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AveGali [126]
3 years ago
14

How many ions are present in 75.0g of magnesium sulfate

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

7.5×10²³ ions

Explanation:

Given data:

Mass of magnesium sulfate = 75.0 g

Number of ions = ?

Solution:

Number of moles of magnesium sulfate:

Number of moles = mass/molar mass

Number of moles = 75.0 g/ 120.36 g/mol

Number of moles = 0.623 mol

Magnesium sulfate formula: MgSO₄

Number of moles of ions in 1 mole of MgSO₄ = 2 mol (Mg²⁺ and SO₄²⁻)

Number of moles of ion in 0.623 moles of MgSO₄:

2 mole ×0.623 = 1.246 mol

Number of ions:

1 mole contain 6.022×10²³ ions

1.246 mol × 6.022×10²³ ions / 1mol

7.5×10²³ ions

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- How many grams are in 4.6 x 1016 molecules of phosphorus trichloride (PCI)?
MariettaO [177]

Answer:

0.0000101 grams.

Explanation:

First of all, use Avogadro's number to convert from molecules to moles.

Avogadro's number is 6.02 x 10^23.

Second of all, to convert from moles to grams, find the substance formula of phosphorus trichloride. After finding the substance formula, then find the molar mass of phosphorus trichloride.

Phosphorus trichloride is PCl3

Now find the molar mass of PCl3

P = 31.0 amu

Cl = 35.5 amu

They are three chlorine atoms and one phosphorus atom. So you do this.

31.0(1) + 35.5(3) = 137.5 g/mol

Now use dimensional analysis to show your work

4.6 x 10^16 grams of PCl3 * 1 mol/ 6.02 x 10^23 * 137.5 g/mol / 1 mol

The grams and the moles cancel out.

So now do [(4.6 x 10^16) / (6.02 x 10^23)] x 137.5

and you will get 0.000010507

4.6 has the fewest digits which is 2.

So use sig figs and the answer rounds to the nearest millionths.

0.000010507 rounds to 0.0000101

So the final answer is 0.0000101 grams(don't forget the units)

Hope it helped!

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