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3 years ago

11

Which equation represents a fission reaction?

2 answers:

g100num [7]3 years ago

5 0

Correct Answer: Option B i.e. 235 92U + 1 0n→ 140 55Cs + 94 37Rb + 2 (1 0n)

Reason:
Nuclear fission<span> is either a nuclear </span>reaction<span> or a radioactive decay process in which the nucleus of a heavy atom splits into smaller atoms. During this process, large amount of heat is given out.
</span>
This condition of nuclear fission reaction is satisfied only by option B.

Rest all the available options are example of nuclear transmutation reactions.

lubasha [3.4K]3 years ago

3 0

Answer:

Option B

Explanation:

On edg

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Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium

pav-90 [236]

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

$k_{c}=3.07\times10^{-4}$

Temperature = 24°C

We know that,

The balance equation is

$2NOBr\Rightarrow 2NO+Br_{2}$

Initial concentration is,

$0.878\Rightarrow 0+0$

Concentration is,

$-2x\Rightarrow 2x+x$

Equilibrium concentration

$0.878-2x\Rightarrow 2x+x$

We need to calculate the value of x

Using formula of concentration

$k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}$

Put the value into the formula

$3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}$

$2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}$

$2x^2=0.0002367+0.001228x^2-0.0010536x$

$2x^2-0.001228x^2+0.0010536x-0.0002367=0$

$1.998772x^2+0.0010536x-0.0002367=0$

$x=0, 0.01062$

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

$concentration\ of\ NO=2x$

Put the value of x

$concentration\ of\ NO=2\times0.01062$

$concentration\ of\ NO=0.02124$

Hence, The equilibrium concentration of NO is 0.02124 M.

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4 years ago

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Natali5045456 [20]

Answer:

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Explanation:

Be sure to friend me and mark me as braniliest if im correct!

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bixtya [17]

Answer:

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Explanation:

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