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3 years ago

This chart show the amount of radioactivity measured from three unknown isotopes.

Chemistry

1 answer:

Ad libitum [116K]3 years ago

5 0

Answer:

We are asked to describe the half-life of the radioactive materials.

Hence the 3rd option is correct

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How many moles of oxygen are produced when 2 mol of potassium chlorate (KClO3) decompose

ioda

Keep in mind that when two moles of potassium chlorate are decomposed, 3 moles of oxygen are produced.Think of the coefficients: since the mole ratio is 2 to 3, it means that 2 mol KClO3 will turn into 3 mol O2.

2kclO3->2kcl+3O2

4 0

3 years ago

ANSWERIT AND YOU WILL BE MARKED THE BRAINLIEST

fgiga [73]

Answer:

The answer is definitely D

Explanation:

3 0

3 years ago

Read 2 more answers

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

An ion is an atom that has has gained or lost:

Elena L [17]

Electrons, specifically valence electrons

5 0

3 years ago

Determine the percent water in MgSO4*7H20?

Semmy [17]

Answer:

$\% H_2O=51.2\%$

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

$MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol$

Thus, the percent water is:

$\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\$

So we plug in to obtain:

$\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%$

Best regards!

5 0

3 years ago