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Agata [3.3K]
3 years ago
6

Please help! need it asap !

Chemistry
1 answer:
BigorU [14]3 years ago
4 0

Answer:

2nd ans

Explanation:

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Consider the following properties of the atmosphere of the planet Mars at a particular measurement point on the surface, as meas
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a. 581.4 Pa

b. 3.33x10⁻⁴ mol/L

c. 3.49x10⁻⁴ mol/L

d. 0.015 g/L

Explanation:

a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:

pA = 0.9532*6.1

pA = 5.81452 mbar

pA = 5.814x10⁻³ bar

1 bar ----- 10000 Pa

5.814x10⁻³ bar--- pA

pA = 581.4 Pa

b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:

PV = nRT

Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.

n = PV/RT

n = (610*1)/(8.314*210)

n = 0.3494 mol

The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):

n = PV/RT

n = (581.4*0.9532)/(8.314*210)

n = 0.3174 mol

cA = n/V

cA = 0.3174/953.2

cA = 3.33x10⁻⁴ mol/L

c. c = ntotal/Vtotal

c = 0.3494/1000

c = 3.49x10⁻⁴ mol/L

d. The molar masses of the gases are:

CO₂: 44 g/mol

N₂: 28 g/mol

Ar: 40 g/mol

O₂: 32 g/mol

CO: 28 g/mol

The molar mass of the mixture is:

M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol

The mass concentration is the molar concentration multiplied by the molar mass:

3.49x10⁻⁴ mol/L * 43.36 g/mol

0.015 g/L

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In a combustion reaction of an alkene, CnH2n+2, identify the products that would form?
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Complete Combustion of any hydrocarbon produces CO2 and H2O ( carbon dioxide and water).

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