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uranmaximum [27]
3 years ago
8

A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is th

e partial pressure of the helium gas?
Chemistry
1 answer:
lara [203]3 years ago
8 0

Answer is: the partial pressure of the helium gas is 0.158 atm.

p(mixture) = 0.48 atm; total pressure.

m(H₂) = 1.0 g; mass of hydrogen gas.

n(H₂) = m(H₂) ÷ M(H₂).

n(H₂) = 1.0 g ÷ 2 g/mol.

n(H₂) = 0.5 mol; amount of hydrogen.

m(He) = 1.0 g; mass of helium.

n(He) = 1 g ÷ 4 g/mol.

n(He) = 0.25 mol; amount of helium.

χ(H₂) = 0.5 mol ÷ 0.75 mol.

χ(H₂) = 0.67; mole fraction of hydrogen.

χ(He) = 0.25 mol ÷ 0.75 mol.

χ(He) = 0.33; mole fraction of helium.

p(He) = 0.33 · 0.48 atm.

p(He) = 0.158 atm; the partial pressure of the helium gas.

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A salt crystal has a mass 0.14mg,How many NaCl formula units does it contain?
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I’m struggling with dimensional analysis in chemistry! Will someone please help me with .74 Kcal/min to cal/sec ? Explain and gi
denis23 [38]

Answer:

            12.33 cal/sec

Explanation:

As we know,

                          1 Kcal  =  1000 cal

So,

                     0.74 Kcal  =  X cal

Solving for X,

                      X  =  (0.74 Kcal × 1000 cal) ÷ 1 Kcal

                      X  =  740 cal

Also we know that,

                      1 Minute  =  60 Seconds

Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,

                                       = 740 cal / 60 sec

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6 0
3 years ago
A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and t
mr Goodwill [35]

Answer : The mass of helium added to the cylinder was, 1.5 grams

Explanation :

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

V\propto n

or,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 3.50 L

n_1 = initial moles of gas = \frac{\text{Mass of He}}{\text{Molar mass of He}}=\frac{2.00g}{4g/mol}=0.5mol

n_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{2.00L}{0.5mol}=\frac{3.50L}{n_2}

n_2=0.875mol

Now we have to calculate the mass of helium were added to the cylinder.

\text{Mass of He}=\text{Moles of He}\times \text{Molar mass of He}

\text{Mass of He}=0.875mol\times 4g/mol=3.5g

Mass of helium added = 3.5 - 2.00 = 1.5 g

Thus, the mass of helium added to the cylinder was, 1.5 grams

8 0
3 years ago
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