Answer:
M HCl sln = 12.0785 M
Explanation:
- molarity (M) [=] mol/L
- %mm = ((mass compound)/(mass sln))*100
∴ mass sln = 100.0 g
∴ δ sln = 1.19 g/mL
∴ % m/m = 37 %
⇒ 37 % =((mass HCl/mass sln))*100
⇒ 0.37 = mass HCl / 100.0 g
⇒ 37 g = mass HCl
∴ molar mass HCl = 36.46 g/mol
⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol
⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L
⇒ M HClsln = 1.015 mol/0.084034 L
⇒ M HCl sln = 12.0785 M
Then we all die because the sun would blow up the earth.
Answer:
The answer to your question is: 6.8 g of water
Explanation:
Data
2.6 moles of HCl
1.4 moles of Ca(OH)2
2HCl + Ca(OH)2 → 2H2O + CaCl2
MW 2(36.5) 74 36 g 111 g
73g
1 mol of HCl ---------------- 36.5 g
2.6 mol -------------- x
x = (2.6 x 36.5) / 1 = 94.9 g
1 mol of Ca(OH)2 -------------- 74 g
1.4 mol --------------- x
x = (1.4 x 74) / 1 = 103.6 g
Grams of water
73 g of HCl ------------------ 36g of H2O
94.9 g ------------------- x
x = (94.9 x 36) / 73 = 46.8 g of water
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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