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3 years ago

When radium-226 (atomic number-88) decays by emitting an alpha particle it becomes _____.

Chemistry

1 answer:

topjm [15]3 years ago

3 0

Answer:

d. Radon-222

Explanation:

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He

Alpha particle is a helium nucleus with mass number 4 and atomic number 2. According to the law of conversation of mass, the sum of the mass number and atomic number must be equal on both side of the reaction.

Since the mass number of Ra is 226 and that of He is 4. The mass number of the unknown element must be 226 - 4 = 222.

Since the atomic number of Ra is 88 and that of He is 2. The atomic number of the unknown element must be 88 - 2 = 86.

Now looking in the periodic table Radon is the only element with atomic number 86.

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4 years ago

Read 2 more answers

In a chemical reaction occurring in an open beaker, the reactants have a mass of 10g and contain 1000J or energy. The products i

Yakvenalex [24]

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3 years ago

A 1.00 x 102mL sample of 0.200 M aqueous hydrochloric acid is added to 1.00 x 102mL of 0.200 M aqueous ammonia in a constant-pre

insens350 [35]

Answer:

ΔH = -98 kJ/mol

Explanation:

To calculate the heat change of the reaction:

HCl(aq) + NH₃(aq) → NH₄Cl(aq)

0.2M 0.2M ΔT=2.34°C

1x10²mL 1x10²mL

We need to use the next equation:

$q = mc \Delta T$ (1)

where q: the amount of heat energy lost or gained, m: the mass of the substance, c: the specific heat capacity of the substance and ΔT: the change in temperature of the substance 

Assuming that the densities of the solutions are the same as for water, we can determine the mass of the solution:

$d = \frac {m}{V}$

where d: density, m: mass and V: volume of solution = 100 + 100 = 200mL

$m = d \cdot V = 1 \frac {g}{mL} \cdot 200mL = 200g$

Now, using the calculated mass in equation (1), and assuming that the specific heats of the solutions are the same as for water, we can find heat change of the reaction:

$q = 200g \cdot 4.184 \frac {J}{g \cdot ^{\circ}C} \cdot 2.34^{\circ}C$

$q = 1.96 \cdot 10^{3}J$

This heat is negative because is the heat lost by the reacting HCl and NH₃ and gained by the water, so:

$q = - 1.96 \cdot 10^{3}J$

To calculate the heat change of the reaction per mole of HCl, we need to divide the heat change by the number of moles, which is called the enthalpy of reaction:

$\Delta H = \frac {q}{moles}$

$\Delta H = \frac {-1.96 \cdot 10^{3} J}{0.2 \frac{mol}{L} \cdot 0.1L}$

$\Delta H = -98 \cdot 10^{3} \frac{J}{mol} = -98 \frac {kJ}{mol}$

So, the heat change of the reaction per mole of HCl reacted, often called enthalpy of reaction, is ΔH = -98 kJ/mol.

Have a nice day!

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4 years ago