Answer: 72L of 30% and 128L of 80%
You can determine the weight of the acid by multiplying the concentration with the volume. Let say v1 is the volume of 30% solution needed and v2 is the volume of 80% solution.
The weight of acid from the used solution should be equal to the product. You can get this equation
final solution= solution1 + solution2
200l * 62%= v1 * 30% + v2*80%
124L= 0.3v1 + 0.8v2
124L- 0.3v1= 0.8v2
v2=155L- 0.375v1
The total volume of both should be 200l. If you use the previous equation, you can calculate:
v1+v2=200L
v1+ (155L- 0.375v1)= 200L
0.625v1= 200L - 155L
v1= 45/ 0.625= 72L
v1+v2=200L
v2= 200L- 72L= 128L
Answer:
1.3 atm
Explanation:
Apply the ideal gas law: PV = nRT
Since we need to find the pressure, solve for pressure:
P = nRT / V
Plug in our given values of n = 4.6 moles, R = 0.08206 (ideal gas law constant using atm), T = 325 K, and V = 96.8 L:
P = 4.6(0.08206)(325) / 96.8 = 1.267 atm -> 1.3 atm (two significant figures)
Answer:
what are we doing......?????