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Margarita [4]
3 years ago
9

What is an energy source produced from nuclear reactions

Chemistry
1 answer:
valkas [14]3 years ago
6 0

Answer:

I believe it would be electricity.

Explanation:

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Determine which law is appropriate for solving the following problem. Carbon dioxide is in a steel tank at 20°C, 10 liters and 1
patriot [66]

<u>Answer:</u>

Law used: Combined Gas Law

<u>Explanation:</u>

We are given the following problem:

Carbon dioxide is in a steel tank at 20°C, 10 liters and 1 atm. What is the pressure on the gas when the tank is heated to 100°C?

To solve this, the most appropriate law that can be used it Combined Gas Law, which is the result of combining the Boyle's law, Charles' law, and Gay-Lussac's law together.

5 0
3 years ago
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Need help answering this question
a_sh-v [17]

Answer:

1st = saturated

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4 0
1 year ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

3 0
3 years ago
Read 2 more answers
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