1. Write out the formula
Pb(NO3)2 (aq) + 2HCl (aq) ----> PbCl2 + 2HNO3
2. Use solubility guidelines (gotta memorize 'em) for the products to see if a solid forms
Nitrates are always soluble so 2HNO3 (aq)
Chlorides (Cl) are always soluble except for when you mix them with copper, lead, mercury, or silver.
Since you mixed it with lead (Pb) it is solid and forms a precipitate. PbCl2 (s)
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
LiOH is soluble. Na2CO3 is soluble. Cu(OH)2 is insoluble.
Oxygen for complete combustion
Answer:
7.5 moles of CaBr2 are produced
Explanation:
Based on the equation:
2AlBr3 + 3CaO → Al2O3 + 3CaBr2
<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>
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Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:
5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =
<h3>7.5 moles of CaBr2 are produced</h3>
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