Compound Z has a MW of 100. Your lab partner weighed 25 grams of compound Z and dissolved it in water to a final volume of 1 liter. What is the concentration of the solution expressed as molarity?

0.25 mol /L

Explanation:

Molarity , is the number of moles present in a liter of solution .

M = n / V

M = molarity ( unit = mol / L or M )

V = volume of solution in liter ( unit = L ),  

n = moles of solute ( unit = mol ),  

Moles is given mass divided by the molecular mass ,  

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

The data given is as follows -

w = 25 g

m = 100 g/mol

V = 1 L

From the above data ,

Moles are given as -

n = w / m

n = 25 / 100 = 0.25 mol ,

Now , the molarity is given as ,

M = n / V

M = 0.25 mol / 1 L

M = 0.25 mol /L

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:

i think that the answer is c