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3 years ago

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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1) Consider this row in the periodic table. What changes can you predict based on what information is provided by the boxes for

bearhunter [10]

Answer:

B) As you move across the row, the number of electrons increases and reactivity also increases.

Explanation:

The periodic table is arranged in a way that if you go across a period, the number of protons, neutrons, and electrons in an element increases. In terms of reactivity, the most reactive elements are the ones which have a high electronegativity. The electronegativity of the elements increases as you travel to the right and upwards on the periodic table.

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2 years ago

In stoichiometric calculations, which quantity MUST be used to convert from one chemical substance to another?

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The answer is

C) mole ratio

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2 years ago

When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit

pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

$\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)$ .

How many moles of this process?

Relative atomic mass from a modern periodic table:

K: 39.098;
N: 14.007;
O: 15.999.

Molar mass of $\text{KNO}_3$ :

$M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of the process = Number of moles of $\text{KNO}_3$ dissolved:

$\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}$ .

What's the enthalpy change of this process?

$Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}$ for $0.212162\;\text{mol}$ . By convention, the enthalpy change $\Delta H$ measures the energy change for each mole of a process.

$\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}$ .

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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3 years ago

A system in chemical equilibrium

swat32

Answer:

its B

Explanation:

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2 years ago

Read 2 more answers

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