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laila [671]
3 years ago
9

A weather balloon containing helium with a volume of 410.0 L rises in the atmosphere and is cooled from 17∘

Chemistry
1 answer:
likoan [24]3 years ago
3 0

Answer:

V₂ = 1500 Liters ( 2 sig. figs.)

Explanation:

Given the following gas law variables:

P₁ = 110.0KPa           P₂ = 25KPa

V₁ = 410 Liters          V₂ = ?

T₁ = 17°C ( = 290K)   T₂ = -27°C  ( = 248K)

P₁V₁/₁T₁ = P₂V₂/T₂  =>  V₂ = V₁(P₁/P₂)(T₂/T₂)

V₂ = 410L(110.0KPa/25KPa)(248k/290K) = 1542 L (calc. ans.)

V₂ = 1500 Liters ( 2 sig. figs.)

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3 years ago
Match each of the following forms of sugar to its best description?
prohojiy [21]

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What activates a convenient current starting to flow of a fluid?
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1 year ago
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it
lions [1.4K]

<u>Answer:</u> The equilibrium concentration of COF_2 is 0.332 M

<u>Explanation:</u>

We are given:

Initial concentration of COF_2 = 2.00 M

The given chemical equation follows:

                2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)

<u>Initial:</u>          2.00

<u>At eqllm:</u>     2.00-2x          x      x

The expression of K_c for above equation follows:

K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}

We are given:

K_c=6.30

Putting values in above expression, we get:

6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M

Hence, the equilibrium concentration of COF_2 is 0.332 M

4 0
3 years ago
A rigid 3.80 L sealed vessel contains 0.650 mol Ne, 0.321 mol Kr, and 0.190 mol Xe. Find the density of the mixture in g/L.
Eddi Din [679]

Answer:

17.09g/L

Explanation:

Density = total mass of elements/ volume

We need to find the mass of each mixture constituents using their molar mass:

mole = mass/molar mass

For Neon (Ne) which contains 0.650mol;

0.650 = mass/20.18

mass = 0.650 × 20.18

mass = 13.12g

For Krypton (Kr) which contains 0.321mol;

0.321 = mass/83.79

mass = 0.321 × 83.79

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For Xenon (Xe) which contains 0.190mol;

0.190 = mass/131.3

mass = 0.190 × 131.3

mass = 24.95g

Total mass = 13.12g + 26.89g + 24.95g = 64.96g

Density = total mass / volume

Density = 64.96g / 3.80L

Density of the mixture = 17.09g/L

7 0
3 years ago
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