Ask question

Ask question

All categories

2 years ago

9

A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

ons per minute after 48.0 hours
1.) What is the half life of 28 Mg in hours?

I have no idea how to work this equation. Please show step by step with correct answers so I can try to understand.

1 answer:

prisoha [69]2 years ago

8 0

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

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

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

$k=9.98\times 10^{-2}hr^{-1}$

Now we have to calculate the half-life.

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

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

You might be interested in

Latent heat of vaporization is used to (1 Point) (a) overcome the forces of attraction between molecules in solid-state. (b) inc

Reil [10]

Answer:

Not sure what the answer is

Explanation:

I did this a while ago and dont remember sorry

4 0

2 years ago

Arsenic produces a blue flame when heated. Calcium produces an orange-red flame. Which of these best explains why this differenc

MrMuchimi

Flame colors are produced from the movement of the electrons in the metal ions present in the compounds. When you heat it, the electrons gain energy and can jump into any of the empty orbitals at higher levels Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular color. As a result of all these jumps, a spectrum of colored lines will be produced. The color you see will be a combination of all these individual colors.

6 0

2 years ago

Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r

Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

$24 \times 3600 sec$

Hence, total charge passed daily is calculated as follows.

$150,000 \times 24 \times 3600 sec$

And, number of Faraday of charge is as follows.

$\frac{150,000 \times 24 \times 3600 sec}{96500}$

= 134300.52 F

The oxidation state of aluminium in $Al_{2}O_{3}$ is +3.

$Al^{3+} + 3e^{-} \rightarrow Al(s)$

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

$\frac{134300.52 F}{3} \times \frac{89}{100}$

= 39842.487 mol

or, = $3.9842 \times 10^{4} mol$

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

Mass in grams = $3.9842 \times 10^{4} mol \times 27$

= $107.57 \times 10^{4} g$

= 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

8 0

2 years ago

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C∘C) has an absolute pressure p1p1p_1. The container

choli [55]

Answer: p2 = 1.06p1

Explanation: pressure increases with temperature increase.

According to Gass law

P1/T1 = P2/T2

T1 = 20°c = 20 +273 = 293k

T2 = 40°c = 40 +373 = 313k

Therefore

P2 = P1T2/T1 = 313P2/293

P2 = 1.06P1

3 0

2 years ago

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction: C(s)+2H2O(g)→2H2

madam [21]

The question is incomplete , complete question is:

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.

Answer:

The ΔH of the reaction is -626 kJ/mol.

Explanation:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

We are given with:

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{H-H}=432 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})$

$=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol$

$\Delta H=-626 kJ/mol$

The ΔH of the reaction is -626 kJ/mol.

5 0

2 years ago