A student has two solutions of a substance;Solution-1:25M , 400ml and Solution-2:30M , 300M. What is the molarity of the final solutions if these two solutions are mixed?

The molarity of the final solution  is 1.7 M

Explanation:

The parameters given are;

First solution = 400 ml of 1.25 M

Second solution = 300 ml of 2.30 M

Therefore, we have;

First solution contains 400/1000 * 1.25 moles = 0.5 moles of the substance

Second solution contains 300/1000 * 2.30 moles = 0.69 moles of the

Hence the sum of the two solutions contains 0.5 + 0.69 = 1.19 moles of the substance

The volume of the sum of the two solutions = 400 ml + 300 ml = 700 ml

Hence we have the concentration of the final solution presented as follows;

700 ml contains 1.19 moles of the substance

Therefore;

1000 ml will contain 1000/700 * 1.19 = 1.7 moles

The molarity of the final solution = The number of moles per 1000 ml = 1.7 M.

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

it is a good conductor

explanation: