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3 years ago

Please answer for me thanks...

Chemistry

1 answer:

tatuchka [14]3 years ago

5 0

Explanation:

Since mole ratio of O2 : NH3 = 7 : 4,

Volume of NH3 = 50dm³ * 4/7 = 28.57dm³.

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A neutral atom of beryllium (Be) has an average mass of 9 amu and 4 electrons. How many neutrons does it have?

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The chemical formula for ferric sulfate is Fe(SO4)3. Determine the following:

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Answer :

(a) The number of sulfur atoms are, $31.61\times 10^{23}$ .

(b) The mass of the mass of $Fe_2(SO_4)_3$ is, 1059.682 grams.

(c) The number of moles of $Fe_2(SO_4)_3$ is, $8.63\times 10^{-3}mole$

(d) The mass of the mass of $Fe_2(SO_4)_3$ is, $19.95\times 10^{-22}g$

Explanation :

(a) As we are given the number of moles of $Fe_2(SO_4)_3$ is, 1.75 mole. Now we have to calculate the number of sulfur atoms.

In the $Fe_2(SO_4)_3$ , there are 2 iron atoms, 3 sulfur atoms, 12 oxygen atoms.

As, 1 mole of $Fe_2(SO_4)_3$ contains $3\times 6.022\times 10^{23}$ number of sulfur atoms.

So, 1.75 mole of $Fe_2(SO_4)_3$ contains $1.75\times 3\times 6.022\times 10^{23}=31.61\times 10^{23}$ number of sulfur atoms.

The number of sulfur atoms are, $31.61\times 10^{23}$

(b) As we are given the number of moles of $Fe_2(SO_4)_3$ is, 2.65 mole. Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$\text{Mass of }Fe_2(SO_4)_3=2.65mole\times 399.88g/mole=1059.682g$

The mass of the mass of $Fe_2(SO_4)_3$ is, 1059.682 grams.

(c) As we are given the mass of $Fe_2(SO_4)_3$ is, 3.45 grams. Now we have to calculate the moles of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$3.45g=\text{Moles of }Fe_2(SO_4)_3\times 399.88g/mole$

$\text{Moles of }Fe_2(SO_4)_3=8.63\times 10^{-3}mole$

The number of moles of $Fe_2(SO_4)_3$ is, $8.63\times 10^{-3}mole$

(d) As we are given the formula unit of $Fe_2(SO_4)_3$ is, 3. Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

As we know that 1 mole of $Fe_2(SO_4)_3$ contains $6.022\times 10^{23}$ formula unit.

Formula used :

$\text{Formula unit of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}$

$3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}$

$\text{Moles of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole$

Now we have to calculate the mass of $Fe_2(SO_4)_3$ .

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

The molar mass of $Fe_2(SO_4)_3$ = 399.88 g/mole

$\text{Mass of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole\times 399.88g/mole=19.95\times 10^{-22}g$

The mass of the mass of $Fe_2(SO_4)_3$ is, $19.95\times 10^{-22}g$

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