To calculate the equilibrium constant (Kc) for the reaction H2(g) + I2(g) \ightarrow 2HI(g), the moles of each reactant and product at equilibrium were calculated and then converted to concentrations. Using the concentrations, the Kc was determined to be approximately 610.97.
The reaction in question is H2(g) + I2(g) \ightarrow 2HI(g). To calculate the equilibrium constant (Kc), we need to determine the moles of HI, H2, and I2 at equilibrium. The molecular weights of H2, I2, and HI are approximately 2 g/mol, 254 g/mol, and 128 g/mol, respectively.
Initial moles of H2 = 0.763 g / 2 g/mol = 0.3815 mol
Initial moles of I2 = 96.9 g / 254 g/mol = 0.3815 mol
At equilibrium, there are 90.4 g of HI:
Equilibrium moles of HI = 90.4 g / 128 g/mol = 0.70625 mol
Since the reaction produces 2 moles of HI for every 1 mole of H2 and I2 that react, 0.70625 mol of HI would require half that amount of H2 and I2 to react:
Moles of H2 and I2 reacted = 0.70625 mol HI / 2 = 0.353125 mol
Moles of H2 and I2 at equilibrium = initial moles - moles reacted
Equilibrium moles of H2 = 0.3815 mol - 0.353125 mol = 0.028375 mol
Equilibrium moles of I2 = 0.3815 mol - 0.353125 mol = 0.028375 mol
We then convert these moles to concentrations by dividing by the volume of the flask:
[H2] = 0.028375 mol / 3.67 L \hickapprox 0.007732 M
[I2] = 0.028375 mol / 3.67 L \hickapprox 0.007732 M
[HI] = 0.70625 mol / 3.67 L \ hickapprox 0.192398 M
Now, we can calculate the equilibrium constant Kc
Kc = [HI]² / ([H2][I2]) = (0.192398 M)² / (0.007732 M × 0.007732 M) \hickapprox 610.97
The vapor pressure of liquid antimony is 400 mm Hg at 1.84×103 K. Assuming that its molar heat of vaporization is constant at 115 kJ/mol, the vapor pressure of liquid Sb is 394.98 mm Hg at a temperature of 1.81×103 K. Find the vapor pressure of liquid Sb?
Answer:
Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg
Explanation:
The vapor pressure can be calculated by using Clausius‐Clapeyron equation.
ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)
Where
p₁ is the vapor pressure at T₁ (Initial Temperature)
p₂ is the vapor pressure at T₂ (final Temperature)
ΔHvap is molar heat of vaporization of the substance
R is the real gas constant = 8.314 x 10⁻³ kJ/mol.K
Data Given:
p₁ = ?
p₂ = 394.98 mm Hg
T₁ = 1.84×10³ K
T₂ = 1.81×10³ K
ΔHvap = 115 kJ/mol
Put the values in the Clausius‐Clapeyron equation
ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)
ln(p₁/394.98 mm Hg) = (-115 kJ/mol / 8.314 x 10⁻³ kJ/mol.K)(1/1.84×10³ K- 1/1.81×10³ K)
ln(p₁ /394.98 mm Hg) = (- 13.8321 x 10³)(-0.5519)
ln(p₁ /394.98 mm Hg) = 7633.936
ln cancel out by E, e is raise to a power x
So,
p₁/394.98 mm Hg = e^7633.936
p₁/ 394.98 mm Hg = 20.75 x 10³
p₁ = 20.75 x 10³ x 394.98 mm Hg
p₁ = 8.19 x 10⁴ mm Hg
Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg
Answer:
338 mm Hg
Explanation:
[tex]ln(\frac{p2}{400mmHg} )=-115*\frac{1}{0.008314} *(\frac{1}{1.8*10^3} -\frac{1}{1.84*10^3}[/tex])
ln(p2/400)=-0.16705
p2/400= e^-0.16705=0.84615
p2=0.84615*400=338
Note that T2 is lower than T1 and that the vapor pressure decreases as the temperature decreases.
A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s. What is the tension in the string?
Answer:
F centripetal force (tension) = 275.9 N
Explanation:
Given data:
Mass = 1.50 kg
Radius = 0.520 m
Velocity of ball = 9.78 m/s
Tension = ?
Solution:
F centripetal force (tension) = m.v² / R
F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m
F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m
F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m
F centripetal force (tension) = 275.9 N
One difference between mixtures and pure substances is that
Answer:
The mixture is made up of different atoms and pure substance is made up of same type of atom.
