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Kay [80]
3 years ago
6

3 Write the alpha decay WITH gamma emission for the following element: 92 AL 27

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

Explanation:

In alpha decay alpha ray comes out . Apha ray is nothing but nucleus of helium atom whose atomic no  is 2 and atomic weight is 4 . Hence in alpha decay , atomic weight reduces by 4 and atomic no reduces by 2. For gamma emission , there is no change in atomic number and atomic mass.

⁹²₂₇AL ⇒ ⁸⁸₂₅X + ⁴₂He + γ

X is Manganese .

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Rubber gloves and other chemical resistant protective clothing are necessary when handling concentrated ammonium hydroxide.

<h3 /><h3>Why it is necessary to handle concentrated solution of ammonium hydroxide with care ?</h3>

  • Ammonia is an irritant and corrosive to the skin, eyes, respiratory tract and mucous membranes.

  • May cause severe chemical burns to the eyes, lungs and skin.

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The extent of injury produced by exposure to ammonia depends on the duration of the exposure, the concentration of the liquid or vapor and the depth of inhalation.

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A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 10
masya89 [10]

<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol

<u>Explanation:</u>

Rate of a gas is defined as the amount of gas displaced in a given amount of time.

\text{Rate}=\frac{V}{t}

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

So,

\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}

We are given:

Volume of unknown gas (X) = 1.0 L

Volume of oxygen gas = 1.0 L

Time taken by unknown gas (X) = 105 seconds

Time taken by oxygen gas = 31 seconds

Molar mass of oxygen gas = 32 g/mol

Molar mass of unknown gas (X) = ? g/mol

Putting values in above equation, we get:

\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol

Hence, the molar mass of unknown gas is 367.12 g/mol

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4 years ago
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artcher [175]

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