A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
2 answers:
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Answer:
1. Yes
2.The solubility of X is 34.55g/L
Explanation:
Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm or 1 Litre of water.
From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved
= 0.019kg*1000 = 19g.
if 19g is required to saturate 550mL at 30°C,
then will saturate 1L
= 34.545g will saturate 1Litre
The solubility thus is 34.55g/L
Answer:
2,57 g of precipitate.
Explanation:
For the reaction:
6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄
The precipitate is Al(OH)₃.
185,5mL of 0,533M NaOH are:
0,1855L × 0,533M = <em>0,0989 moles NaOH</em>
Moles of Al₂(SO₄)₃ are:
15,8g × = <em>0,0462 moles Al₂(SO₄)₃</em>
For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:
0,0989moles NaOH × = <em>0,0165 moles Al₂(SO₄)₃</em>
As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.
0,0989 moles of NaOH produce:
0,0989moles NaOH × = <em>0,0330 moles of Al(OH)₃</em>
These moles are:
0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>
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Answer:
274.08 g of CO2
Explanation:
please see paper for work!