The main difference is that mixture can be separated into its component by physical mean while pure substances can not be separated by physical process
Explanation:
Mixture:
The properties of the mixture are not same and contains the properties of all those component present in it.it is a combination of one or more Pure substances and can be separated by simple physical methods.it have varying boiling and melting pointExamples are:
mixture of salt and sandSalt water is mixture of water and NaCl and can be separated by physical mean.Alloys: its a mixture of different metalAir: mixture of gasesPure Substance:
Pure substances are those made of same type of atoms all elements and compounds are pure substances.
it can not be separated by simple physical meanit have very constant and consistent melting and boiling pointExamples are:
Water : contain only water moleculeAll elements: all elements are pure substance made of same atomsAll compounds: can not be separated by physical mean.A food chemist determines the concentration of acetic acid in a sample of apple vinegar by acid base titration. The density of the sample is 1.01 g/mL. The titrant is 1.024 M NaOH. The average volume of titrant required to titrate 25.00 mL subsamples of the vinegar is 20.78 mL. What is the concentration of acetic acid in the vinegar?
Answer:
The concentration of acetic acid in the vinegar is 7,324 (%V/V)
Explanation:
The titration equation of acetic acid with NaOH is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
The moles required were:
1,024M×0,02500L = 0,02560 moles NaOH. These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:
0,02560 moles CH₃COOH×[tex]\frac{60,052g}{1mol}[/tex]= 1,537g of CH₃COOH
As density is 1,01g/mL:
1,537g CH₃COOH×[tex]\frac{1mL}{1,01g}[/tex]= 1,522mL of CH₃COOH
As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
[tex]\frac{1,522mLCH_{3}COOH}{20,78mL}[/tex]×100= 7,324 (%V/V)
I hope it helps!
Considering the definition, molarity, density and volume percentage, the concentration of acetic acid in the vinegar is 7.32% (v/v).
The balanced reaction is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
Molarity is the number of moles of solute that are dissolved in a given volume. Molarity is determined by:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case you know for NaOH:
Molarity= 1.024 Mnumber of moles of solute= ?Volume= 25 mL= 0.025 LSo, the moles required are calculated as:
[tex]1.024 M=\frac{number of moles of solute}{0.025 L}[/tex]
Solving:
1.024 M× 0.025 L= number of moles of solute
0.0256 moles= number of moles of solute
By observing the balanced reaction, you can see that, since the ratio of CH₃COOH to NaOH is 1: 1, these moles are equivalent to the number of moles of CH₃COOH.
On the other hand, since the molar mass of CH₃COOH is 60.052 g/mol, this is the mass that contains one mole of the compound, the mass in 0.0256 moles of CH₃COOH is:
[tex]0.0256 molesx\frac{60.052 grams}{1 mole} =[/tex] 1.537 grams
Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
[tex]density=\frac{mass}{volume}[/tex]
In this case, being the density 1.01 [tex]\frac{g}{mL}[/tex], the volume can be calculated as:
[tex]1.01 \frac{g}{mL} =\frac{1.537 grams}{volume}[/tex]
Solving:
1.01 [tex]\frac{g}{mL}[/tex]× volume= 1.537 grams
[tex]volume=\frac{1.537 grams}{1.01 \frac{g}{mL} }[/tex]
volume= 1.522 mL of CH₃COOH
Volume Percentage (%v/v) is a measure of concentration that indicates the volume of solute per 100 volume units of the solution. In other words, the volume percent of a component in the solution is defined as the ratio of the volume of the component to the volume of the solution, expressed as a percentage.
The volume percentage of a solution is determined by the following expression:
[tex]volume percentage=\frac{volume of solute}{volume of solution}x100[/tex]
Then, as volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
[tex]volume percentage=\frac{1.522 mL}{20.78 mL}x100[/tex]
Solving:
volume percentage= 7.32%
Finally, the concentration of acetic acid in the vinegar is 7.32% (v/v).
Learn more about:
molarity with this example: brainly.com/question/15406534?referrer=searchResultsdensity: brainly.com/question/952755?referrer=searchResults brainly.com/question/1462554?referrer=searchResultsvolume percentagebrainly.com/question/1713808An apparatus consists of a 3 L flask containing nitrogen gas at 17◦C and 822 kPa, joined by a valve to a 10 L flask containing argon gas at 17◦C and 59 kPa. The valve is opened and the gases mix. a) What is the partial pressure of nitrogen after mixing?
b) What is the partial pressure of argon aftermixing?
c) What is the total pressure of the gas mixture?
Answer:
a) 189.7 kPa
b) 45.4 kPa
c) 235 kPa
Explanation:
After the mixture, the temperature must remain the same, because both gases are at 17°C. Besides, the gases will not react because both nitrogen and argon are non-reactive. So, the partial pressure can be calculated by Boyle's Law:
P1*V1 = P2*V2
Where P is the pressure, V is the volume, 1 is the initial state, and 2 the final state.
