g Molar mass of NaCl = 58.5 g/molMass of NaOH from the balanced equation = 1 × 58.5 = 58.5 gTheoretical yield of sodium chloride =?From the balanced equation above,
36.5 g of HCl reacted to produce 58.5 of NaCl
Therefore,
3.3 g of HCl will react to produce = [tex]\frac{3.3\ \times\ 58.5}{36.5}[/tex] = 5.3 g of NaCl
In conclusion, theoretical yield of sodium chloride is 5.3 g
Complete question:
Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water.
What is the theoretical yield of sodium chloride formed from the reaction of 3.3 g of hydrochloric acid and 6.7 g of sodium hydroxide.? Round your answer to significant figure
The bond between carbon and hydrogen is one of the most important types of bonds in chemistry. The length of an H—C bond is approximately 1.1 Å. Based on this distance and differences in electronegativity, do you expect the dipole moment of an individual H—C bond to be larger or smaller than that of an H—I bond?
Answer:
The dipole moment of H-C bond will be smaller than that of an H-I bond.
Explanation:
The electronegativity of iodine is greater than that of hydrogen.As a result the iodine atom tends to attract the bond electron pair of H-I bond towards itself creating a bond dipole which does not occur in case of H-C bond as the electronegativity of carbon and hydrogen are almost same.
That"s why dipole moment of H-C bond is smaller than that of H-I bond.
A sample of gas has a mass of 0.555g. volume is 117 mL at a temperature of 85 degrees Celsius and a pressure of 0.99 atm. What is the molar mass of the gas?
Answer:
The molar mass of the gas is 140.86 g/mol
Explanation:
Step 1: Data given
Mass of the gas = 0.555 grams
Volume of the gas = 117 mL = 0.117 L
Temperature = 85 °C
Pressure = 0.99 atm
Step 2: Calculate the number of moles
p*V = n*R*T
⇒ with p = the pressure of the gas = 0.99 atm
⇒ with V = the volume = 0.117 L
⇒ with n = the number of moles
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature of the gas = 85 °C = 358.15 Kelvin
n = (p*V)/(R*T)
n = (0.99 * 0.117)/(0.08206 * 358.15)
n = 0.00394 moles
Step 3: Calculate molar mass of the gas
Molar mass = mass / moles
Molar mass = 0.555 grams / 0.00394 moles
Molar mass = 140.86 g/mol
The molar mass of the gas is 140.86 g/mol
Answer: 0.04 moles
Explanation:
pV = nRT
p=pressure = 0.99atm
v = volume = 117ml = 0.117L
n = number of moles = n
T = temperature = 85 +273 = 358k
R = rate constant = 0.082 L.atm/kmol
n = pV/RT = 0.99 X 0.117/(0.082 X 358)
= 0.004 moles
Sulfur dioxide, SO2(g), can react with oxygen to produce sulfur trioxide, SO3(g), by the following reaction
2SO2+O2=2SO3
The standard enthalpies of formation for SO2(g) and SO3(g) are
deltaH= SO2(g)= -296.8 kj
dH= SO3(g)= -395.7 kJ
Calculate the amount of energy in the form of heat that is produced when a volume of 3.75 L of SO2(g) is converted to 3.75 L of SO3(g) according to this process at a constant pressure and temperature of 1.00 atm and 25.0
Answer:
The heat produced is -15,1kJ
Explanation:
For the reaction:
2SO₂+O₂ → 2SO₃
The enthalpy of reaction is:
ΔHr = 2ΔHf SO₃ - 2ΔHf SO₂
As ΔHf SO₃ = -395,7kJ and ΔHf SO₂ = -296,8kJ
ΔHr = -197,8kJ
Using n=PV/RT, the moles of reaction are:
[tex]n = \frac{1,00atm*3,75L}{0,082atmL/molK*298,15K}[/tex] = 0,153 moles of reaction
As 2 moles of reaction produce -197,8kJ of heat, 0,153moles produce:
0,153mol×[tex]\frac{-197,8kJ}{2mol}[/tex] = -15,1kJ
I hope it helps!
14.8 KJ/mol of heat energy is produced when a volume of 3.75 L of SO2(g) is converted to 3.75 L of SO3(g).
The equation of the reaction is;
2SO2(g) + O2(g) ⇄ 2SO3(g)
We have the following information;
ΔHf SO2(g) = -296.8 KJ/mol
ΔHf O2(g) = 0 KJ/mol
ΔHf SO3(g) = -395.7 kJ/mol
We can calculate the heat of reaction ΔHrxn from;
ΔHrxn = [ΔHf (products) - ΔHf (reactants)]
ΔHrxn = [2( -395.7 kJ/mol)] - [2(-296.8 KJ/mol) + 0 KJ/mol]
ΔHrxn = (-791.4 kJ/mol) + 593.6 KJ/mol
ΔHrxn = -197.8 KJ/mol
We can find the number of moles of SO2 reacted using the ideal gas equation;
P = 1.00 atm
T = 25.0°C + 273 = 298 K
n = ?
