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3 years ago

When K-37 undergoes positron emission, what is the resulting atom?

Chemistry

1 answer:

natita [175]3 years ago

4 0

Answer:

C) Ar-37

Explanation:

A positron is a positively charged electron. It has the same mass as that of an electron, but opposite charge. It is the counterpart of an electron.

The reaction involving the emission of a positron by k-37 would produce a Ar-37, which is an unstable isotope of Argon. In this emission, there must not be a particle loss or creation, the addition of the mass number and proton number on LHS must be the sane as that on RHS of the equation.

An isotope is ability of an element to exist in two or more forms with the same atomic number but different mass number.

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Which reaction follows the law of conservation of matter?

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3 years ago

The solubility of CO2 in water is 0.161 g/100 mL at 20oC and a partial pressure of CO2 of 760 mmHg. What partial pressure of CO2

Schach [20]

<u>Answer:</u> The partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

The equation given by Henry's law is:

$C_{CO_2}=K_H\times p_{CO_2}$ ......(1)

where,

$C_{CO_2$ = solubility of carbon dioxide in water = 0.161 g/100 mL

$K_H$ = Henry's constant = ?

$p_{CO_2}$ = partial pressure of carbon dioxide = 760 mmHg

Putting values in equation 1, we get:

$760mmHg=K_H\times 0.161g/100mL\\\\K_H=\frac{760mmHg}{0.161g/100mL}=4720.5g.mmHg/100mL$

Now, calculating the pressure of carbon dioxide using equation 1, we get:

$C_{CO_2$ = solubility of carbon dioxide in water = 0.886 g/100 mL

$K_H$ = Henry's constant = 4720.5 g.mmHg/100 mL

$p_{CO_2}$ = partial pressure of carbon dioxide = ?

Putting values in equation 1, we get:

$p_{CO_2}=4720.5g.mmHg/100mL\times 0.886g/100mL=4182.4mmHg$

Hence, the partial pressure of carbon dioxide having solubility 0.886g/100mL is 4182.4 mmHg

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3 years ago

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3 years ago

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3 years ago

21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4

bija089 [108]

The balanced equation for the above reaction is
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