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1 year ago

It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?

Chemistry

1 answer:

Mnenie [13.5K]1 year ago

4 0

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present = 1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9 = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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you have been given 250.0 mL of an aqueous solution containing 18.7grams of AgNO3. What is the molarity of solution?

Reil [10]

Hey there!:

Molar mass AgNO3 = 169.87 g/mol

Number of moles:

moles of solution = mass of solute / molar mass

moles of solution = 18.7 / 169.87

moles of solution = 0.110084 moles of AgNO3

Volume in liters:

250.0 mL / 1000 => 0.25 L

Therefore:

Molarity = moles of solution / Volume of solution ( L )

Molarity = 0.110084 / 0.25

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