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melamori03 [73]

2 years ago

12

Calculate the number of atoms in 4.2 moles of sulfur atoms.

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

1 mole = 6.022*10^23 atoms
4.2 moles sulfur = 4.2* 6.022 * 10^23 sulfur atoms = 2.421*10^24 sulfur atoms

1 mole sulfur = <span>32.065 g sulfur
4.2 moles sulfur = 4.2 * </span>32.065 g sulfur = 128.901 g sulfur

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The following question appears on a quiz: ""You fill a tank with gas at 60Â°C to 100 kPa and seal it. You decrease the temperatu

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Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K$

Putting values in above equation, we get:

$\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa$

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2 years ago

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2 years ago

What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised

musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

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Initial volume (V₁): 3.65 L

Final pressure (P₂): 861.0 mmHg

Final volume (V₂): ?

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Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

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2 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

lions [1.4K]

<u>Answer:</u> The equilibrium concentration of $COF_2$ is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

<u>Initial:</u> 2.00

<u>At eqllm:</u> 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of $COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M$

Hence, the equilibrium concentration of $COF_2$ is 0.332 M

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