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Nostrana [21]
2 years ago
12

A gas mixture at 0°C and 1.0atm contains 0.010mol of H2, 0.015mol of O2, and 0.025mol of N2. Assuming ideal behavior, what is th

e partial pressure of hydrogen gas (H2) in the mixture?
A. About 0.010atm, because there is 0.010mol of H2 in the sample.

B. About 0.050atm, because there is 0.050mol of gases at 0°C and 1.0atm.

C. About 0.20atm, because H2 comprises 20% of the total number of moles of gas.

D. About 0.40atm, because the mole ratio of H2:O2:N2 is 0.4:0.6:1.
Chemistry
1 answer:
Olegator [25]2 years ago
7 0

Answer:

PH₂ = 0.2 atm

C) About 0.20atm, because H2 comprises 20% of the total number of moles of gas.

Explanation:

To determine the partial pressure of hydrogen gas (H2) in the mixture,

Partial pressure H₂ = Ptotal * xH₂

xH₂ = Mole fraction of H₂ = ∩H₂ / ( ∩H₂ + ∩O₂ + ∩N₂)

xH₂ = 0.01 / (0.01 + 0.015 + 0.025)

xH₂ = 0.01/0.05

xH₂ = 0.2

therefore

PH₂ = pT * xH₂

PH₂ = 1.0 atm * 0.2

PH₂ = 0.2 atm

so the correct option is C)  About 0.20atm, because H2 comprises 20% of the total number of moles of gas.

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A titration is performed to determine the amount of sulfuric acid, H2SO4, in a 6.5 mL sample taken from car battery. About 50 mL
Orlov [11]

Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL

Putting values in above equation, we get:

2\times M_1\times 56.5=1\times 0.5824\times 43.37

M_1=0.2235

Now to calculate the molarity of original solution:

M_1\times 6.5=0.2235\times 56.5

M_1=1.943

Thus the molar concentration of sulfuric acid in the original sample is 1.943 M

5 0
2 years ago
How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?
Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}

\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}

\text{Volume of solution (in L)}=2.96L

Therefore, the volume of solution will be 2.96 liters.

3 0
1 year ago
A sample of 22K gold contains the following: 22 g gold, 1.0 g silver, and 1.0 g copper. What is the percent gold in the sample?
iris [78.8K]

Answer:

35 percent

Explanation:

8 0
2 years ago
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21  = 13.1 g

3 0
1 year ago
What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?
Ilya [14]

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

4 0
1 year ago
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