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Ronch [10]
3 years ago
14

An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the

molar mass (g/mol) of Q
Chemistry
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

The correct answer is 199.66 grams per mole.

Explanation:

Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,  

R1/R2 = √ M2/√ M1

Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.  

Rate Q/Rate N2 = √M of N2/ √M of Q

The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67

Now putting the values we get,  

rate of N2/2.67/rate of N2 = √28/ √M of Q

√M of Q = √ 28 × 2.67

M of Q = (√ 28 × 2.67)²

M of Q = 199.66 grams per mole

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Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

learn more about pH value:

brainly.com/question/12942138

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1 year ago
Elements ending in the electron configurations ns^1 are highly reactive metals . What family does these elements belong to ?
myrzilka [38]

Answer:

<h3>A . Alkali metals</h3>

Explanation:

The highlighted elements of the periodic table belong to the alkali metal element family. The alkali metals are recognized as a group and family of elements. These elements are metals. Sodium and potassium are examples of elements in this family.

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6 0
3 years ago
2. List and explain the three steps for solving a numeric problem.
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Answer:

The steps for solving a numeric word problem are analyze, calculate, and evaluate. To solve a word problem, you must first determine where you are starting from (identify what is known) and where you are going (identify the unknown).

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An aqueous solution has a hydroxide-ion concentration of 1.0 x 10-3 M. What is the pH of the solution?
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Explanation:

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Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =
Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be \frac{-196}{2}\times 1=-98kJ

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol

  • To calculate the heat of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released

n = number of moles = 0.0735 moles

\Delta H_{rxn} = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ

Hence, the amount of heat released is -7.203 kJ

8 0
3 years ago
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