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Which statements correctly describe the decay rates of radioactive isotopes?

storchak [24]

Answer: b} The exact time when an individual atom will decay can be accurately predicted.

c} After each half-life, the amount of radioactive material is reduced by half.

Explanation:

All radioactive decay follows first order kinetics.

Rate law expression for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant

a - x = amount left after decay process

Expression for calculating half life, which is the time taken by the half of the reactants to decompose is:

$t_{1/2}=\frac{0.693}{k}$

4 0

3 years ago

Read 2 more answers

What is the nucleus of an atom?

Leno4ka [110]

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

5 0

2 years ago

What is the Nobel gas notation for selenium

seraphim [82]

Answer:

Se =[Ar] 3d¹⁰ 4s² 4p⁴

Explanation:

The noble gas notation is used for the shortest electronic configuration of other periodic table elements.

For example:

The atomic number of Argon is 18, and its electronic configuration is,

Ar₁₈ = 1s² 2s² 2p⁶ 3s² 3p⁶

The atomic number of selenium is 34, its electronic configuration is,

Se₃₄ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴

By using the noble gas notation, electronic configuration of selenium can be written is shortest form.

Se =[Ar] 3d¹⁰ 4s² 4p⁴

This electronic configuration is also called abbreviated electronic configuration.

5 0

3 years ago

How many reactant molecules and product gas molecules are in this equation?

Mice21 [21]

Answer:

N₂ = 6.022 × 10²³ molecules

H₂ = 18.066 × 10²³ molecules

NH₃ = 12.044 × 10²³ molecules

Explanation:

Chemical equation;

N₂ + 3H₂ → 2NH₃

It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

Number of molecules of nitrogen gas:

1 mol = 6.022 × 10²³ molecules

Number of molecules of hydrogen:

3 mol × 6.022 × 10²³ molecules/ 1 mol

18.066 × 10²³ molecules

Number of molecules of ammonia:

2 mol × 6.022 × 10²³ molecules/ 1 mol

12.044 × 10²³ molecules

6 0

3 years ago

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g

luda_lava [24]

Answer:

18.75 g of NH3.

Explanation:

The balanced equation for the reaction is given below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.

This can be obtained as follow:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160 g

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Next, we shall determine the excess reactant. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.

From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.

Therefore, O2 is the limiting reactant and NH3 is the excess reactant.

Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, Xg of NH3 will react with 50 g of O2 i.e

Xg of NH3 = (68 × 50)/160

Xg of NH3 = 21.25 g

Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.

Finally, we shall determine mass of the remaining excess reactant as follow:

Mass of excess reactant = 40 g

Mass of excess reactant that reacted = 21.25 g

Mass of excess reactant remainig =?

Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)

Mass of excess reactant remainig

= 40 – 21.25

= 18.75 g

Therefore, the mass of excess reactant remaining is 18.75 g of NH3.

8 0

3 years ago