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Leya [2.2K]
2 years ago
15

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

Chemistry
1 answer:
navik [9.2K]2 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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The amount of which subatomic particle is different between an atom and its ion?
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What is the order of increasing rate of eusion for the following gases, Ar, CO2, H2, N2?
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3 years ago
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If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?
Y_Kistochka [10]

Answer:

-26.125 kj

Explanation:

Given data:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

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ΔT = -25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m.c. ΔT

Q = 250.0 g × 4.18 j/g.°C × -25°C

Q = -26125 j

J to kJ

-26125 j ×1 kj /1000 j

-26.125 kj

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3 years ago
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