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irina1246 [14]
3 years ago
13

Tritium 3 1H decays to 3 2He by beta emission. Find the energy released in the process. Answer in units of keV.

Chemistry
1 answer:
-Dominant- [34]3 years ago
7 0

Answer:

The energy released in the decay process = 18.63 keV

Explanation:

To solve this question, we have to calculate the binding energy of each isotope and then take the difference.

The mass of Tritium = 3.016049 amu.

So,the binding energy of Tritium =  3.016049 *931.494 MeV

= 2809.43155 MeV.

The mass of Helium 3 = 3.016029 amu.

So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV

= 2809.41292 MeV.

The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV

1 MeV = 1000keV.

Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.

So, the energy released in the decay process = 18.63 keV.

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The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro
Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

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An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

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Cause their are both have difference s

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