Im bored in this quarentine and have a lot of work to catch up on. One of my math problems is "How much water should be added to 20 ounces of a 20% acid solution to reduce it to a 10% solution?" Can you help me?

20/3  ounces of water is added.

Step-by-step explanation:

how many ounces of pure water should be added to 20 ounces of 40% solution of muriatic acid to obtain a 30% solution of muriatic acid?

The acid is 40% of 20 ounces = 8 ounces.

0.4*20 = 8

That same 8 ounces is to be 30%. The water will be the other 70%.

8/30 = x/100

30x = 800

x = 80/3, the total amount of the solution.

We started with 20, so 80/3 - 20 = 20/3

20/3 ounces of water is added.

step-by-step explanation:

1. the cylinder is a. 54

2.   a.108 /

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.