The circle on the left shows a magnified view of a very small portion of liquid water in a closed container. What would the magn
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Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.
The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.
Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles
We can obtain the mass of 3-methyl-1-butanol from its density.
Mass = density × volume
Density of 3-methyl-1-butanol = 0.810 g/mL
Volume of 3-methyl-1-butanol = 9 mL
Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL
Mass of 3-methyl-1-butanol = 7.29 g
Number of moles of 3-methyl-1-butanol = mass/molar mass = 7.29 g/88 g/mol = 0.083 moles
Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol
Mass of product 1-bromo-3-methylbutane = number of moles × molar mass
Molar mass of 1-bromo-3-methylbutane = 151 g/mol
Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol
= 12.53 g
Recall that % yield = actual yield/theoretical yield × 100
Actual yield of product = 6.48 g
Theoretical yield = 12.53 g
% yield = 6.48 g/12.53 g × 100
% yield = 51.7%
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Answer:
There was 450.068g of water in the pot.
Explanation:
Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L
Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s
Let m = x g be the weight of water in the pot.
Energy required to vaporise water = mL = 2260x
Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x
Total energy required =
Hence, there was 450.068g of water in the pot.