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ludmilkaskok [199]
3 years ago
8

At constant temperature and pressure, 2.05 g of oxygen gas O2 is added to a 1.0 L balloon containing 1.00 g of O2. What is the n

ew volume of the balloon?
Chemistry
1 answer:
leonid [27]3 years ago
5 0

Answer:

1.50 L.

Explanation:

  • From the general gas law:

<em>PV = nRT,</em>

Where, P is the pressure if the gas,

V is the volume if the gas container,

n is the no. of gas moles,

R is the general gas constant,

T is the temperature of the gas.

  • At constant P and T:

n₁V₂ = n₂V₁.

V₁ = 1.0 L, V₂ = ??? L.

n₁ = mass/molar mass = (2.05 g)/(32.0 g/mol) = 0.064 mol.

  • n₂ is the no. of moles of the total gas (2.05 g + 1.0 g).

n₂ = n₁ + (1.00 g)/(32.0 g/mol) = 0.0953 mol.

<em>∴ V₂ = n₂V₁/n₁ </em>= (0.0953 mol)(1.0 L)/(0.064 mol) = <em>1.489 L ≅ 1.50 L.</em>

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A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba ( OH ) 2 for complete neutralization. What is the c
oksano4ka [1.4K]

Answer: 0.17M

Explanation:

The equation for the reaction is :

2HNO3 + Ba(OH)2 —> Ba(NO3)2 + 2H2O

From the balanced equation, we obtain :

nA = mole of acid = 2

nB = mole of the base = 1

From the question, we obtain:

Va = Vol. Of acid = 0.105L

Ma = conc. Of acid =?

Vb = Vol of base = 35.7 mL = 0.0357L

Mb = conc. of base = 0.25M.

We solve for the conc. of the acid using:

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(Ma x 0.105) / (0.25x0.0357) = 2

Cross multiply to express in linear form. We have:

Ma x 0.105 = 0.25 x 0.0357 x 2

Divide both side by 0.105. We have

Ma = (0.25 x 0.0357 x 2) / 0.105

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