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3 years ago

Complete the following radioactive decay problem.222 86 RN to 4 2 HE + ?

Chemistry

1 answer:

Andreas93 [3]3 years ago

8 0

Answer:

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Explanation:

The given nuclear reaction shows alpha decay.

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Properties of alpha radiations:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy

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State and Explain the trend in the atomic radius down a group

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Explanation:

<h2>The number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.</h2>

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1 year ago

What volume (L) will 3.56 mol NH3 occupy at STP. (1 point)

mart [117]

Answer:

Explanation:

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3 years ago

A sample of gas has an initial volume of 3.62 L at a pressure of 0.987 atm. If the volume of the gas decreases to 2.50 L, what w

statuscvo [17]

Answer:

0.033 atm

Explanation:

PV=nRT

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P= nRT/V

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3 years ago

Select the pair that consists of a base and its conjugate acid in that order. CO32−/CO22−

lana [24]

Answer: The pair that consists of a base and its conjugate acid in that order. $NH_3/NH_4^+$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$H_3PO_4\rightarrow H_PO_4{2^-}+2H^+$

$H_2CO_3\rightarrow HCO_3^-+H^+$

$NH_3+H^+\rightarrow NH_4^+$

$HCO_3^-\rightarrow CO_3^{2-}+H^+$

$NH_3$ is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms $NH_4^+$ which is a conjugate acid.

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3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.