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2 years ago

A

Chemistry

1 answer:

meriva2 years ago

6 0

Answer:

Option B is correct. A nuclear alpha decay

Explanation:

Step 1

This equation is a nuclear reaction. So it can be an alpha decay or a beta decay

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation that is showed.

A Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.

Option B is correct. A nuclear alpha decay

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3 years ago

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

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Explanation:

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

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$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

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Putting values in above equation, we get:

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$Q_p=\frac{(p_{NOCl})^2}{p_{Cl_2}\times (p_{NO})^2}$

We are given:

$p_{NOCl}=1.76atm$

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$p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{0.42\times (1.01)^2}=7.23$

We are given:

$K_p=1583.43$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium

As, $K_p>Q_p$ , the reaction will be favoring product side.

Hence, the reaction proceeds in the forward direction

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Explanation:

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