It is in the ten thousands place
B=3
Hope this helps.
Correct me if I’m wrong.
Are you rounding to the nearest 100’s place?
Answer:
the chemist should use 60 liters of 55% solution and 40 litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.
Step-by-step explanation:
From the given information,
Let x be the litres of 55% pure solution
Let y be the litres of 30% pure solution
Also;
Given that our total volume of solution is 100 litres
x+y =100 ---- (1)
The total solution of pure by related by the sum of the individual pure concentrations to make up the concentration of final solution.
(0.55)(x)+(0.30)(y) = 0.45(100) ---- (2)
From equation (1)
Let ; y = 100 - x
Replacing the value for y = 100 - x into equation (2)
(0.55)(x)+(0.30)(100-x) = 0.45(100)
0.55x + 30 - 0.30x = 45
0.55x - 0.30x = 45 - 30
0.25x = 15
x = 15/0.25
x = 60 liters of 55% solution
From ; y = 100 - x
y = 100 - 60
y = 40 litres of 30% solution.
Therefore, the chemist should use 60 liters of 55% solution and 40 litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.
