Answer:
a. 174 mL
Explanation:
Let's consider the following reaction.
2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)
We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:
0.1550 L × 0.112 mol/L = 0.0174 mol
The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:
2 × 0.0174 mol = 0.0348 mol
The volume of a 0.200 M KI solution that contains 0.0348 moles is:
0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL
Answer:
the complete question is found in the attachment
Explanation:
the complete explanation is found in the attachment
<em>V = 151 mL = 151 cm³</em>
<em>d = 0,789 g/mL = 0,789 g/cm³</em>
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d = m/V
m = d×V
m = 0,789×151
<u>m = 119,139g</u>
Place the mixture in hot water and stir well.
<span>The KNO3 is very soluble in hot water. </span>
<span>Use a fine filter paper and filter off the sand. </span>
<span>The sand will be separated from the KNO3 solution. </span>
<span>The water can now be evaporated from the solution by further, gentle heating leaving the solid in the container.</span>
I don’t know but look up the ph of ammonia and if it is below 7 then it is yellow and if it is above 7 then blue
