Ask question

Ask question

All categories

3 years ago

15

A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life

1 answer:

mixas84 [53]3 years ago

8 0

T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>

<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>

<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>

<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>

<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>

<span>HL = 14.4 s / 2 = 7.2 seconds </span>

You might be interested in

Which statement describes the change that occurs during the reaction of tin with lead(II) nitrate? A. Tin atoms give electrons t

timama [110]

Answer is: <span>D. Tin atoms give electrons to lead(II) ions and are oxidized to tin(II) ions.
Chemical reaction: Sn</span>⁰ + Pb²⁺ → Sn²⁺ + Pb.
Tin atom (oxidation number 0) give two electrons to led ions, oxidation number of tin is greater now (oxidation number +2).
<span>Oxidation is loss of electrons.</span>

3 0

3 years ago

Thank uuuu so much!!!!!

kow [346]

Answer:

A. 0.83

Explanation:

Hopefully this helps!

7 0

3 years ago

15<br> Question 56 (1 point)<br> What is the name of the compound with the chemical formula SnSnO3?

muminat

Answer: sodium sulfite

Explanation:

4 0

3 years ago

How many particles are in one mole of copper (II) sulfate, CuSO4?

Kobotan [32]

2. Avogadro’s number

3 0

3 years ago

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

4 0

3 years ago