V = 3.75 L
R = 0.082 atmLK-1mol-1
So,
PV = nRT
n = PV/RT
n = 1.00 atm × 3.75 L/0.082 atmLK-1mol-1 × 298 K
n = 0.15 moles
If 2 moles of SO2 produced -197.8 KJ/mol
0.15 moles of SO2 will produce 0.15 moles × (-197.8 KJ/mol)/ 2 moles
= -14.8 KJ/mol
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ATP hydrolysis, ATP + H2O → ADP + Pi, is the exothermic chemical reaction that provides the energy for many of the processes that take place in a cell. The reaction is described as arising from "breaking a phosphate bond in ATP" that is often described as a "high energy bond." The reaction also forms an OH-P bond to create Pi. Just consider this reaction in isolation. Which of the following statements would you therefore expect are true about the reaction? It only takes a little energy to break the O-P bond in ATP. The OH-P bond that is formed in the reaction is a weak bond. The OH-P bond that is formed in the reaction is a strong bond. It takes a lot of energy to break the O-P bond in ATP. The formation of the OH-P bond is the part of the reaction responsible for releasing energy. The breaking of the O-P bond releases energy that is stored in the bond.
In the ATP hydrolysis reaction, the O-P bond in ATP is broken, requiring a lot of energy. The OH-P bond formed in the reaction is strong and does not release energy.
Explanation:In the ATP hydrolysis reaction, the O-P bond in ATP is broken, which requires a lot of energy. This breaking of the bond releases energy that is stored in the bond. Therefore, it takes a lot of energy to break the O-P bond in ATP.
The OH-P bond that is formed in the reaction is a strong bond. The formation of this bond is not responsible for releasing energy in the reaction.
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Final answer:
The breaking of the high-energy phosphate bond in ATP during hydrolysis releases energy. The OH-P bond formed in the reaction is strong, but its formation is not the source of energy release. The correct statement is 1) "The OH-P bond that is formed in the reaction is a strong bond."
Explanation:
ATP hydrolysis is an exothermic reaction where energy is released from the breaking of the high-energy phosphate bond in ATP, producing ADP and inorganic phosphate (Pi). The bond between the phosphates in ATP (phosphoanhydride bonds) is considered high-energy, not because it takes a lot of energy to break, but because the products of the reaction (ADP and Pi) have considerably lower free energy than ATP and a water molecule. Hence, the breaking of the O-P bond in ATP releases the energy that was stored in the bond.
The OH-P bond that forms when Pi is created during ATP hydrolysis is actually a strong bond, and the formation of this bond is not responsible for the release of energy, rather it is the breaking of the high-energy bond in ATP that releases energy.
In part D of the experiment, a student begins with a beaker containing 10.0 mL of 0.010 M HCl solution and adds different volumes of 0.010 M NaOH. (a) How many millimoles of acid are initially in the beaker? (If in doubt, look back at your calculation from experiment 9).
Answer:
There are initially 0.1 millimoles of acid (HCl) in the beaker.
Explanation:
Step 1: Data given
volume of 0.010 M HCl = 10.0 mL = 0.01 L
Step 2: Calculate the number of moles HCl
Number of moles = Molarity HCl * volume of solution
Number of moles HCl = 0.010 M * 0.01 L = 0.0001 moles HCl = 0.1 *10^-3 mol = 0.1 mmol HCl
There are initially 0.1 millimoles of acid (HCl) in the beaker.
The initial beaker contains 0.1 millimoles of HCl. The process of adding NaOH to HCl is called titration, which helps measure the concentration of a solution and involves achieving an equivalence point where all reactant molecules have reacted.
Explanation:The question is asking about the quantity of acid initially present in a beaker before the addition of sodium hydroxide. The millimoles of acid can be calculated using the formula: Molarity (M) = Moles/Liter. For this experiment, you have an initial volume of 10.0 mL (or 0.010 L) of a 0.010 M HCl solution. Therefore, you can calculate the millimoles of acid in the beaker by multiplying the molarity by the volume in liters: 0.010 moles/L * 0.010 L = 0.0001 moles or 0.1 millimoles of HCl.
Titration, mentioned in the references, is a technique used to find the concentration of a solution, where a reaction is created that can easily be observed when the amount of reactant in a solution is exactly sufficient to react with another compound. In your experiment, NaOH is used to titrate HCl. The point when all the molecules of one reactant have reacted with another is called the equivalence point, which you can determine from a titration curve.
