Which has more energy water at 100°C or water vapor at 100°C
1 answer:
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The theoretical yield of will be 4.20 grams while the percent yield will be 7.93%
From the equation of the reaction, the mole ratio of to
is 1:2.
Mole of 48.1 mL, 0.478 M = 0.478 x 48.1/100 = 0.023 mols
Mole of 48.8 mL, 0.160 M = 0.160 x 48.8/1000 = 0.0078 moles
is the limiting reactant.
Mole ratio of and
= 1:1
Equivalent mole of = 0.023 moles
Mass of 0.023 noles = 0.023 x 182.81 = 4.20 grams
With 0.0333 g of recovered, percent yield = 0.333/4.2 x 100 = 7.93%
More on yields of reactions can be found here: brainly.com/question/17042787
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Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.
SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)
Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.
1) Calculate the theoretical yield of SnS2. ;
2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.
Answer:
The volume of the gas at a pressure of 65.0 kPa would be 363 mL
Explanation:
Boyle's Law is a gas law that relates the pressure and volume of a certain amount of gas, without temperature variation, that is, at constant temperature.
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. In other words, the product P · V remains constant at the same temperature:
P*V=k
Being P1 and V1 the pressure and volume in state 1 and P2 and V2 the pressure and volume in state 2 are fulfilled:
P1*V1=P2*V2
In this case:
- P1= 45 kPa= 45,000 Pa (being 1 kPa=1,000 Pa)
- V1= 525 mL= 0.525 L (being 1 L=1,000 mL)
- P2= 65 kPa= 65,000 Pa
- V2= ?
Replacing:
45,000 Pa* 0.525 L= 65,000 Pa*V2
Solving:
V2=0.363 L=363 mL
<u><em>The volume of the gas at a pressure of 65.0 kPa would be 363 mL</em></u>