Answer:
1.95g of Mg(OH)2 are needed
Explanation:
Mg(OH)2 reacts with HCl as follows:
Mg(OH)2 + 2 HCl → MgCl2 + 2H2O
<em>Where 1 mole of Mg(OH)2 reacts with 2 moles of HCl</em>
To solve this question we must find the moles of acid. Then, with the chemical equation we can find the moles of Mg(OH)2 and its mass:
<em>Moles HCl:</em>
158mL = 0.158L * (0.106mol / L) = 0.01675 moles HCl
<em>Moles Mg(OH)2:</em>
0.01675 moles HCl * (2mol Mg(OH)2 / 1mol HCl) = 0.3350 moles Mg(OH)2
<em>Mass Mg(OH)2 -Molar mass: 58.3197g/mol-</em>
0.3350 moles Mg(OH)2 * (58.3197g / mol) =
<h3>1.95g of Mg(OH)2 are needed</h3>
Answer:
The answer to your question is 1.1 moles of water
Explanation:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
0.45 mol 0.55 mol ?
Process
1.- Calculate the limiting reactant
Theoretical proportion
Al(OH)₃ / H₂SO₄ = 2/3 = 0.667
Experimental proportion
Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81
From the proportions, we conclude that the limiting reactant is H₂SO₄
2.- Calculate the moles of H₂O
3 moles of H₂SO₄ ---------------- 6 moles of water
0.55 moles of H₂SO₄ ----------- x
x = (0.55 x 6) / 3
x = 3.3 / 3
x = 1.1 moles of water
Answer:
A. Producing certain synthetic materials could have a greater environmental impact than disposing of them.
Explanation
I just did this question and got it right.
Metals present in municipal waste water may still be present in treated sewage sludge IN CONCENTRATIONS THAT MAY AFFECT THE PUBLIC HEALTH. Sewage sludge is an end product of municipal waste water treatment and it contains many of the pollutant that are removed from the waste water.
The third option because the elements mentioned in that selection are in the same group, they must have similar properties.
