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alexandr1967 [171]
3 years ago
10

Which notation is used to represent gamma decay?

Chemistry
2 answers:
natali 33 [55]3 years ago
7 0
Gamma decay is a type of radioactivity in which an unstable atomic nucleus dissipates energy by gamma<span> emission, producing </span>gamma<span> rays. </span>gamma decay<span> in Science Expand. </span>gamma decay. A radioactive process in which an atomic nucleus loses energy by emitting a gamma<span> ray (a stream of high-energy photons).</span>
masya89 [10]3 years ago
7 0

<u>Answer:</u> The gamma decay is represented below.

<u>Explanation:</u>

Gamma decay is defined as the process in which an unstable nuclei gives off excess energy by a spontaneous electromagnetic process and thus releases \gamma -radiations. These radiations does not carry any charge and are electrically neutral.

The chemical equation representing gamma decay process follows:

_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma

where,

_Z^A\textrm{X}^* = unstable nuclei

_Z^A\textrm{X} = stable nuclei

_0^0\gamma = gamma particle

Hence, the gamma decay reaction is given below.

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Mg = 24.3
Cl = 35.5

24.3 + 35.5 x 2 = 95.3 ~ 95.21 ( all periodic tabes have different accuracies)

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The scientific name for water is H2O because it contains:_________
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Write a decay equation for neon-19
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Why does electronegativity increase as you move through a period on the periodic table?
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2 years ago
A total of 663 cal 663 cal of heat is added to 5.00 g 5.00 g of ice at − 20.0 °C . −20.0 °C. What is the final temperature of th
Serjik [45]

Answer:

Final Temperature = 305.9 K or 32.9°C

Explanation:

Information given;

Mass (m) = 5g

Amount of Heat (H) = 663 calories

Initial Temperature = -20°C + 273 = 253K (Converting to Kelvin Temperature)

Final Temperature?

Specific Heat capacity of Water (c) =?

One might probably rush to use H = m * c * (T2-T1) to calculate the fianl temperature. That would have been wrong because Ice would melt when heat is being applied to it. And using that formular directly would lead to not considering the amount of heat required to melt it.

First let us check the amount of heat required to raise the temperature of the ice to 0°C

In this case now; Final temperature = 0°C + 273 = 273 K (Converting to Kelvin)

H = m * c * (T2-T1)

H = 5 * 1 * (273 - 253)

H = 5 * 1 * 20 = 100 cal

This shows the heat supplied is enough (663 cal is more than 100 cal) to bring the ice to its melting point.

Let's see if it would be sufficient to melt it.

Amount of Heat required to Melt ice;

H = mL;

where L = heat of fusion = 79.7 cal/g

H = 5 * 79.7 = 398.5 cal

Again, the heat is sufficient to melt it; the remaining heat would be used in raising the temperature of the liquid water.

In this case, initial temperature = 0°C + 273 = 273 K (Converting to Kelvin)

Amount of Heat left = 663 - 398.5 - 100 = 164.5 cal

Final temperature is given as;

H = m * c * (T2-T1)

164.5 = 5 * 1 * (T2 - 273)

T2 - 273 = 164.5 / 5

T2 = 273 + 32.9 = 305.9 K

Final Temperature = 305.9 K or 32.9°C

4 0
3 years ago
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