When the valve is opened, the gases mix and occupy all the two vases. So, V2 = 13 L.
a) For nitrogen, P1 = 822 kPa, V1 = 3L, V2 = 13 L
822*3 = P2*13
P2 = 189.7 kPa
b) For argon, P1 = 59 kPa, V1 = 10 L, V2 = 13 L
59*10 = P2*13
P2 = 45.4 kPa
c) By Dalton's law, the total pressure of a gas mixture is the sum of the partial pressures of the components, so:
P = 189.7 + 45.4
P = 235 kPa
When 0.0400 mol of HCl(aq) is reacted with 0.0400 mol of NaOH(aq) in 50.0 mL of water, the temperature of the solution increases by 9.82°C. What is the enthalpy of reaction for the following thermochemical equation?HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Assume that the heat capacity of the solution and calorimeter is 227.4 J/°C.
The enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.
What is thermochemical equation?Thermochemical equation is defined as a chemical equations that are properly balanced and take into account the physical conditions of all reactants, products, and energy change. A chemical equation is an equation that shows the beginning molecule, reactants, and final products separated by arrows, but a thermochemical equation is a balanced stoichiometric chemical process that also includes the enthalpy change.
Given heat capacity = 227.14 J/°C
Temperature = 9.82 °C
Number of moles of HCl and NaOH is 0.04 moles
q (solution) = -C (solution) x ΔT
= 227.14 J/°C x 9.82 °C
= 2.2305 kJ
For enthalpy of reaction we have to divide it by mole of reactant
= 2.2305 / 0.04
= 55.4 kJ/ mole
Thus, the enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.
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To determine the enthalpy of reaction for HCl reacting with NaOH, the heat released by the reaction is calculated using the mass of the solution, the specific heat capacity, and the change in temperature. The result is then divided by the number of moles reacted to express the enthalpy change in kJ per mole.
Explanation:The problem here concerns the measurement of an enthalpy change during the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O). To find the enthalpy of reaction for the provided thermochemical equation, we use the temperature change and the heat capacity of the solution to calculate the total heat released by the reaction, and then express this heat change in kJ per mole of reactants.
First, calculate the total heat (q) released using the formula q = mc∆T, where m is the mass of the solution, c is the specific heat capacity, and ∆T is the change in temperature. As the reaction takes place in 50.0 mL of water, and assuming the density of water is 1.00 g/mL, the mass of the solution is 50.0 g. Therefore, q = 50.0 g × 227.4 J/°C × 9.82°C.
After calculating q, divide this value by the number of moles of HCl reacted to obtain the enthalpy of reaction per mole. Remember to convert from joules to kilojoules since enthalpy is usually expressed in kJ/mol. Please note that the reaction is exothermic, so the enthalpy change should be a negative value, indicating heat is released.
The atomic number of an element is the same as the number of its
Answer: The atomic number of an element is the same as the number of its protons.
Explanation: For example, the atomic number of oxygen is 8. So, this would mean that the number of protons would be 8.
Hope this helps^
Which of the following is not true for an exothermic reaction?
A. the products have a higher enthalpy than reactants
B. the temperature of the surroundings rise
C. the enthalpy change for the reaction is negative heat flows from the rxn system to the surroundings
Answer:
The false statement regarding an exothermic reaction is: A. the products have a higher enthalpy than reactants
Explanation:
An exothermic reaction is a type of chemical reaction that involves the release of energy from the system to the surroundings. Thus increasing the temperature of the surroundings.
In this reaction, the enthalpy or energy of the reactants is greater than the enthalpy or energy of the products. ([tex]\Delta H_{f} (Products) < \Delta H_{f} (reactants)[/tex])
As the enthalpy change of a reaction: [tex]\Delta H_{r} = \sum \Delta H_{f} (Products) - \sum \Delta H_{f} (reactants)[/tex]
Therefore, the enthalpy change for an exothermic reaction is negative ([tex]\Delta H_{r} < 0[/tex])
A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.
To find the pKa of X-281, you prepare a 0.089 M test solution of X-281 at 25.0 ∘C. The pH of the solution is determined to be 2.40.
a. What is the pKa of X-281? Express your answer numerically.
At 25∘C, for any conjugate acid-base pair
pKa + pKb = 14.00
b. What is pKb of the conjugate base of X-281? (Assume 25 ∘C.) Express your answer numerically.
Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
[tex]ka = \frac{[H_3O^+][X^-]}{[HX]}[/tex]
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
[tex][H_3O^+]} = 10^{-2,40}[/tex]
[H₃O⁺] = 0,004M
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = 0,085M
[tex]ka = \frac{[0,004][0,004]}{[0,085]}[/tex]
ka = 1,88x10⁻⁴
And pka = 3,73
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
pkb = 10,27
I hope it helps!
