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Shalnov [3]
2 years ago
5

What is the mass of 6.3 moles of Sulfur

Chemistry
2 answers:
Sidana [21]2 years ago
7 0

Answer:

201.6 grams is the mass of 6.3 moles of sulfur.

Explanation:

Moles(n) =\frac{\text{Mass of the compound}(m)}{\text{molar mass of the compound}(M)}

Moles of sulfur ,n= 6.3 moles

Atomic mass of sulfur = M = 32 g/mol

Mass of the sulfur = m = ?

m=M\times n

=6.3 mol\times 32 g/mol=201.6 g

201.6 grams is the mass of 6.3 moles of sulfur.

bixtya [17]2 years ago
6 0
The molar mass of sulfur is 32 g/mol.
The mass of 6.3 moles of Sulfur is 32×6.3=201.6 grams

<span>♡♡Hope I helped!!! :)♡♡
</span>
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* Need ASAP * Explain the relationship between percent composition, Empirical formula, and molecular formula.
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By dividing the percentage composition with the molar mass of that element we will get the empirical formula. Then using that empirical formula and formula mass we can find the molecular formula.

<u>Explanation:</u>

The chemical properties of any substance are defined obviously by the different types and relative amounts of atoms constituting its primary entities (in case of covalent compounds the primary entities are molecules and ions in the event of ionic compounds).

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5 0
3 years ago
How many atoms of hydrogen (h) are present in 200 molecules of ammonia (nh3)? express your answer numerically?
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6 0
3 years ago
If 5.12 g of oxygen O2 gas occupies a volume of 6.21L at a certain temperature and pressure, how many grams of oxygen gas will o
ddd [48]

Answer : The mass of O_2 occupy 30.3 L under the same conditions will be, 24.9 grams.

Explanation :

First we have to calculate the moles of O_2

\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=\frac{5.12g}{32g/mol}=0.16mol

Now we have to calculate the moles of O_2 in 30.3 L by using Avogadro's law.

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

V\propto n

or,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1 = initial volume of gas = 6.21 L

V_2 = final volume of gas = 30.3 L

n_1 = initial moles of gas = 0.16 mol

n_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{6.21L}{0.16mol}=\frac{30.3L}{n_2}

n_2=0.781mol

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

Molar mass of O_2 = 32 g/mol

\text{ Mass of }O_2=(0.781moles)\times (32g/mole)=24.9g

Therefore, the mass of O_2 occupy 30.3 L under the same conditions will be, 24.9 grams.

6 0
3 years ago
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