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2 years ago

a chemistry teacher adds 50.0 ml of 1.50 m h2so4 solution to 200 ml of water. What is the concentration of the final solution?

Chemistry

1 answer:

nevsk [136]2 years ago

5 0

Answer:0.300M

Explanation:1) Data:

a) Initial solution

M = 1.50M

V = 50.0 ml = 0.050 l

b) Solvent added = 200 ml = 0.200 l

2) Formula:

Molarity: M = moles of solute / volume of solution is liters

3) Solution:

a) initial solution:

Clearing moles from the molarity formula: moles = M × V

moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol

b) final solution:

i) Volumen of solution = 0.050 l + 0.200l = 0.250l

ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer

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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.

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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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