Answer:
0.4 moles
Explanation:
To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.
40gCaCO3* 1mol CaCO3/100.09gCaCO3 = 0.399640 mol CaCO3
This rounds to 0.4 moles CaCO3
Answer:
<h2>18 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
volume = final volume of water - initial volume of water
volume = 62 - 52 = 10 mL
From the question we have
We have the final answer as
<h3>18 g/mL</h3>
Hope this helps you
Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
When Ni(OH)₂ starts precipitate :
Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]
5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]
[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18
[ OH⁻ ] = 5.5 × 10⁻⁸ M
pOH = 7.2
therefore , pH = 14 - 7.2
pH = 6.8
Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.
To learn more about pH here
brainly.com/question/15289741
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The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g
<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
- Volume = 250 mL = 250 / 1000 = 0.25 L
- Molarity = 0.3 M
Mole = Molarity x Volume
Mole of Na₂SO₄ = 0.3 × 0.25
Mole of Na₂SO₄ = 0.075 mole
<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
- Molar mass of Na₂SO₄ = 142.05 g/mol
- Mole of Na₂SO₄ = 0.075 mole
Mass = mole × molar mass
Mass of Na₂SO₄ = 0.075 × 142.05
Mass of Na₂SO₄ = 10.65 g
Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.
Learn more about molarity:
brainly.com/question/15370276
