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steposvetlana [31]
3 years ago
15

Two different gases are stored in separate closed containers. If both gases have a temperature of 50°C, which statement is true?

Chemistry
2 answers:
Taya2010 [7]3 years ago
7 0
D is the correct answer to your problem
maw [93]3 years ago
6 0

Answer: Option (D) is the correct answer.

Explanation:

When a gas contains vibrational, translational, and rotational kinetic energies then their sum is known as average kinetic energy.

Therefore, when two molecules are mixed together at a certain temperature then the molecules of respective gases collide in each of the containers. Thus, these particles will have vibrational, translational, and rotational kinetic energies, that is, average kinetic energy.

Hence, we can conclude that the gas molecules in both containers have the same average kinetic energy.

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Which represents a balanced nuclear equation?
shtirl [24]

Answer: second answer

Explanation:

The thing is that every new nuclear cycle a new element forms and reduces an electron. I may be wrong but this is the most logical and scientifically correct answer

3 0
3 years ago
What do blizzards consist of? (Select all that apply)
Maru [420]

Answer:

Heavy Snowfall

High Winds

Extremely Low Temperatures

Reasoning:

Lots of snow, kinda self explanatory

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2 years ago
Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found
Igoryamba

Answer:

The concentrations are :

[HAsc^-]=0.000702 M

[Asc^{2-}]=5.92\times 10^{-8} M

The pH of the solution is 3.15.

Explanation:

H_2Asc\rightleftharpoons HAs^-+H^+         K_{a1}=1.0\times 10^{-5}

Initial

c                0              0

Equilibrium

c-x                x          x

K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}

1.0\times 10^{-5}=\frac{x\times x}{(c-x)}

1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}

Solving for x:

x = 0.000702 M

[HAsc^-]=0.000702 M

HAsc^-\rightleftharpoons As^{2-}+H^+        K_{a2}=5\times 10^{-12}

Initially

x                0          0

At equilibrium ;

(x - y)            y         y

K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}

5\times 10^{-12}=\frac{y\times y}{(x-y)}

5\times 10^{-12}=\frac{y^2}{(x-y)}

Putting value of x = 0.000702 M

5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}

y=5.92\times 10^{-8} M

[Asc^{2-}]=5.92\times 10^{-8} M

Total concentration of [H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M

The pH of the solution :

pH=-\log[H^+]

pH=-\log[7.0206\times 10^{-4} M}=3.15

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ZnBr2. hopes this helps
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