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Archy [21]
3 years ago
12

What happens to isotopes that are unstable?

Chemistry
1 answer:
Lesechka [4]3 years ago
5 0

<u>Explanation:</u>

Isotopes are defined as the chemical species of the same element which differs in the number of neutrons. The isotopes which are unstable are known as radioactive isotope. A radioactive (unstable )isotope can undergo 3 decay process:

1. Alpha Decay: In this decay process, a larger nuclei decays into smaller nuclei by releasing alpha particle. The particle released has a charge of +2 and a mass of 4 units.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha

2. Beta-minus decay: In this decay process, a neutron gets converted into a proton and an electron. the particle released during this process is a beta-particle.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

3. Beta-plus decay: In this decay process, a protons gets converted into a neutron and electron-neutrino particle. The particle released during this process is a positron particle.

_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0\beta

Isotopes which are unstable in nature can undergo these 3 decay processes.

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This results in a precipitate within a liquid or aqueous solution

Hope this helps

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You have a cup of hot water and you dissolve as much salt in it as possible. You then put the cup into the refrigerator and you
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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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63.499%

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