The correct answers are:
a. The pKa of X-281 is 2.40.
b. The pKb of the conjugate base of X-281 is 11.60.
a. To find the pKa of X-281, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the concentration of the acid and its conjugate base. For a monoprotic weak acid HA, the equation is:
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Given that the pH of the solution is 2.40 and the concentration of X-281 (HA) is 0.089 M, we can assume that the concentration of the conjugate base A^- is negligible compared to the concentration of the acid HA because the pH is close to the pKa. This means that the ratio[tex]\([\text{A}^-] / [\text{HA}]\)[/tex]is approximately zero, and thus the log term is approximately zero. Therefore, the pH is approximately equal to the pKa of the acid:
[tex]\[ \text{pH} \approx \text{pKa} \] \[ \text{pKa} \approx 2.40 \][/tex]
b. At 25°C, the sum of pKa and pKb for any conjugate acid-base pair is 14.00, which is the pKw of water:
[tex]\[ \text{pKa} + \text{pKb} = 14.00 \][/tex]
We have already determined that the pKa of X-281 is 2.40. Therefore, we can solve for the pKb of the conjugate base of X-281:
[tex]\[ 2.40 + \text{pKb} = 14.00 \] \[ \text{pKb} = 14.00 - 2.40 \] \[ \text{pKb} = 11.60 \][/tex]
Thus, the pKb of the conjugate base of X-281 is 11.60.
Barium can be analyzed by precipitating it as BaS04 and weighing the precipitate. When a 0.713-g sample of a barium compound was treated with excess H2S04, 0.5331 g of BaSO4 formed. What is the percentage of barium in the compound?
Molecular mass of BaSO4 233.39 g/mol
Molecular mass of Ba = 137.327 g/mol
Answer:
31.37%
Explanation:
For this case, you should consider the following reaction:
Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O
For which you obtain the precipitate of BaSO₄
In order to obtain the mas of Barium on the precipitate, you may use the following formula:
gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)
Where:
gBa= mass of Barium
M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate
M₍Ba₎= mass of Barium from the original sample
M₍BaSO₄₎= mass of BaSO₄ from the precipitate
gBa= (0.5331)x(137.327/233.39)= 0.3136 g
Then we ontain the percentage of Barium multiplying by 100:
% Ba on the original sample= 31.36%
Final answer:
The percentage of barium in the compound is calculated by finding the moles of BaSO₄, converting it to moles of Ba, and then finding the mass of Ba before expressing it as a percentage of the initial sample mass. The resulting percentage of barium is approximately 43.99%.
Explanation:
To determine the percentage of barium in the compound, we must calculate the moles of BaSO₄ and then use the molar mass of barium to find the mass of barium in the initial sample. We can calculate the moles of BaSO₄ using the following steps:
First, use the mass of the BaSO₄ precipitate formed (0.5331 g) and the molar mass of BaSO₄ (233.39 g/mol) to find the moles of BaSO₄.Moles of BaSO₄ = mass of precipitate / molar mass of BaSO₄
= 0.5331 g / 233.39 g/mol
= 0.002284 mol (rounded to six decimal places)
Since the molar ratio of Ba to BaSO₄ is 1:1, the moles of Ba will be the same as the moles of BaSO₄.Next, multiply the moles of Ba by the molar mass of Ba to find the mass of Ba in the sample.Mass of Ba = moles of Ba × molar mass of Ba= 0.002284 mol × 137.327 g/mol
= 0.3137 g (rounded to four decimal places)
Finally, we calculate the percentage of barium in the sample using the initial sample mass (0.713 g).
Percentage of barium = (mass of Ba / initial sample mass) × 100%
= (0.3137 g / 0.713 g) × 100%
= 43.99%
Therefore, the percentage of barium in the compound is approximately 43.99%.
Which of the following is true?
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
b. A reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.
c. A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
d. A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.
Answer:
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
Explanation:
The change in free energy (ΔG) that is, the energy available to do work, of a system for a constant-temperature process is:
[tex]ΔG = ΔH - TΔS[/tex]
When ΔG < 0 the reaction is spontaneous in the forward direction.When ΔG > 0 the reaction is nonspontaneous. The reaction isspontaneous in the opposite direction.
When ΔG = 0 the system is at equilibrium.If both ΔH and ΔS are positive, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.