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Toluene, C7H8, is used in the manufacture of explosives such as TNT (trinitrotoluene). A 1.500 g sample of liquid toluene was placed in a bomb calorimeter along with excess oxygen. When the combustion of the toluene was initiated, the temperature of the calorimeter rose from 25.000 °C to 26.413 °C. The products of the combustion are CO2(g) and H2O(l), and the heat capacity of the calorimeter was 45.06 kJ °C-1. Write the balanced chemical equation for the reaction in the calorimeter. How many joules were liberated by the reaction? How many joules would be liberated under similar conditions if 1.000 mol of toluene was burned?
Answer:
There were 63.67 kJ liberated
If 1 mol of toluene was burned, there would be liberated 3906 kJ / mole
Explanation:
Step 1: Data given
Mass of the toluene sample = 1.500 grams
The temperature of the calorimeter rose from 25.000 °C to 26.413 °C
Write the balanced chemical equation for the reaction in the calorimeter.
C7H8(l) + 9O2(g) ⇒ 7CO2(g) + 4H2O(l)
Step 2: Calculate Q of the calorimeter
Q = C * Δt
Q = 45.06 kJ/ °C * ( 26.413 - 25)
Q = 63.67 kJ
Step 3: Calculate moles of toluene
Moles toluene = mass toluene / molar mass toluene
Moles toluene = 1.500 grams /92.14 g/mol
Moles toluene = 0.0163 moles
Step 4: Calculate energy per mole
energy per mole = 63.67 kJ / 0.0163 moles = 3906 kJ / mole
Final answer:
The balanced chemical equation for the combustion of toluene is 2 C7H8 + 21 O2 -> 14 CO2 + 8 H2O. The heat liberated by the reaction in the calorimeter is 63.30 kJ. If 1.000 mol of toluene is burned, 0.507 kJ of heat will be liberated.
Explanation:
The balanced chemical equation for the combustion of toluene can be written as:
2 C7H8 + 21 O2 -> 14 CO2 + 8 H2O
To calculate the heat liberated by the reaction, we can use the formula:
q = Ccal * ΔT
where q is the heat liberated, Ccal is the heat capacity of the calorimeter, and ΔT is the change in temperature.
Plugging in the values, we get:
q = (45.06 kJ/°C) * (26.413 °C - 25.000 °C) = 63.30 kJ
To calculate the heat liberated when 1.000 mol of toluene is burned, we can use the molar mass of toluene to convert grams to moles.
The molar mass of toluene is 92.14 g/mol, so:
1.500 g * (1 mol / 92.14 g) = 0.016 mol
Using stoichiometry, we know that for every 2 moles of toluene burned, 63.30 kJ of heat is liberated.
Therefore, for 0.016 mol of toluene:
q = (63.30 kJ / 2 mol) * 0.016 mol = 0.507 kJ
When you exercise, the burning sensation that sometimes occurs in your muscles represents the buildup of lactic acid (HC3H5O3). In a .20M aqueous solution, lactic acid is 2.6% dissociated. What is the value of Ka for this acid?
A.) 4.3x10^-6
B.) 8.3x10^-5
C.) 1.4x10^-4
D.) 5.2x10^-3
Answer:
The Ka for this acid is 1.4 * 10^-4 (option C)
Explanation:
Step 1: Data given
In a 0.20M aqueous solution, lactic acid is 2.6% dissociated.
Step 2: The equation for the dissociation of HAc is:
HAc ⇌ H+ + Ac¯
The Ka expression is:
Ka = ([H+] [Ac¯]) / [HAc]
1) [H+] using the concentration and the percent dissociation:
(0.026) (0.2) = 0.0052 M
2) Calculate [Ac-] and [H+]
[Ac¯] = [H+] = x = 0.0052 M
3) Calculate [HAc]
[HAc] = 0.20M - x ( since x << 0.20, we can assume [HAc] = 0.20 M
4) Calculate Ka
Ka= [( 0.0052) ( 0.0052)] / 0.20
Ka = 0.0001352 = 1.4 * 10^-4
The Ka for this acid is 1.4 * 10^-4
Final answer:
The value of Ka for lactic acid, given that a 0.20 M solution is 2.6% dissociated, can be calculated using the concentration of dissociated ions; the closest value to our calculation is 1.4x10^-4.
Explanation:
When you exercise, the burning sensation in your muscles is often due to the buildup of lactic acid. To find the acid dissociation constant (Ka) for lactic acid given a 0.20 M solution that is 2.6% dissociated, we use the dissociation equation of a weak acid:
HC₃H₅O₃ → H⁺ + C₃H₅O₃⁻
Initial concentration of lactic acid = 0.20 MPercent dissociation = 2.6%Dissociation of lactic acid (α) = 2.6% of 0.20 M = 0.0026 × 0.20 = 0.0052 MThe concentration of H⁺ and C₃H₅O₃⁻ will also be 0.0052 M, since it's a one-to-one dissociation. Thus, the expression for Ka becomes:
Ka = [H⁺][C₃H₅O₃⁻] / [HC₃H₅O₃]The closest answer to our calculated Ka value is C.) 1.4x10^-4.
