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Svet_ta [14]
3 years ago
15

2.At 35°C, a small sample of methane gas (CH4) has a volume of 1.5 liters. The temperature of the methane gas is slowly cooled t

o 17°C. What is the new volume of the methane sample?​
Chemistry
1 answer:
defon3 years ago
5 0

Answer:

V₂ = 1.41 L

Explanation:

Given data:

Initial temperature = 35°C (35 +273.15 K = 308.15 K)

Initial volume = 1.5 L

Final temperature = 17°C (17+273.15 K = 290.15 K)

Final volume = ?

Solution:

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

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1.5 L × 290.15 K / 308.15 k

V₂ = 435.23 L.K / 308.15 k

V₂ = 1.41 L

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Explanation:

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3 years ago
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Answer:

Average atomic mass = 17.5 amu.

Explanation:

Given data:

X-17 isotope = atomic mass17.2 amu, abundance:78.99%

X-18isotope =  atomic mass 18.1 amu, abundance 10.00%

X-19isotope = atomic mass:19.1 amu, abundance: 11.01%

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Solution:

Average atomic mass  = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass)  / 100

Average atomic mass  = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100

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8 0
3 years ago
Before we can use this equation for
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Answer:

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Explanation:

Chemical equation:

C₂H₆ +  O₂     →      CO₂ + H₂O

Balanced chemical equation:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Step 1:

2C₂H₆ +  O₂     →      CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C = 1

H = 12                                    H = 2

O = 2                                     O = 3

Step 2:

2C₂H₆ +  O₂     →      4CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 2

O = 2                                     O = 9

Step 3:

2C₂H₆ +  O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 2                                     O = 14

Step 4:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 14                                     O = 14

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