All categories

2 years ago

What is the empirical formula of a compound that is 41.4% Strontium, 13.24%

Chemistry

1 answer:

Pepsi [2]2 years ago

6 0

Answer:

Empirical Formula N2O6Sr Strontium Nitrate

Explanation:

N=13.2% O=45.4% Sr=41.4%

You might be interested in

A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element pre

nordsb [41]

Answer:

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

6 0

3 years ago

Read 2 more answers

What is a sign that a person may be abusing drugs?

lidiya [134]

Answer:

answer is a because drugs do so to the person.

6 0

3 years ago

Read 2 more answers

What is the percent composition by mass of sulfur in the compound mgso4

Margaret [11]

Answer:

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

Explanation:

Molar mass of compound = 120 g/mol

Number of sulfur atom = 1

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

$=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100$

Sulfur :

$=\frac{1\times 32 g/mol}{120 g/mol}\times 100=26.7\%$

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

6 0

3 years ago

How many milliseconds are there in 3.5 seconds

maw [93]

There are 3,500 milliseconds in a second.

One second contains 1,000 milliseconds. Three seconds contain 3,000 milliseconds. Half of three hours, therefore, would contain 3,500 milliseconds.

6 0

2 years ago

Determine the boiling point of a 3.70 m solution of phenol in benzene. Benzene has a boiling point of 80.1°C and a boiling point

xeze [42]

Answer: The boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

Explanation:

Elevation in boiling point:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = change in boiling point

i= vant hoff factor = 1 (for benzene which is a non electrolyte )

$k_b$ = boiling point constant = $2.53^0C/kgmol$

m = molality = 3.70

$T_{solution}-T_{solvent}=i\times k_b\times m$

$T_{solution}-80.1^0C=1\times 2.53\times 3.70$

$T_{solution}=89.5^0C$

Thus the boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

8 0

2 years ago