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Hitman42 [59]
3 years ago
13

For the following incomplete combustion reaction, what are the coefficients necessary to balance the equation:

Chemistry
1 answer:
Alexxandr [17]3 years ago
6 0

Answer: A

Explanation:

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How many grams of helium must be released to reduce the pressure to 65 atmatm assuming ideal gas behavior? Express the answer in
lyudmila [28]

The question is incomplete, here is the complete question:

How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.

<u>Answer:</u> The mass of helium released is 1.6 grams

<u>Explanation:</u>

We are given:

Mass of helium in the cylinder = 5.209 g

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{w}{M}RT

where,

P = Pressure of the gas  = 65 atm

V = Volume of the gas  = 334 mL = 0.334 L   (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = ?

M = Molar mass of helium gas  = 4 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gas = 23^oC=[23+273]K=296K

Putting values in above equation, we get:

65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g

Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g

Hence, the mass of helium released is 1.6 grams

5 0
3 years ago
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