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Hitman42 [59]
3 years ago
14

A gas has a volume of 5.00 L at 20 degrees Celsius. What is the volume of gas at 150 degrees Celsius.

Chemistry
1 answer:
WARRIOR [948]3 years ago
5 0

Answer:

7.22 L

Explanation:

From the question,

Applying Charles law,

V₁/T₁ = V₂/T₂...................... Equation 1

Where V₁ = Initial volume of gas, V₂ = Final volume of gas, T₁ = Initial Temperature of gas in Kelvin, T₂ = Final Temperature of gas in Kelvin

Make V₂ the subject of the equation

V₂ = (V₁×T₂)/T₁................. Equation 2

Given: V₁ = 5.00 L, T₁ = 20 °C = (20+273) K = 293 K, T₂ = 150 °C = (150+273) K = 423 K

Substitute these value into equation 2

V₂ = (5.00×423)/293

V₂ = 7.22 L

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So, number of moles in 5.4\times 10^{24} molecules are :

n = \dfrac{5.4\times 10^{24}}{6.022\times 10^{23}}\\\\n = 8.97 \ moles

Therefore, number of moles are 8.97 .

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3 years ago
When placed in a 100mL cube shaped box a substance fills the cube and has a volume of 100 mL. When places in a 50mL tube that sa
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the state of the substance D- Gas.

8 0
3 years ago
The number 4012 has how many significant figures?​
grandymaker [24]

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3 0
3 years ago
The heaviest element to be created by exothermic nuclear fusion is:
Lunna [17]

Answer:

The heaviest element to be created by exothermic nuclear fusion is Iron

Explanation:

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8 0
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Read 2 more answers
Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
3 years ago
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