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Leya [2.2K]
3 years ago
15

When 38^88Sr decays to 34^84Kr, a(n) ______________ is emitted

Chemistry
1 answer:
navik [9.2K]3 years ago
3 0

Answer:

The correct answer is - alpha particle and positron.

Explanation:

In this question, it is given that, 38^88Sr decays to 34^84Kr, which means there is an atomic number decrease by 4, 38 to 34, and atomic mass decreases by 4 as well 88 to 84.

A decrease in the atomic mass is possible only when there is an emission of the alpha particle as an alpha particle is made of 2 protons and 2 neutrons. If an atom emits an alpha particle, there is a change in atomic number as it decreases by two, and its mass number decreases by four.

So after the emission of an alpha particle, the new atom would be

38^88Sr=> 36^84X => 34^84Kr

so there is also two positron emission that leads to decrease in atomic number by one with each emission:

38^88Sr=> 2^4He+ 36^84X => 36^84X + 2(1^0β+) => 34^84Kr

Positron decay is the conversion of a proton into a neutron with the emission of a positron that causes the atomic number is decreased by one, which causes a change in the elemental identity of the daughter isotope.

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Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of
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The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}

For the oxygen:

81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

\frac{0.81}{2}=0.41

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\frac{2.54}{19}=0.13

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}

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