Which of the following best defines conductivity?

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?

Mnenie [13.5K]

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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4 0

1 year ago

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0

3 years ago

What do scientists use to deterimine the temperature of a star

lidiya [134]

Measure the brightness of a star through two filters and compare the ratio of red to blue light.

7 0

3 years ago

If 28 grams of N reacts completely with 12 grams of H2, then how many

Bogdan [553]

Answer:

Mass of NH₃ produced = 34 g

Explanation:

Given data:

Mass of nitrogen = 28 g

Mass of Hydrogen = 12 g

Mass of NH₃ produced = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 28 g/ 28 g/mol

Number of moles = 1 mol

Moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 12 g/ 2 g/mol

Number of moles = 6 mol

Now we will compare the moles of hydrogen and nitrogen with ammonia.

H₂ : NH₃

3 : 2

6 : 2/3×6 = 4 mol

N₂ : NH₃

1 : 2

Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.

Mass of ammonia produced:

Mass = number of moles × molar mass

Mass = 2 mol × 17 g/mol

Mass = 34 g

5 0

3 years ago

Read 2 more answers

The temperature of a sample of water increases from 20c to 46.6c as it absorbs 5650 Joules of heat. What is the mass of the samp

WINSTONCH [101]

Answer: 51 g

Explanation:

5 0

2 years ago