Match the chromatography term with its definition. The volume of solvent traveling through the column per unit time. The elapsed time between sample injection and detection. The time required by a retained solute to travel through the column beyond the time required by the unretained solvent. The distance traveled by the solvent per unit time. Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio. Volume of the mobile phase required to elute a solute from the column. Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor. The ratio of the solute concentrations in the mobile and stationary phases.
Answer:
The statements are definitions to chromatography terms which have been highlighted below.
Explanation:
Match the chromatography term with its definition.
Volumetric Flow Rate = The volume of solvent traveling through the column per unit time.
Retention time = The elapsed time between sample injection and detection.
Adjusted Retention Time = The time required by a retained solute to travel through the column beyond the time required by the un -retained solvent.
Linear Flow Rate = The distance traveled by the solvent per unit time.
Retention factor = Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio.
Relative Volume = Volume of the mobile phase required to elute a solute from the column.
Relative Retention = Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor.
Partition coefficient = The ratio of the solute concentrations in the mobile and stationary phases.
Chromatography is a laboratory technique used to separate and analyze the components of a mixture based on their different affinities for a stationary phase and a mobile phase. The different terms in chromatography can be defined as given below-
The volume of solvent traveling through the column per unit time - Flow rate
The elapsed time between sample injection and detection - Retention time
The time required by a retained solute to travel through the column beyond the time required by the unretained solvent - Adjusted retention time
The distance traveled by the solvent per unit time - Velocity
Describes the amount of time that a sample spends in the stationary phase relative to the mobile phase. It is sometimes also called the capacity factor or capacity ratio - Retention factor
Volume of the mobile phase required to elute a solute from the column - Elution volume
Ratio of the adjusted retention times or retention factors of two solutes. It is sometimes also called the separation factor - Selectivity factor
The ratio of the solute concentrations in the mobile and stationary phases - Distribution coefficient
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Which of the following does the internal energy of an ideal gas depend upon?
a. The internal energy depends upon its temperature.
b. The internal energy depends upon its volume.
c. The internal energy depends upon its pressure.
Answer:
The correct option is: a. The internal energy depends upon its temperature.
Explanation:
Ideal gas is a hypothetical gas that obeys the ideal gas law. The equation for the ideal gas law:
P·V=n·R·T
Here, V- volume of gas, P - total pressure of gas, n- total mass or number of moles of gas, T - absolute temperature of gas and R- the gas constant
Also, according to the Joule's second law, the internal energy (U) of the given amount of ideal gas depends on the absolute temperature (T) of the gas only, by the equation:
[tex]U = c_{V}nRT[/tex]
Here, [tex]c_{V}[/tex] is the specific heat capacity at constant volume
The internal energy of an ideal gas depends on its temperature and is independent of its volume and pressure. For non-ideal systems, internal energy can be influenced by volume and pressure but only through their effect on temperature.
The internal energy of an ideal gas depends primarily on its temperature. This is because, for an ideal gas, the internal energy is a function of state and is determined by the kinetic energy of its particles, which is directly related to temperature. The volume and pressure of an ideal gas do not directly determine its internal energy, although changes in these properties can indirectly affect temperature and thus internal energy. Option A
In the case of an ideal gas, the internal energy does not change with volume or pressure shifts alone; it is the temperature that directly impacts internal energy. However, for other types of systems or under non-ideal conditions, both volume and pressure can have more complex relationships with internal energy.
What is the pH of 0.30 M ethanolamine, HOCH2CH2NH2, (Kb = 3.2 x 10−5)?
Answer:
pH= 11.49
Explanation:
Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.
From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5
Using the formula below;
[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)
[OH^-] =√(3.2×10^-5 × 0.30M)
[OH^-]= √(9.6×10^-6)
[OH^-]=3.0984×10^-3
pOH= -log[OH^-]
pOH= -log 3.1×10^-3
pOH= 3-log 3.1
pH= 14-pOH
pH= 14-(3-log3.1)
pH= 11+log 3.1
pH= 11+ 0.4914
pH= 11.49
The pH of the ethanolamine solution has been 11.5.
The pH of the solution has been defined as the concentration of hydrogen ion in the solution.
The pH has been the difference of pOH from 14.
Computation for the pH of ethanolamineThe ethanolamine has been a weak base. The OH concentration of ethanolamine has been given as:
[tex]\rm OH^-=\sqrt{Kb\;\times\;Molarity}[/tex]
The Kb of sample has been given as [tex]3.2\;\times\;10^-^5[/tex]
The molarity of sample has been, 0.3 M.
Substituting the values for hydroxide ion concentration:
[tex]\rm OH^-=\sqrt{3.2\;\times\;10^-^5\;\times\;0.3}\\OH^-=3.098\;\times\;10^-^3[/tex]
The pOH has been the negative logarithmic values of hydroxide ion concentration. The pOH has been given as:
[tex]\rm pOH=-log\;OH^-\\pOH=-log\;3.098\;\times\;10^-^3\\pOH=2.5[/tex]
The pOH of the sample has been 2.5.
