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3 years ago

What is the mass of 1.75 moles of HF?

Chemistry

1 answer:

olganol [36]3 years ago

7 0

Answer:

Mass = 35.02 g

Explanation:

Given data:

Number of moles = 1.75 mol

Mass of HF = ?

Solution:

Formula:

Mass = number of moles × molar mass

Molar mass of HF = 20.01 g/mol

by putting values,

Mass = 1.75 mol × 20.01 g/mol

Mass = 35.02 g

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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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3 years ago

Check the approach to obtain the correct conversion equation. Use the temperature in °F to find the temperature in °C.

Maru [420]

Answer:

( °F − 32) × 5/9 = °C

Explanation:

Also there is a mental calculation to convert from Fahrenheit to Celsius. The ratio 5/9 is approximately equal 0.55555….

Subtract 32º to adapt the equivalent in the Fahrenheit scale.

Divide the degrees Celsius by 2 (multiply by 0.5).

Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it to the number obtained previously.

Example: Convert 98.6º F to Centigrade.

98.6 - 32 = 66.6

66.6 * 1/2 = 33.3

33.3 * 1/10 = 3.3

33.3 + 3.3 = 36.6 which is an approximation in degrees Centigrade

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