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jolli1 [7]
3 years ago
12

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)

Chemistry
1 answer:
Blizzard [7]3 years ago
8 0
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
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Which of the following is a possible set of quantum numbers for an electron (n,l,m0,ms) ?
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Answer is: (3, 2, 0, -1/2).

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3 years ago
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What is the percentage strength (w/w) of a solution made by dissolving 62.5 g of potassium chloride in 187.5 ml of water?
kiruha [24]

Percent strength (% w/w) of a solution is defined as the amount of solute present in 100 g of the solution.

Given data:

Mass of the solute, potassium chloride = 62.5 g

Volume of water (solution) = 187.5 ml

We know that the density of water = 1 g/ml

Therefore, the mass corresponding to the given volume of water

= 187.5 ml * 1 g/1 ml = 187.5 g

We have a solution of 62.5 g of potassium chloride in 187.5 g water

Therefore, amount of solute in 100 g of water= 62.5 * 100/187.5 = 33.33

The percentage strength = 33.33 %

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3 years ago
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