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vlabodo [156]
2 years ago
11

In a mixture of argon and hydrogen, occupying a volume of 1.18 l at 894.6 mmhg and 44.1oc, it is found that the total mass of th

e sample is 1.25 g. what is the partial pressure of argon?
Chemistry
1 answer:
alexandr1967 [171]2 years ago
6 0

To solve the question, there is a need to use the equation:

PV = nRT

(894.6/760) × 1.18 = n × 0.0821 × (273 + 44.1)

By solving we get:

Total moles, n = 0.053

Assume, the moles of argon as a and of hydrogen as b,

So,

40 × a + 2 × b = 1.25 --------- (i)

a + b = 0.053 ------- (ii)

By solving i and ii we get:

a = 0.03,

Thus, mole fraction of Ar = XAr = 0.03/0.053 = 0.57

So,

Partial pressure of Ar = 894.6 × XAr = 894.6 × 0.57 = 509.92 mm Hg



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Explanation:

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Iron has a density of 7.87 g/cm3. What is the volume in cm3 of 3.729 g of iron?
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If iron has a density of 7.87g/cm³ and a mass of 3.729g, then the volume of iron is 0.474cm³

HOW TO CALCULATE VOLUME:

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Volume (mL) = mass (g) ÷ density (g/mL)

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6 0
2 years ago
What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?
Vadim26 [7]

Answer:

158 L.

Explanation:

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Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

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Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

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Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

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V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}

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The SI unit for temperature is Kelvin.

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