The pH of ethanolamine sample has been given as:
[tex]\rm pH=14-pOH\\pH=14-2.5\\pH=11.5[/tex]
The pH of the ethanolamine solution has been 11.5.
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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 379 MPa causes the specimen to plastically elongate to a length of 51.7 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 49.7 mm to a length of 57.1 mm.
Answer:
The true stress required = 379 MPa
Explanation:
True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,
σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent
True strain is given by
Epsilon t =㏑ (l/l₀)
Substitute㏑(l/l₀) for ε(t)
σ(t) = K(㏑(l/l₀))ⁿ
Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa
379 x 10⁶ = K (㏑(51.7/49.7))^0.2
K = 379 x 10⁶/(㏑(51.7/49.7))^0.2
K = 723.48 MPa
Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.
σ(t) = K(㏑(l/l₀))ⁿ
l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa
σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2
σ(t) = 379 MPa
The true stress necessary to plastically elongate the specimen is 379 MPa.
Answer:i ea
Explanation:
If 60.0 grams of carbonic acid are sealed in a 2.00 L soda bottle at room temperature (298 K) and decompose completely via the equation below, what would be the final pressure of carbon dioxide assuming it had the full 2.00 L in which to expand? H₂CO₃(aq) → H₂O(l) + CO₂(g)
Answer:
The final pressure of the carbon dioxide gas will 11.84 atm.
Explanation:
Moles of carbonic acid = [tex]\frac{60.0 g}{62 g/mol}=0.9677 mol[/tex]
[tex]H_2CO_3(aq)\rightarrow H_2O(l) + CO_2(g)[/tex]
According to reaction, 1 mol of carbonic acid gives 1 mole of carbon dioxide gas.
Then 0.9677 moles of carbonic acid will give :
[tex]\frac{1}{1}\times 0.9677 mol=0.09677 mol[/tex] of carbon dioxide
Moles of carbon dioxide gas = n = 0.09677 mol
Volume of soda bottle = [tex]V_1=2 L[/tex]
Pressure of the carbon dioxide gas = P
Temperature of the carbon dioxide gas = T = 298 K
[tex]PV=nRT[/tex] (ideal gas law)
[tex]P=\frac{nRT}{V}=\frac{0.9677 mol\times 0.0821 atm L/mol K\times 298 K}{2 L}=11.84 atm[/tex]
The final pressure of the carbon dioxide gas will 11.84 atm.
By using the ideal gas law equation PV = nRT, calculating the number of moles of CO₂ formed from 60.0 grams of decomposed carbonic acid, and inputting the given values for the volume (V) and temperature (T), we calculated that the final pressure of the CO₂ in the soda bottle would be approximately 1.18 atm.
Explanation:To calculate the final pressure of carbon dioxide in the soda bottle we must utilize the ideal gas law equation, which is PV = nRT. Where P stands for pressure, V is volume, n is the number of moles of gas, R is the gas constant and T is the temperature.
First, we need to determine the number 'n' of moles of CO₂ formed. Knowing that 60.0 grams of carbonic acid decompose completely into water and CO₂, and knowing that the molar mass of carbonic acid H₂CO₃ is approximately 62.03 g/mol, we can find that the number of moles (n) is equal to the mass (60g) divided by the molar mass (62.03 g/mol), so n equals about 0.967 moles.
Because it specifies that we should assume CO₂ to 'have the full 2.00 L in which to expand', the volume V in the ideal gas law equation is 2.00 L. Similarly, the temperature T is given as 298 K. We will now plug these values into the ideal gas law equation, using 0.0821 as the value for R (the ideal gas constant in units of L*atm/(mol*K)). This gives us P = (nRT) / V, so P is about 1.18 atm, which is the final pressure of CO₂.
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When 1.010 of sucrose (C12H22O11) undergoes combustions in a bomb calorimeter, the temperature rises from 24.92 ◦ C to 28.33 ◦ C. Find DErxn for the combustion of sucrose in kJ/mol sucrose. The heat capacity of the bomb calorimeter determined in a separate experiment is 4.90 kJ/◦ C. (Answer: -5.66 x 103 kJ/mole).
Answer:
Heat of combustion = 5.6 ×10³ kj/mol
Explanation:
Given data:
Mass of sucrose = 1.010 g
Initial temperature = 24.92 °C
Final temperature = 28.33 °C
Heat capacity of calorimeter = 4.90 KJ/°C
Heat of combustion = ?
