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gavmur [86]
3 years ago
6

HELP PLEASE, I NEED THIS BY TODAY!

Chemistry
1 answer:
labwork [276]3 years ago
6 0

Answer:

2.48 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 10 g

Time (t) = 1407.6 million years

Amount remaining (N) =?

Next, we shall determine the rate of decay (K) of uranium-235. This can be obtained as follow:

NOTE: Uranium-235 has a half life of 700 million years.

Decay constant (K) =?

Half life (t½) = 700 million years

K = 0.693/t½

K = 0.693/700

K = 9.9×10¯⁴ / year

Therefore, Uranium-235 decay at a rate of 9.9×10¯⁴ / year.

Finally, we shall determine the amount of Uranium-235 remaining after 1407.6 million years as follow:

Original amount (N₀) = 10 g

Time (t) = 1407.6 million years

Decay constant (K) = 9.9×10¯⁴ / year

Amount remaining (N) =?

Log (N₀/N) = kt /2.3

Log (10/N) = (9.9×10¯⁴ × 1407.6) /2.3

Log (10/N) = 0.60588

10/N = antilog (0.60588)

10/N = 4.04

Cross multiply

10 = 4.04 × N

Divide both side by 4.04

N = 10/4.04

N = 2.48 g

Therefore, 2.48 g of uranium-235 is remaining after 1407.6 million years.

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In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌
Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of  CO = 0.35 mole

Volume of container = 1 L

Initial concentration of CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M

Initial moles of  H_2O = 0.40 mole

Volume of container = 1 L

Initial concentration of H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M

equilibrium concentration of CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M [/tex]

The given balanced equilibrium reaction is,

                            CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.            0.35 M       0.40M       0     0

At eqm. conc.    (0.35-x) M   (0.40-x) M   (x) M    (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}

K_c=\frac{x\times x}{(0.40-x)(0.35-x)}

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}

K_c=0.70

Thus the value of the equilibrium constant is 0.70.

5 0
3 years ago
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