Using the relationship M1V1 = M2V2 where M1 and M2 are the molar concentrations (mol/L or mmol/ml) and V1 and V2 are the volumes of the solutions, we can arrive at the following answer for the given problem:
<span>15.0M (L of stock solution) = 2.35M (0.25L) *all volumes were converted to liters.
L of stock solution = (2.35*0.25)/15.0
Therefore, 0.0392L or 39.17 ml of stock solution is needed. </span>
D. the activation energy is the displacement of energy needed for the reaction, so the distance underneath the maximum point
To prepare 350 mL of 0.100 M solution from a 1.50 M
solution, we simply have to use the formula:
M1 V1 = M2 V2
So from the formula, we will know how much volume of the
1.50 M we actually need.
1.50 M * V1 = 0.100 M * 350 mL
V1 = 23.33 mL
So we need 23.33 mL of the 1.50 M solution. We dilute it
with water to a volume of 350 mL. So water needed is:
350 mL – 23.33 mL = 326.67 mL water
Steps:
1. Take 23.33 mL of 1.50 M solution
<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M
solution</span>
Answer:
Any kind, as long as there is an action.
Holy I'm not that smart lol but I'll get back to you