Solution:
ΔT = 28.33 °C - 24.92 °C = 3.41 °C
Q = - c. ΔT
Q = 4.90 KJ/°C . 3.41 °C
Q = - 16.7 kj
Number of moles of sucrose :
Number of moles of sucrose = mass/ molar mass
Number of moles of sucrose = 1.010 g / 342.3 g/mol
Number of moles of sucrose = 0.003 mol
Heat of combustion:
Heat of combustion = Q/n
Heat of combustion = - 16.7 kj/0.003 mol
Heat of combustion = -5.6 ×10³ kj/mol
The combustion of sucrose in a bomb calorimeter can be calculated using the formula for heat transfer: q = C * ΔT. Given the heat capacity is 4.90 kJ/°C and the change in temperature is 3.41 °C, the energy released per gram is 16.709 kJ. By multiplying this by the molecular weight of sucrose, over 342.3 g/mol, yields approximately -5.66 x 103 kJ/mole.
Explanation:The combustion of sucrose in a bomb calorimeter can be determined through use of the formula for heat transfer, q = C x ΔT. Here, C represents the heat capacity of the calorimeter, and ΔT is the change in temperature. Given the heat capacity (C) is 4.90 kJ/°C, and the change in temperature (ΔT) is 28.33 °C - 24.92 °C = 3.41 °C, we plug these values into the equation yielding q = 4.90 kJ/°C x 3.41° C = 16.709 kJ. This is the energy released per gram. To find the energy released per mole, we need to multiply this value by the molecular weight of sucrose which is around 342.3 g/mole. This gives us an answer of -5.66 x 103 kJ/mole, and the negative sign indicates this is an exothermic reaction, meaning heat is released.
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Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of k = 1.6x10^−3yr−1. By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k = 0.011 day−1.
(a) What are the half-lives of these two isotopes?
(b) Which one decays at a faster rate?
(c) How much of a 1.00-mg sample of each isotope remains after three half-lives?
(d) How much of a 1.00-mg sample of each isotope remains after 4 days?
Answer:
A. 433 years and 63 days
B. Iodine
C. 0.125mg
D. 1.00mg
Explanation:
A. Americium
The formula of a radioactive decay constant half life is t = 0.693/k
Where k is the decay constant.
For americium, k = 0.0016
t = 0.693/0.0016 = 433.125 apprx 433 years
For iodine, k = 0.011
Half life = 0.693/0.011 = 63 days.
B. Iodine decays at a faster rate.
C. After three half lives
For both, first half life yields a mass of 0.5mg, next yields 0.25mg, next yield 0.125mg
D. 1.00mg still remains.
Half life is high for both so the decay after one day is insignificant.
a) The half-lives of Americium-241 and Iodine-125 are 433 years and 63 days, respectively.
b) Iodine-125 decays faster.
c) After three half-lives, 0.125 mg remains.
d) After 4 days, nearly 1.00 mg of Americium-241 and 0.957 mg of Iodine-125 remain.
To answer the given questions regarding the radioactive decay of Americium-241 and Iodine-125:
(a) Half-lives
For a first-order reaction, the half-life (t1/2) can be calculated using the equation:
t1/2 = 0.693 / k
For Americium-241:
k = 1.6 × 10⁻³ yr⁻¹
t1/2 = 0.693 / (1.6 × 10⁻³) = 433 years
For Iodine-125:
k = 0.011 day⁻¹
t1/2 = 0.693 / 0.011 = 63 days
(b) Decay Rate
Since the half-life of Iodine-125 is shorter (63 days) compared to Americium-241 (433 years), Iodine-125 decays at a faster rate.
(c) Remaining Sample After Three Half-lives
After three half-lives, the remaining amount of a radioactive isotope can be calculated using:
Remaining Amount = Initial Amount / (23)
Remaining Amount = 1.00 mg / 8 = 0.125 mg
Therefore, for both isotopes, 0.125 mg remains after three half-lives.
(d) Remaining Sample After 4 Days
For Americium-241 (with t1/2 about 433 years), 4 days is negligible, so nearly the entire 1.00 mg remains.
For Iodine-125, using the first-order decay formula:
N = N₀ e-kt
t = 4 days, k = 0.011 day⁻¹
N = 1.00 mg × e-(0.011 × 4) ≈ 1.00 mg × e-0.044 ≈ 1.00 mg × 0.957 = 0.957 mg
Thus, after 4 days, 0.957 mg of Iodine-125 remains.
Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which value is closest to the percentage mass of water lost from the total mass of salt in the crucible when the crucible undergoes repetitive heatings until a constant mass is reached?36%26%13%25%
Answer:
The water lost is 36% of the total mass of the hydrate
Explanation:
Step 1: Data given
Molar mass of CuSO4*5H2O = 250 g/mol
Molar mass of CuSO4 = 160 g/mol
Step 2: Calculate mass of water lost
Mass of water lost = 250 - 160 = 90 grams
Step 3: Calculate % water
% water = (mass water / total mass of hydrate)*100 %
% water = (90 grams / 250 grams )*100% = 36 %
We can control this by the following equation
The hydrate has 5 moles of H2O
5*18. = 90 grams
(90/250)*100% = 36%
(160/250)*100% = 64 %
The water lost is 36% of the total mass of the hydrate
In a particular electroplating process, the metal being plated has a +4 charge. If 861.8 C of charge pass through the cell, how many moles of metal should be plated? Useful information: F = 96,500 C/mol e- Provide your response to four digits after the decimal.
