how many grams of pure water must be added to 40 grams of a 10% saline aokution to make a saline solution tha is 5% salt?

40

Step-by-step explanation:

1) the basic is: the mass of the salt before pure water is added = the mass of the salt after the pure water is added;

2) the mass of the salt before pure water is added was:

40*0.1;

3) the mass of the salt after pure water is added is:

(40+w)*0.05, where w - the required mass of pure water;

4) if the mass of the salt is the same, then:

40*0.1=(40+w)*0.05;

5) if to evaluate the equation:

(w+40)*0.05=4; ⇔ w+40=80; ⇔ w=40 gr.

b) well because you get well

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2