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2 years ago

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How many moles are in 182 grams of BC13?

1 answer:

spayn [35]2 years ago

5 0

Answer:

?

Explanation:

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A. There must be a change in chemical properties.

At least 2 or 3 atoms must change for it to be considered chemical. There must be some type of reactants that will change this. Otherwise, it's not a chemical reaction.

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1 year ago

What are elements and how many elements discovered

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2 years ago

How many grams of water will form when 10.54g of h2o react with 95.10g of O2

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3 years ago

Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

Marat540 [252]

Answer : The partial pressure of helium is, $1.815\times 10^4KPa$

Solution : Given,

Molar mass of $O_2$ = 32 g/mole

Molar mass of helium = 4 g/mole

Molar mass of $N_2$ = 28 g/mole

Total pressure of gas = $2.07\times 10^4KPa$

As we are given gases in percent, that means 10 g of oxygen gas, 50 g of helium gas and 40 g of nitrogen gas present in 100 g of mixture.

First we have to calculate the moles of oxygen, helium and nitrogen gas.

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{10g}{32g/mole}=0.3125moles$

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{50g}{4g/mole}=12.5moles$

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{40g}{28g/mole}=1.428moles$

Now we have to calculate the total number of moles of gas mixture.

$\text{Total number of moles of gas}=\text{Moles of oxygen gas}+\text{Mole of helium gas}+\text{Moles of nitrogen gas}$

$\text{Total number of moles of gas}=0.3125+12.5+1.428=14.24moles$

Now we have to calculate the moles fraction of helium gas.

$\text{Mole fraction of He gas}=\frac{\text{Moles of He gas}}{\text{Total number of moles of gas}}=\frac{12.5}{14.25}=0.877$

Now we have to calculate the partial pressure of helium.

$p_{He}=X_{He}\times P_T$

where,

$p_{He}$ = partial pressure of helium

$P_T$ = total pressure

$X_{He}$ = mole fraction of helium

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

$p_{He}=(0.877)\times (2.07\times 10^4KPa)=1.815\times 10^4KPa$

Therefore, the partial pressure of helium is, $1.815\times 10^4KPa$

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2 years ago