Answer: [tex]2.2326\times 10^{-3}[/tex] moles
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]
The metal being plated has a +4 charge, thus the equation will be:
[tex]M^{4+}+4e^-\rightarrow M[/tex]
[tex]4\times 96500C[/tex] of electricity deposits = 1 mole of metal
Thus 861.8 C of electricity deposits =[tex]\frac{1}{4\times 96500}\times 861.8=2.2326\times 10^{-3}[/tex] moles of metal
Thus [tex]2.2326\times 10^{-3}[/tex] moles of metal should be plated
The number of moles of the metal deposited is 0.0022 moles of metal.
What is electroplating?The term electroplating refers to the use of one metal to cover the surface of another metal. Let the metal in question be M, the equation of the reaction is; M^4+(aq) + 4e -----> M(s).
1 mole of the metal is deposited by 4( 96,500) C
x moles will be deposited by 861.8 C
x = 1 mole * 861.8 C/ 4( 96,500) C
x = 0.0022 moles of metal
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Ethylene, C2H4, and tetrafluoroethylene, C2F4, are used to make the polymers polyethylene and polytetrafluoroethylene (Teflon), respectively. (a) Draw the Lewis structures for C2H4 and C2F4, and give the ideal H-C-H and F-C-F bond anges. (b) The actual H-C-H and F-C-F bond angles are 117.4∘ and 112.4∘ respectively. Explain these deviations.
Explanation:
The Lewis structures can be seen in the Figure below.
The ideal bond angle for both H-C-H and F-C-F is 120° because the have a triangular geometry.
The H-C-H has 117.4° angles due to the repulsion caused by the electrons of the double bond and the electrons of the H-C bond. This repulsion brings closer the H and decreases the angle.
On the other hand, in the F-C-F the same phenomenom happens but given that the F have lone pairs of electrons the repulsion is bigger, the F get closer and the angle is smaller than the H-C-H.
Final answer:
The Lewis structure for tetrafluoroethylene (C2F4) shows each carbon atom double-bonded to the other and single-bonded to two fluorine atoms, with an sp2 hybridization. The ideal bond angles for sp2 hybridized carbon are 120 degrees, but the actual angles differ due to variations in electronegativity and substituent size.
Explanation:
Polymerization and Lewis Structures
To draw the Lewis structure for the monomer tetrafluoroethylene (C2F4) used to make the polymer polytetrafluoroethylene (Teflon), we start by considering that carbon and fluorine both follow the octet rule, meaning that they both seek to be surrounded by eight electrons to reach a stable configuration. In C2F4, each carbon atom is double-bonded to the other carbon atom and single-bonded to two fluorine atoms. The hybridization of carbon in C2F4 is sp2 because it has three sigma bonds (one with the other carbon and two with fluorine atoms) and one pi bond (part of the double bond with the other carbon).
The ideal bond angles for an sp2 hybridized atom are 120 degrees. However, in the case of C2H4 and C2F4, the actual bond angles are 117.4 degrees for H-C-H and 112.4 degrees for F-C-F respectively. These deviations from the ideal angles can be attributed to the differences in electronegativity and size of the substituents, causing variations in electron cloud repulsions and thus slightly adjusting the bond angles.
Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1 ΔG∘2 for second step=−35.8 kJ⋅mol−1 ΔG∘3 for third step=−24.3 kJ⋅mol−1 Calculate the overall formation constant (Kf) for the complex [Ni(en)3]2+.
Answer:
kf = 1.16 x 10¹⁸
Explanation:
Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹
Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹
Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹
________________________________________________________
Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r
ΔG°r = ΔG°1 + ΔG°2 + ΔG°3
ΔG°r = -42.9 - 35.8 - 24.3
ΔG°r = -103.0 kJmol⁻¹
ΔG°r = -RTlnKf
-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf
kf = e ^(-103,000/-8.31x298)
kf = e ^41.59
kf = 1.16 x 10¹⁸
A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank
To calculate the mole fraction and partial pressure of each gas, we calculate the number of moles of each gas using the given masses and molar masses. The mole fraction is then determined by dividing the moles of each gas by the total moles. The partial pressure of each gas and the total pressure in the tank are found using the mole fraction and the given total pressure.
Explanation:To calculate the mole fraction and partial pressure of each gas, we first need to calculate the number of moles of each gas. For dinitrogen difluoride (N2F2), we divide the given mass by the molar mass of N2F2 to get the number of moles.
