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Bezzdna [24]
2 years ago
7

If 0.213 moles of argon occupies a volume of 652 mL at a particular temp and pressure, what volume would it occupy if 0.162 mole

s of argon were added at the same conditions?
Chemistry
1 answer:
Andreas93 [3]2 years ago
3 0

Answer:

V₂  = 495.89 mL

Explanation:

Given data:

Initial number of moles = 0.213 mol

Initial volume = 652 mL

Final number of moles = 0.162 mol

Final volume = ?

Solution:

V₁/n₁    =    V₂/n₂

By putting values,

652 mL/0.213 mol   =  V₂ /0.162 mol

V₂  = 652 mL 0.162 mol /0.213 mol

V₂  = 105.62 mL.mol /0.213 mol

V₂  = 495.89 mL

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Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr
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2 years ago
One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal
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Answer:

Lithium oxide, Li₂O.  

Explanation:

Hello!  

In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:

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Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:

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Thus, we perform the operations in order to obtain:

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So we solve for x as shown below:

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