The enthalpy of combustion of acetylene C2H2 is described by
C2H2 (g) + (5/2)O2 (g) >>>>>>>CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol
Calculate the enthalpy of formation of accetylene, given the following enthalpies of formation
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol
Answer: The enthalpy of formation of accetylene is, 226.2 kJ/mol
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The chemical equation for the combustion of acetylene follows:
(1) [tex]C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)[/tex]
[tex]\Delta H^o_{rxn}=-1299kJ/mol[/tex]
The formation of [tex]CO_2[/tex] will be,
(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=-393.5kJ/mol[/tex]
The formation of [tex]H_2O[/tex] will be,
(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
The formation of [tex]C_2H_2[/tex] will be,
(4) [tex]2C(s)+H_2(g)\rightarrow C_2H_2(g)[/tex] [tex]\Delta H_f_{(C_2H_2)}=?[/tex]
Now we are reversing equation 1, multiplying equation 2 by 2 and then adding equation 1, 2 and 3, we get:
Reaction (1) :
[tex]2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)[/tex]
[tex]\Delta H^o_{rxn}=1299kJ/mol[/tex]
Reaction (2) :
[tex]2C(s)+2O_2(g)\rightarrow 2CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=2\times -393.5kJ/mol=-787kJ/mol[/tex]
Reaction (3) :
[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
[tex]\Delta H_f_{(C_2H_2)}=\Delta H^o_{rxn}+\Delta H_f_{(CO_2)}+\Delta H_f_{(H_2O)}[/tex]
[tex]\Delta H_f_{(C_2H_2)}=(1299kJ/mol)+(-787kJ/mol)+(-285.8kJ/mol)[/tex]
[tex]\Delta H_f_{(C_2H_2)}=226.2kJ/mol[/tex]
Therefore, the enthalpy of formation of accetylene is, 226.2 kJ/mol
To calculate the enthalpy of formation of acetylene, we can use the enthalpy change of the combustion reaction and the enthalpies of formation of CO2 and H2O. By rearranging the equation and applying Hess's law, we can determine the enthalpy of formation of acetylene.
Explanation:The enthalpy of formation of a compound can be calculated using Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken. To calculate the enthalpy of formation of acetylene (C2H2), we can use the equation:
C2H2 (g) + (5/2)O2 (g) → CO2 (g) + H2O (l)
The enthalpy change of this reaction is -1299 kJ/mol. By rearranging the equation, we can compare it to the standard formation equations for CO2 (g) and H2O (l) to determine the enthalpy of formation of acetylene.
The enthalpy of formation of acetylene is:
ΔHf(C2H2) = -1299 kJ/mol - [(-393.5 kJ/mol) + (-285.8 kJ/mol)]
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To determine the moles of ammonium chloride produced, we need to find the limiting reactant. H2O is present in an unlimited amount, so it cannot be the limiting reactant. Let’s then determine the amount of each remaining reactant present (in moles). NaCl is already given in moles (0.250 mol). NH3 and CO2 are not. They are given in liters. How many moles are present in 3.08 L of NH3 and 2.09 L of CO2?
Answer:
0.094 mols of CO2 are present in 2.09 L.
0.132 mols of NH3 are present in 3.08 L
The moles of ammonium chloride produced are 0.094
Explanation:
First of all, think the reaction
NH3 + 2H2O + NaCl + CO2 → NH4Cl + NaOH + H2CO3
To get the moles with volume, you must need density
Density = mass / volume
Density CO2 = 0,001976 g/ml
Density NH3 = 0,00073 g/ml
Volume of CO2 = 2.09L
2.09 L . 1000 = 2090 mL
0,001976 g/ml = mass / 2090 mL
4.13 g = mass
Molar weight = 44 g/m
mass / molar weight = 4.13 g / 44 g/m = 0.094 moles
Volume of NH3 = 3.08 L
3.08 L . 1000 = 3080 mL
0,00073 g/ml = mass/ 3080 mL
2.25 g = mass
Molar weight = 17 g/m
mass / molar weight = mols → 2.25 g / 17g/m = 0.132 mols
In my reagents, the least amount I have is CO2. This is my limiting reactant. Take account the reaction.
1 mol of NH3 reacts with 1 mol of CO2, so as I have 0.132 moles of NH3 I need 0.132 moles of CO2; I only have 0.094.
Relation between CO2 and NH4Cl is 1:1 so, 1 mol of CO2 is needed to produce 1 mol e chloride, so 0.094 mols are needed to produce the same amount of chloride.
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ ⋅ mol − 1 calculate the average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule.
Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
[tex]\frac{1}{2}[/tex]Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is [tex]\frac{1134.9}{4}[/tex] = 283.725 kJ ⋅ mol − 1
The average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule is 283.725 kJ ⋅ mol − 1
Calculation of average molar bond:C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3)
⇒ 2Br2(g) + C(s)
⇒ 4 Br(g) + C(g) , Δ H ∘
= 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4)
⇒ 4 Br(g) + C(g)
⇒ CBr4(g) , Δ H ∘
= -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is = 283.725 kJ ⋅ mol − 1
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A .115 L sample of dry air has a pressure of 1.0895633 atm at 377 K. What is the volume of the sample if the temperature is increased to 483 K when the pressure is kept constant?
Answer:
V= 0.147 L
Explanation:
This is simply the application of combined gas law twice, to find the unknowns.
Combined gas law states that: [tex]PV = nRT[/tex]
P= pressure of air
V= volume of air
n= moles of air
R= Universal gas constant ( 0.08205 L atm mol⁻¹ K⁻¹)
T= Absolute temperature in kelvin.
