Molar mass = 305.42 g/mol
C = ( 12 x 18 / 305.42 ) x 100 => 70.72 % of C
H = ( 1 x 27 / 305.42 ) x 100 => 8.84 % of H
N = ( 14 x 1 / 305.42 ) x 100 => 4.58 % of N
O = ( 16 x 3 / 305.42) x 100 => 15.71% of O
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If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
Answer:
The answer is below
Explanation:
one gallon of gasoline produces 9.50 kg of carbon.
The total number of cars = 40 million
Distance covered by each car = 7930 miles
Consumption rate of the cars per miles traveled is 23.6 miles per gallon.
Hence the annual gasoline consumption by all the cars in the United States of America = (total number of cars × Distance covered by each car) ÷ Consumption rate of the cars per miles
annual gasoline consumption by all the cars = (40000000 × 7930 miles) ÷ 23.6 miles/gallon = `1.344067797 × 10¹⁰ gallons
1.344067797 × 10¹⁰ gallons = 
To determine the masses of each component, we need to know the concentration first in terms of molality which is mol per mass of solution. Assuming the density of the solution is equal to that of water we would find:
molality = 0.160 mol KBr / L ( 1 L / 1 kg solution ) = 0.160 mol KBr / kg solution
mass KBr = 0.160 mol KBr / kg solution (.0750 kg solution ) ( 119 g KBr / mol KBr ) = 1.428 g KBr
mass solvent = 75 - 1.428 = 73.572 g water
Answer:
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Explanation:
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