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marshall27 [118]
3 years ago
7

What happens to the volume of a gas in a closed container if the temperature increases, but the pressure remains the same? Why?

Chemistry
1 answer:
labwork [276]3 years ago
3 0

Answer:

Volume will goes to increase.

Explanation:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.

Now we will put the suppose values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 4.5 L × 348 K / 298 k

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

Hence prove that volume increase by increasing the temperature.

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Decide whether a chemical reaction happens in either of the following situations. If a reaction does happen, write the chemical
ValentinkaMS [17]

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1) Yes

2) No

Explanation:

A strip of solid magnesium metal is put into a beaker of 0.053M ZnSO4 solution-YES

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Mg(s) + ZnSO4(aq) ----------> MgSO4(aq) +Zn(s)

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A strip of solid zinc metal is put into a beaker of 0.058M MgBr2 solution. NO

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8 0
3 years ago
2 H2+O2=2 H2O
tia_tia [17]

Answer:

The percent yield is about 66.6%, but with sig figs it's 70%

Explanation:

The formula for percent yield is: (Actual Yield/Theoretical Yield) * 100

1. In order to solve for percent yield, you need both an actual and theoretical yield. This problem gives us an actual yield for H2O, which is 75 grams. Great, you're halfway there!

2. Next, you need to solve for the theoretical yield. In order to do so, you need to set up your stoichiometric ratio.

100g O_{2}  * (\frac{1 mol O_{2} }{31.998g O_{2} } )*(\frac{2 mol H_{2}O }{1 mol O_{2} } )*(\frac{18.015gH_{2}O }{1 mol H_{2}O } ) = 112.6g H_{2}O

a) Take your given 100g of O2 and first convert it to moles

b) Then take the O2 in moles and create your mole-to-mole ratio. You're only given 2 values (75g of H2O and 100g of O2), so you know to use O2 and H2O in your ratio. The coefficients are essentially your ratio, so for H2O to O2 it's 2:1

c) Then convert the value from the mole-to-mole ratio back into grams to get your theoretical yield (as values from mole ratios are in moles)

3. Now that you have both your actual and theoretical yields, you can plug it into the equation and solve.

\frac{Yield_{Actual} }{Yield_{Theoretical} } * 100 = \frac{75}{112.6} = 66.60403646

Which is 66.6% when rounded. With sig figs, again, it is 70%

5 0
3 years ago
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