425 / 50 = 8,5
There you go.
Answer:
Therefore the concentration of salt in the incoming brine is 1.73 g/L.
Step-by-step explanation:
Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.
Let the concentration of salt be a gram/L
Let the amount salt in the tank at any time t be Q(t).

Incoming rate = (a g/L)×(1 L/min)
=a g/min
The concentration of salt in the tank at any time t is =
g/L
Outgoing rate =



Integrating both sides

[ where c arbitrary constant]
Initial condition when t= 20 , Q(t)= 15 gram


Therefore ,
.......(1)
In the starting time t=0 and Q(t)=0
Putting t=0 and Q(t)=0 in equation (1) we get









Therefore the concentration of salt in the incoming brine is 1.73 g/L
Answer:
<h2>which question dear..maybe you forgot to add the attachment...</h2>
I think the answer is 8.5 h
Answer:
Step-by-step explanation:
The solution of the problem has been solved on paper and attached in the attachment section. Kindly refer to that and feel free to ask any doubt.