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2 years ago

12.2 g of N₂ gas at 1132 torr and 91°C has a volume of

Chemistry

1 answer:

Bingel [31]2 years ago

6 0

Answer:

8.73 L

Explanation:

First, you need to convert grams to moles using the molar mass.

Molar Mass (N₂): 2(14.009 g/mol)

Molar Mass (N₂): 28.018 g/mol

12.2 grams N₂ 1 mole
---------------------- x ------------------------ = 0.435 moles N₂
28.018 grams

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (torr)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 torr*L/mol*K)

-----> T = temperature (K)

After converting the temperature from Celsius to Kelvin, you can plug the given values into the equation.

P = 1132 torr R = 62.36 torr*L/mol*K

V = ? L T = 91 °C + 273.15 = 364.15 K

n = 0.435 moles

PV = nRT

(1132 torr)V = (0.435 moles)(62.36 torr*L/mol*K)(364.15 K)

(1132 torr)V = 9888.015

V = 8.73 L

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Helium is mixed with oxygen gas for deep-sea divers. Calculate the percent by volume of oxygen gas in the mixture if the diver h

kvasek [131]

Answer : The percent by volume of oxygen gas in the mixture is, 5.3 %

Explanation :

According to the Raoult's law,

$p^o_A=X_A\times p_A$

where,

$p_A$ = total partial pressure of solution = 3.8 atm

$p^o_A$ = partial pressure of oxygen = 0.20 atm

$X_A$ = mole fraction of oxygen = ?

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

$p^o_A=X_A\times p_A$

$0.20atm=X_A\times 3.8atm$

$X_A=0.0526$

Now we have to calculate the percent by volume of oxygen gas in the mixture.

The mole percent of oxygen gas = $0.0526\times 100=5.3\%$

As we know that, there is a direct relation between the volume of moles.

So, mole percent of oxygen gas = volume percent of oxygen gas

Volume percent of oxygen gas = $0.0526\times 100=5.3\%$

Therefore, the percent by volume of oxygen gas in the mixture is, 5.3 %

5 0

4 years ago

Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?

dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

C₄H₁₀: 2 moles
O₂: 13 moles
CO₂: 8 moles
H₂O: 10 moles

Being:

C: 12 g/mole
H: 1 g/mole
O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
O₂: 2*16 g/mole= 32 g/mole
CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

C₄H₁₀: 2 moles* 58 g/mole= 116 g
O₂: 13 moles* 32 g/mole= 416 g
CO₂: 8 moles* 44 g/mole= 352 g
H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

$mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10} }{116 grams of C_{4}H_{10}}$

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0

3 years ago

A 2uc charge is at rest in a magnetic field of 2 t pointing along the x-axis. what is the force acting on this charge in the mag

sp2606 [1]

Zero...bcz force which acts on this charge is f=qvB q=2uc ,B=2T and v=o
so,f=zero...

4 0

4 years ago

A soft drink contains 63 g of sugar in 378 g of H2O. What is the concentration of sugar in the soft drink in mass percent

Doss [256]

Answer:

$\% m/m= 14.3\%$