A 50% antifreeze solution is to be mixed with a 90% antifreeze solution to get 200 liters of a 80% solution. How many liters of
the 50% solution and how many liters of the 90% solution will be used?
1 answer:
Answer:
50 ltr 150 ltr
Explanation:
this problem can be solved by the mixture and allegation concept which can be clearly understand from bellow figure in which the concentration of solution 1 is 50% and concentration of solution 2 is 90% before mixing after mixing with help bellow concept the ratio of concentration become 10:30
ratio of solution 1 and solution 2 =10:30
=1:3
total mixture is 200 liters
part of solution 1= ×200
=50 liters
part of solution 2= ×200
=150 liters
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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.