[tex]n=\frac{P * 0.115}{R * 377}[/tex]
Now, applying the same gas law at 483K and substituting for n
[tex]V = \frac{P * 0.115}{R * 377} * \frac{R* 483}{P}[/tex]
V= [tex]\frac{0.115 * 483}{377}[/tex]
V= 0.147 L
The reaction is found to be first order in IO3−, first order in SO32−, and first order in H+. If [IO3−]=x, [SO32−]=y, and [H+]=z, what is the rate law for the reaction in terms of x, y, and z and the rate constant k?
Answer:
rate = kxyz
Explanation:
It is worth knowing that the rate low can only be determined by experimentation only not by just balancing equations. So here we are told that all the reactants x , y and z are all first order. This is important because we use this as exponents. That is why the exponents of all the reactants will be 1.
rate = kxyz
Assuming that the container is completely full, that the temperature is 22.0 ∘C, and that the atmospheric pressure is 1.2 atm , calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.)
Answer:
The % of displaced volume of nitrogen is 29.06%.
Explanation:
Volume of nitrogen = 1.2 L = 1200 mL
Density of nitrogen = 0.807 g/ml
Mass of nitrogen = [tex]Density \times Volume= 0.807 \times 1200 = 1044 g[/tex]
Molar mass of nitrogen = 28 g/mol
[tex]Number\,of\,moles\,nitrogen=\frac{Mass}{Molar\,mass}[/tex]
[tex]= \frac{1044}{28}=37.28[/tex]
The ideal gas equation is as follows
[tex]PV = nRT[/tex]
Rearrange the equation is as follows.
[tex]V= \frac{nRT}{P}...............(1)[/tex]
n= Number of moles = 37.28
R = Gas constant = 0.0820
T = Temperature = 22+ 273 = 295
P = Pressure = 1.2 atm
Substitute the all values in equation (1)
[tex]V= \frac{37.28 \times 0.0820 \times 295 }{1.2}= 751.5L= 0.751 \,m^{3}[/tex]
[tex]0.751 \,m^{3}[/tex] of nitrogen will displace same amount of air.
[tex]Volume \,\,of\,\, closet= 1 \times 1.3 \times 2 = 2.6\,m^{3}[/tex]
[tex]%\,displaced\,volume=\frac{0.751}{2.6}=28.8%[/tex]
Therefore, The % of displaced volume of nitrogen is 29.06%.
Final answer:
The percent by volume of air displaced by the evaporation of liquid nitrogen is calculated using the density of liquid nitrogen, the molar mass of nitrogen, and the Ideal Gas Law, accounting for the specific temperature and pressure conditions.
Explanation:
To calculate the percent by volume of air displaced if all the liquid nitrogen evaporated, we need to consider the Ideal Gas Law as well as the density of the liquid nitrogen. Since liquid nitrogen has a density of 0.807 g/mL, we can first find the mass of the nitrogen and then use the molar mass to determine the number of moles of nitrogen gas at the given temperature and pressure when it evaporates.
Using the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature in Kelvin. At standard temperature and pressure, the density of nitrogen gas can be calculated using the given molar mass and standard gas equation factors.
Once the volume of the gaseous nitrogen is determined, that volume will represent the amount of air volume displaced. To find the percent by volume displaced, divide the volume of gaseous nitrogen by the volume of the container and multiply by 100%.
The pH of 0.50 M benzoic acid is 2.24. Calculate the change in pH when 2.64 g of C6H5COONa is added to 38 mL of 0.50 M benzoic acid, C6H5COOH. Ignore any changes in volume. The Ka value for C6H5COOH is 6.5 x 10-5.
Answer:
1.93
Explanation:
Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH = 38/1000 × 0.50 = 0.019mol
Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol
Final pH = pKa + log([C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/[C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH]
= -log Ka + log(mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH
= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17
change in pH = final - initial pH
= 4.17 - 2.24
=1.93
pH is the depiction of the acidity and basicity of the solution and can range from 0 to 14 on the pH scale. The change in pH of benzoic acid is 1.93.
What is pH?pH is the concentration of the hydrogen ion in the solution depicting the acidity and the basicity of the solution.
Calculate moles of Benzoic acid as:
[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{38}{1000}\times 0.50\\\\&= 0.019\;\rm mol\end{aligned}[/tex]
Calculate moles of Sodium benzoate as:
[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{2.64}{144.10}\\\\&= 0.01832\;\rm mol\end{aligned}[/tex]
pH can be calculated as:
[tex]\begin{aligned}\rm pH &= \rm pKa + log\dfrac{([C_{6}H_{5}COONa]}{[C_{6}H_{5}COOH]}\\\\&= \rm -log(6.5 \times 10^{-5}) + log (\dfrac{0.018321}{0.019})\\\\&= 4.17\end{aligned}[/tex]
Change in the pH can be calculated as:
[tex]\begin{aligned} \rm pH &= \rm final - initial \; pH\\\\&= 4.17 - 2.24\\\\&=1.93\end{aligned}[/tex]
Therefore, 1.93 is the change in the pH of benzoic acid.
