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3 years ago

7

A gas effuses 4. 0 times faster than oxygen (O2). What is the molecular mass of the gas? 1. 0 g/mol 2. 0 g/mol 4. 0 g/mol 7. 67

g/mol.

1 answer:

egoroff_w [7]3 years ago

5 0

Answer:

2.0

Explanation:

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Answer:

Earth takes in thermal energy from the Sun in a process called. ... Sunlight strikes Earth's surface at different angles. This angle is called the angle of. ⇒ insolation.

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4 0

3 years ago

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

<u>Step 1</u>: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

<u>Step 2</u>: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

<u>Step 3:</u> Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

<u>Step 4:</u> Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

<u>Step 5:</u> Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

<u>Step 6:</u> Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

<u>Step 7: </u>Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

<u>Step 8:</u> Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

<u>Step 9</u>: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

<u>Step 10</u>: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

3 years ago

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$

From the reaction, we conclude that

As, 1 mole of $KI$ react to give 1 mole of $AgI$

So, 0.0178 moles of $KI$ react to give 0.0178 moles of $AgI$

Thus,

Moles of AgI = Moles of $I^-$ anion = Moles of $Ag^+$ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

$\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}$

$\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M$

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

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After leaving the cockroaches in their metabolic chamber for one hour the gases in the chamber was measured. The fraction of oxy

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