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Viefleur [7K]
1 year ago
6

A mixture of 14.2 g of H2 and 36.7 g of Ar is placed in a 100.0 L container at

Chemistry
1 answer:
TiliK225 [7]1 year ago
3 0

a) The total pressure of the system is  1.79 atm

b) The mole fraction and partial pressure of hydrogen is  0.89 and 1.59 atm respectively

c) The mole fraction and the partial pressure  of argon is 0.11 and 0.19 atm.

<h3>What is the total pressure?</h3>

We know tat we can be able to obtain the total pressure in the system by the use of the ideal gas equation. We would have from the equation;

PV = nRT

P = pressure

V = volume

n = Number of moles

R = gas constant

T = temperature

Number of moles of hydrogen = 14.2 g/2g = 7.1 moles

Number of moles of Argon = 36.7 g/40 g/mol

= 0.92 moles

Total number of moles =  7.1 moles + 0.92 moles = 8.02 moles

Then;

P = nRT/V

P = 8.02 * 0.082 * 273/100

P = 1.79 atm

Mole fraction of hydrogen = 7.1/8.02 = 0.89

Partial pressure of hydrogen =  0.89 * 1.79 atm

= 1.59 atm

Mole fraction of argon = 0.92 / 8.02

= 0.11

Partial pressure of argon = 0.11  *  1.79 atm

= 0.19 atm

Learn more about partial pressure:brainly.com/question/13199169

#SPJ1

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3 years ago
Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h
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Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

k=\frac{0.693}{t_1_/_2}

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

k=\frac{0.693}{1.5 hrs}

The expression for the rate constant is :

k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0
3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
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2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

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Answer:

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