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Norma-Jean [14]
2 years ago
12

The composition of dry air at sea level is 78.03% N2, 20.99% O2, and 0.033% CO2 by volume. (a) calculate the average molar mass

of this air sample (b) calculate the partial pressures of N2, O2, and CO2 in atm. (at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas). Also what should be used for constant temperature and pressure?
Chemistry
1 answer:
levacccp [35]2 years ago
7 0

Answer:

the average molar mass of this air sample can be calculated as

addition of the product of the average molar weights of the component gases and their percentage compositions

1. Average Molar mass of Air = 0.7803 x 28 + 0.2099 x 32 + 0.00033 x 44 = 28.58g/mol

2. The partial pressures of N2, O2, and CO2 in atm.

From Ideal gas law, at stp, Volume of air V=22.4L/mol

PV =nRT

Since, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas

Total Pressure P=1atm

Partial Pressure p = mol fraction x P

Volume of N2 = 0.7803 x 22.4L = 17.47L, Partial Pressure = 0.7803atm

Volume of O2 = 0.2099 x 22.4L = 4.68L  Partial Pressure = 0.209atm

Volume of CO2 = 0.00033 x 22.4L = 0.00739L, Partial Pressure = 0.033atm

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If 0.255 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ co
Ber [7]

Answer: 6.162g of Ag2SO4 could be formed

Explanation:

Given;

0.255 moles of AgNO3

0.155 moles of H2SO4

Balanced equation will be given as;

2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)

Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,

Therefore the number of moles of Ag2SO4 produced is given by,

n(Ag2SO4) = 0.255 mol of AgNO3 ×

[0.155mol H2SO4 ÷ 2 mol AgNO3] x

[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]

= 0.0198 mol of Ag2SO4.

mass = no of moles x molar mass

From literature, molar mass of Ag2SO4 = 311.799g/mol.

Thus,

Mass = 0.0198 x 311.799

= 6.162g

Therefore, 6.162g of Ag2SO4 could be formed

4 0
2 years ago
The half life of a certain radioactive element is 1,250 years what percent of the atoms remain after 7500 years
andrezito [222]
Each half-life results in ~50% (1/2) of the original element remaining.

7500/1250 = 6 half-lives, so 100(1/2)^6
= 100(0.015625)
= 1.5625% of the original element would remain
7 0
2 years ago
Read 2 more answers
What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c
olya-2409 [2.1K]

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

pOH=pKb+log\frac{[salt]}{[base]}

pH = 14- pOH

Let us calculate pOHpOH=3.43+ (-0.397)=3.03

pH=14-pOH=14-3.03=10.97

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M

[base] = [Methylamine]=0.10

After mixing with salt

[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M

pKb= -log[Kb]= 3.43

Putting values

pOH = 3.43+log(\frac{[0.0286]}{0.0714}

4 0
2 years ago
Analyze the example of this door knob wheel and axle.
bulgar [2K]

Answer:

28

Explanation:

Velocity ratio= Radius of wheel/radius of axle

Radius of wheel= 4.125 inches

Radius of axle= 0.125 inches

Velocity ratio = 4.125/0.125 = 33

Then;

Efficiency = mechanical advantage/velocity ratio × 100

Since the efficiency of the system = 85%

85 = mechanical advantage/33 × 100

Mechanical advantage = 85 × 33/100 = 28

4 0
2 years ago
A substance is 89.2% carbon by mass. how much of the substance would be needed to recover 34.6 mol of pure carbon?
kari74 [83]
1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon, 
then, (415.2/89.2) ×100 =  465.47 g of the substance will be required to yield 34.6 moles of carbon.
8 0
2 years ago
Read 2 more answers
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