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Evgen [1.6K]
3 years ago
15

How many atoms are found in 350.8 moles aluminum? _____________ atoms Al

Chemistry
1 answer:
Ksju [112]3 years ago
8 0

Answer:

2.11x10^26 atoms in 350.8 moles of Aluminum

Explanation:

I converted moles to atoms by using Avogadro's number which is 6.022x10^23

The solution is:

350.8 moles of Al x (6.022x10^23/1 mole) = 2.11x10^26 atoms

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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e
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The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

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In one of his experiments, Lavoisier placed 10.0 grams of mercury (II) oxide into a sealed container and heated it. The mercury
inysia [295]

Oxygen gas produced : 0.7 g

<h3>Further explanation</h3>

Given

10.0 grams HgO

9.3 grams Hg

Required

Oxygen gas produced

Solution

Reaction⇒Decomposition

2HgO(s)⇒2Hg(l)+O₂(g)

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

mass of reactants = mass of products

mass  HgO = mass Hg + mass O₂

10 g = 9.3 g + mass O₂

mass O₂ = 0.7 g

4 0
3 years ago
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