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An aqueous mixture containing starch (a colloid), Nacl, glucose, and albumin (a colloid) is placed in a dialyzing bag and in distilled water. Which of the following correctly describes the location of the indicated substance after dialysis?
a) starch outside
b) water inside only
c) albumin inside
d) albumin inside and outside
e) starch inside and outside
Answer:
c) albumin inside
Explanation:
Let us see look closer at each answer and see why it is wrong/correct.
a) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside
b) There cannot be water inside only because it would mean starch and albumin diffused out and they cannot pass through the bag
c) This is correct. The bag isn't permeable to albumin so it stays inside
d) Albumin cannot be outside the bag because the bag isn't permeable to albumin. The albumin stays inside
e) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside
After dialysis, albumin - a colloid, will remain inside the dialyzing bag because the dialysis membrane is designed to let only small molecules and ions pass through, not large colloidal particles. Starch, another colloid, will remain inside for the same reason. Water will reach equilibrium, moving in and out of the membrane.
Explanation:The question relates to the process of dialysis, which is used to separate different substances in a mixture based upon their size. An aquatic mixture containing both colloids like starch and albumin and small molecules like NaCl (sodium chloride) and glucose is placed in a dialyzing bag to see where the substances will be after the process is complete. A characteristic of dialysis is that the dialyzing membrane allows small molecules and ions to pass through, while retaining larger colloidal particles.
Referring to the given options, the correct descriptions after dialysis are as follows:
Therefore, the accurate statement concerning the location of the substances post-dialysis is
c) albumin inside
How many atoms of fluoride are there in one molecule of nitrogen trifluoride?
Answer:
there are three Fluorine atom. This is because Nitrogen give 3 electron away but fluorine can only take 1. So nitrogen gives 1 neutron to 3 fluorine each which makes it a stable compound
Answer : The number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.
Explanation :
Molecule : It is defined as the smallest particle in an element or a compound which has chemical properties of that element or a compound.
Molecules are made up of atoms and the atoms are bonded by the sharing of electrons.
The given molecules is, [tex]NF_3[/tex]
The nitrogen trifluoride molecule is made up of two elements which are nitrogen and fluorine.
In the given molecules [tex]NF_3[/tex], there are 1 nitrogen atom and there are 3 fluorine atoms.
Hence, the number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.
On Earth, the white colored paper represents areas that are A) covered with snow and reflect sunlight. B) heavily populated and use the sunlight. Eliminate C) covered with water and reflect sunlight. D) covered with concrete and absorb sunlight.
In an ecosystem model of earth , white colored paper represents areas that are covered with snow and reflect sunlight.
What is an ecosystem?Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
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Photophosphorylation produces ___ , which stores chemical energy that can be used for the Calvin cycle.
Photophosphorylation of sugar moiety produces ATP (adenosine triphosphate) which stores chemical energy that can be used for the Calvin cycle.
What is Calvin cycle?The Calvin cycle, often known as the Calvin-Benson cycle after its discoverers, is a series of chemical processes that collect carbon dioxide.
It is also referred to as the C3 cycle alone. C3 plants are defined as those that only fix carbon through the Calvin cycle. In the stroma of chloroplasts, carbon dioxide diffuses and mixes with the five-carbon sugar ribulose 1, 5-biphosphate (RuBP).
Rubisco, a large molecule that may be the most prevalent organic compound on Earth, is the name of the enzyme that catalyzes this reaction.
The three-carbon chemical 3-PGA is changed into another three-carbon compound termed G3P by ATP and NADPH using their stored energy. A reduction reaction is the name for this kind of reaction.
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A researcher is performing a gram stain on a specimen in a medical laboratory. She gets distracted by a phone call and accidentally uses water instead of alcohol to decolorize. What may happen as a result of this mistake?
Answer:
Explanation:
It wouldn't be able to separate negative gram from positive. it would not wash away some of the crystal violet stain effectively. Gram - Cells will appear to be the same hue. Water can also dilute cells or wash them off the slide
Answer:
Explanation:
Would not be able to differentiate gram - from + it would not efficiently wash some of the crystal violet stain away. gram - cells will still remain the same color. also water may dilute or wash cells off the slide
What is the full name of NI3
Answer:
Nitrogen triiodide is NI3.
Explanation:
Dalton's experiments advanced our understanding of the atom withoutA.carefully observing and measuring events.B.discovering the internal structure of the atom.C.examining the ratios of elements involved in chemical reactions.D.testing a hypothesis.
Answer:
B: Discovering the internal structure of the atom.
