Predict the products of the following nuclear reactions...

Chemistry

1 answer:

koban [17]3 years ago

6 0

1. According to Rutherford and soddy's law of Beta emission:

${K}^{42} \: ---> {Ca}^{40} + {2e}^{0}$

3. According to Rutherford and soddy's law of Alpha emission:

${U}^{238} \: ---> {Th}^{234} + {He}^{4}$

4. The process in which two or more light nuclei are combined to form a single nucleus with the release of tremendous amount of energy is called a nuclear fusion. The inner core of Sun is at very high temperature and is suitable for fusion. In fact the source of energy of Sun and other stars is nuclear fusion reactions.

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Atoms and elements are examples of A. molecules. B. mixtures. C. compounds. D. pure substances.

UkoKoshka [18]

Answer:

atom are very tiny partical

3 0

2 years ago

1 point

Bumek [7]

Answer:

1 mole of C2H6.

Explanation:

The balanced equation for the reaction is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

We can determine the number of mole of C2H6 that reacted to produce 2 moles of CO2 as follow:

From the balanced equation above,

2 moles of C2H6 reacted to produce 4 moles of CO2.

Therefore, Xmol of C2H6 will react to produce 2 moles of CO2 i.e

Xmol of CO2 = (2 x 2)/4

Xmol of CO2 = 1 mole.

Therefore, 1 mole of C2H6 is required to produce 2 moles of CO2.

6 0

2 years ago

A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.