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ArbitrLikvidat [17]
2 years ago
8

A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

essure of the mixture is 3.9 bar?A. 0.67 barB. 1.33 barC. 0.98 barD. 0.39 barE. 2.5 bar
Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
7 0

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

P=p_{N_2}+p_{O_2}+p_{Ar}

p_{N_2}=P\times \chi_{N_2}

p_{O_2}=P\times \chi_{O_2}

p_{Ar}=P\times \chi_{Ar}

where,

P = total pressure = 3.9 bar

p_{N_2} = partial pressure of nitrogen gas  

p_{O_2} = partial pressure of oxygen gas  

p_{Ar} = partial pressure of argon gases  

\chi_{N_2} = Mole fraction of nitrogen gas  = 0.25

\chi_{O_2} = Mole fraction of oxygen gas  = 0.65

\chi_{Ar} = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar

Partial pressure of oxygen gas :

p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar

Partial pressure of argon gas :

p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar

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If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.
Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water =  Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0
2 years ago
What is molarity of 3 mole of hydrochloric acid in 3L of water
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1

Explanation:

6 0
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How many ethyne molecules are contained in 84.3 grams of ethyne (C2H2)?
ololo11 [35]
(~26grams/mole) and Avogadros # (6.022x10^23) 84.3grams x 1mole/26grams x 6.022x10^23 molecules/mole = 1.95x10^24 molecules of C2H2
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2 years ago
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An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine
nekit [7.7K]

Answer:

The molecular formula of the compound is C_{7}H_{6}O_{2}.

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

Carbon

m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)

m_{C} = 84.067\,g

Hydrogen

m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)

m_{H} = 6.057\,g

Oxygen

m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)

m_{O} = 31.995\,g

Now, the number of moles (n), measured in moles, of each element are calculated by the following expression:

n = \frac{m}{M}

Where:

m - Mass of the element, measured in grams.

M- Molar mass of the element, measured in grams per mol.

Carbon (m_{C} = 84.067\,g, M_{C} = 12.011\,\frac{g}{mol})

n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }

n = 7

Hydrogen (m_{H} = 6.057\,g, M_{H} = 1.008\,\frac{g}{mol})

n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }

n = 6

Oxygen (m_{O} = 31.995\,g, M_{O} = 15.999\,\frac{g}{mol})

n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }

n = 2

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

C_{7}H_{6}O_{2}

8 0
2 years ago
Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base
Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

Explanation :

First we have to calculate the value of K_b.

As we know that,

K_a\times K_b=K_w

where,

K_a = dissociation constant of an acid = 1.8\times 10^{-5}

K_b = dissociation constant of a base = ?

K_w = dissociation constant of water = 1\times 10^{-14}

Now put all the given values in the above expression, we get the dissociation constant of a base.

1.8\times 10^{-5}\times K_b=1\times 10^{-14}

K_b=5.5\times 10^{-10}

Now we have to calculate the concentration of hydroxide ion.

Formula used :

[OH^-]=(K_b\times C)^{\frac{1}{2}}

where,

C is the concentration of solution.

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

[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}

[OH^-]=1.3\times 10^{-5}M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1.3\times 10^{-5})

pOH=4.9

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1

Therefore, the pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

4 0
2 years ago
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