Explanation:
Dalton’s experiments with gases marked the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton’s atomic theory is based can be
summarized as follows:
Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass, and chemical properties. The atoms of one element are different from the atoms of all other elements. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction.Dalton made no attempt to describe the structure or composition of atom —he had no idea what an atom is really like. But he did realize that the different properties shown by elements such as hydrogen and oxygen can be explained by assuming that hydrogen atoms are not the same as oxygen atoms.
The charge of B is +3 and the charge of O is -2. What is its formula?
A) B3O2
B) B2O3
C) O2B3
Hydrogen gas produced in the laboratory by the reaction of zinc and hydrochloric acid was 23) collected over water at 25°C. The barometric pressure at the time was 742.5 mm Hg. What is the pressure of dry hydrogen gas if the vapor pressure of water at 25'C is 23.8 mm Hg?
A) 760.0 mm Hg
B)718.7 mm Hg
C)742.5 mm Hg
D) 766.3 mm Hg
E) none of the above
Answer:
Explanation:
The balanced chemical reaction is:
[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2[/tex]
As the hydrogen is collected over water, the total pressure will be sum of pressure of water and pressure of dry hydrogen.
[tex]p_{total}=p_{H_2}+p_{H_2O}[/tex]
[tex]p_{total}[/tex] = 742.5 mm Hg
[tex]p_{H_2}[/tex] = ?
[tex]p_{H_2O}[/tex] = 23.8 mm Hg
Putting in the values:
[tex]742.5=p_{H_2}+23.8[/tex]
[tex]742.5-23.8=p_{H_2}[/tex]
[tex]p_{H_2}=718.7mmHg[/tex]
Thus the pressure of dry hydrogen gas is 718.7 mm Hg
An apparatus consists of a 4.0 dm^3 flask containing nitrogen gas at 25°c and 803 kPa, it is joined by a valve to a 10.0dm^3 flask containing argon gas 25°c and 47.2KPa. The valve is opened and the gases mix , what is the partial pressure of each gas and calculate the total pressure of the gas mixture
Answer:
so total pressure is = 261.92 kPa
partial pressure of Ar gas= 33.50 Kpa
partial pressure of N2 gas = 227 kPa
Explanation:
PV = nRT
n = PV / RT
as we know that 1dm3 = 1 Liter
smaller flask before mixing
ideal gas constant
n = (803 kPa) x (4L) / ((8.3144621 L kPa/K mol) x (25 + 273 K))
n = (3212 kPa L) / ((8.3144621 L kPa/K mol) x (298 K))
n = (3212 kPa L) /2477.7
n= 1.29 mol
In the larger flask:
n = (47.2 kPa) x (10 L) / ((8.3144621 L kPa/K mol) x (25 + 273 K)) =
n = (472 kpaL/ 2477.7
n= 0.19 mol
PV = nRT
P = nRT / V
After mixing:
P = (1.29 mol + 0.19 mol) x (8.3144621 L kPa/K mol) x (25+ 273 K) / (4 L + 10 L)
= (1.48) x (2477.7 / (14 L)
= 261.92 kPa total pressure
so total pressure is = 261.92 kPa
a)
(261 kPa) x (0.19 mol Ar) / (1.29 + 0.19mol) =
= 49.59/1.48 = 33.50 Kpa for Arg
b)
(261 kPa total) - (33.50 kPa Ar) = 227 kPa for N2
partial pressure of Ar gas= 33.50 Kpa
partial pressure of N2 gas = 227 kPa
A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container at 20 ∘C. A quantity of O2 gas originally at 5.21 atm and 26 ∘C in a 5.10 −L container is transferred to this same container.
To find the final pressure of the gas mixture after transferring the gases, we can use the ideal gas law equation. The final pressure is calculated based on the total number of moles of gas and the final volume of the container. The final pressure of the gas mixture is 0.229 atm.
Explanation:To find the final pressure of the gas mixture after the transfer, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We can calculate the number of moles of N2 gas using the information given:
n = PV/RT = (5.23 atm)(1.20 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 0.287 mol
Similarly, we can calculate the number of moles of O2 gas:
n = PV/RT = (5.21 atm)(5.10 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 1.05 mol
To find the final pressure, we need to find the total number of moles of gas and the final volume of the container:
Total moles of gas = moles of N2 + moles of O2 = 0.287 mol + 1.05 mol = 1.34 mol
Final volume of the container = sum of the initial volumes = 1.20 L + 14.5 L = 15.7 L
Now we can use the ideal gas law again to calculate the final pressure:
P = nRT/V = (1.34 mol)(0.0821 atm·L/mol·K)(20 + 273 K)/(15.7 L) = 0.229 atm
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Which of the following were found to destroy ozone in the upper atmosphere?A. sulfur oxides.B. CFCs.C. nitrogen oxides.D. carbon dioxide.
Answer:
sulfur oxides
Explanation: