How much water would I need to add to a 175 mL of a 0.45 M KOH solution to make a 0.2 M solution?

393.75mL

Explanation:

We know that

C1V1= C2V2

where C = centration, V= volume

C1= 0.45 M

V1= 175 mL = 0.175 L

C2 = 0.2M

⇒V2= C1V1/C2

⇒V2 = 0.175×0.45/0.2

=0.39375L

Therefore, amount of water needed= 393.75mL

there all right good

explanation:

the tides are influenced by both the moon and the sun, it's easy to see that when the sun lines up with the moon and the earth, as during a new moon or full moon, the tidal effect is increased. these are known as spring tides, named not for the season, but for the fact that the water "springs" higher than normal. on the other hand, if the sun and the moon are 90 degrees apart in relation to an observer on earth as during the first quarter moon or third quarter moon, then high tides are not as high as they normally would be. this is because despite its greater distance, the sun's mass allows it to exert enough gravitational force on the oceans that it can negate some of the effects of the moon's pull. this phenomenon of lower high tides is called a neap tide. if you need more information on this topic visit moonconnection.com.