Moles of N2F2 = 5.53 g / (28.01 g/mol) = 0.1977 mol
Similarly, for sulfur hexafluoride (SF6), the moles are calculated as follows:
Moles of SF6 = 17.3 g / (146.06 g/mol) = 0.1185 mol
The mole fraction of each gas is then calculated by dividing the moles of the gas by the total moles:
Mole fraction of N2F2 = 0.1977 mol / (0.1977 mol + 0.1185 mol) = 0.6258
Mole fraction of SF6 = 0.1185 mol / (0.1977 mol + 0.1185 mol) = 0.3742
To calculate the partial pressure of each gas, we use the formula:
Partial pressure = Mole fraction * Total pressure
Given that the total pressure is 26.9 C, we substitute the values to find:
Partial pressure of N2F2 = 0.6258 * 26.9 C = 16.82 C
Partial pressure of SF6 = 0.3742 * 26.9 C = 10.08 C
Finally, the total pressure in the tank is the sum of the partial pressures:
Total pressure = Partial pressure of N2F2 + Partial pressure of SF6
Total pressure = 16.82 C + 10.08 C = 26.9 C
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The mole fraction of N₂F₂ is 0.4141 with a partial pressure of 0.256 atm, while the mole fraction of SF₆ is 0.5859 with a partial pressure of 0.361 atm. The total pressure in the tank is 0.617 atm.
To solve this problem, we need to use the Ideal Gas Law and the concept of partial pressures.
Step 1: Calculate Moles of Each Gas
First, we calculate the number of moles for each gas:
Molar mass of dinitrogen difluoride N₂F₂ = 2(14) + 2(19) = 66 g/molMoles of N₂F₂ = 5.53 g / 66 g/mol = 0.0838 molMolar mass of sulfur hexafluoride SF₆ = 32 + 6(19) = 146 g/molMoles of SF₆ = 17.3 g / 146 g/mol = 0.1185 molStep 2: Total Moles in the Tank
Total moles = 0.0838 mol + 0.1185 mol = 0.2023 molStep 3: Calculate the Mole Fraction
Mole fraction of N₂F₂ = 0.0838 mol / 0.2023 mol = 0.4141Mole fraction of SF₆ = 0.1185 mol / 0.2023 mol = 0.5859Step 4: Calculate Partial Pressures
We use the Ideal Gas Law: PV = nRT
R (Ideal Gas Constant) = 0.0821 atm·L/mol·KTemperature (T) = 26.9°C = 300.05 K (T = 26.9 + 273.15)First, calculate the total pressure:
P = (nRT) / V = (0.2023 mol × 0.0821 atm·L/mol·K × 300.05 K) / 8.00 L = 0.617 atmThen, calculate the partial pressures:
Partial pressure of N₂F₂ = 0.4141 × 0.617 atm = 0.256 atmPartial pressure of SF₆ = 0.5859 × 0.617 atm = 0.361 atmSummary
In summary, the mole fraction of dinitrogen difluoride is 0.4141, and its partial pressure is 0.256 atm. The mole fraction of sulfur hexafluoride is 0.5859, and its partial pressure is 0.361 atm. The total pressure in the tank is 0.617 atm.A chemistry student needs 65.0g heptane for an experiment. He has available 20.0g of a 38.1% w/w solution of heptane in chloroform.
Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Round your answer to 3 significant digits.
Answer : The correct answer is "No solution".
Explanation :
As we are given that 38.1 % w/w solution of heptane in chloroform. That means 38.1 gram of heptane present in 100 gram of solution.
Now we have to determine the mass of solution for 65.0 grams of hepatne.
As, 38.1 grams of heptane present in 100 grams of solution
So, 65.0 grams of heptane present in [tex]\frac{65.0}{38.1}\times 100=171[/tex] grams of solution
Since, there is only 20.0 grams of solution available. That means, there is not enough solution.
Thus, the correct answer is "No solution".
Final answer:
The student would need 170.6g of the 38.1% w/w heptane solution to obtain 65.0g of heptane. However, the student only has 20.0g of solution, which is not enough to provide the required amount of heptane, so the student does not have a viable solution.
Explanation:
The student needs to calculate the mass of solution to use to obtain 65.0g of heptane from a 38.1% w/w solution of heptane in chloroform. The weight/weight percentage (w/w%) indicates that for every 100g of solution, there are 38.1g of heptane. To find the mass of the solution needed to obtain 65.0g of heptane, the following calculation can be performed:
m = (desired mass of heptane) / (percentage of heptane in solution as a decimal)
m = 65.0g / 0.381
m = 170.6g
The student would need to use 170.6g of the 38.1% w/w heptane solution to obtain the required amount of pure heptane. However, since the student only has 20.0g of solution, there is not enough to obtain 65.0g of heptane without obtaining more solution. Therefore, the student cannot proceed with the experiment as is and must either get more solution or adjust the experiment.