a solution of 0.170 mol/l ion strength is prepared using nano3 (solid) and na2so4 (solid). calculate the mass of each salt that is required to prepare 250 ml of the said solution, knowing that the final na+ concentration was 0.130 mol/l.

1,06 g of NaNO₃ and 1,42 g of Na₂SO₄

Explanation:

In water, NaNO₃ and Na₂SO₄ dissociates as:

NaNO₃ → Na⁺ + NO₃⁻

Na₂SO₄ → 2 Na⁺ SO₄²⁻

The ionic strength is difined as:

∑ Where Ci and Zi are concentration and charge of each ion in solution. Thus:

= 0,170 mol/L

Knowing = 0,130 mol/L you can obtain:

0,210 mol/L = (1)

The 0,130 mol/L of Na⁺ comes from  , thus:

0,130 mol/L = (2)

Replacing (2) in (1)

And:

The 0,050 M of NO₃⁻ comes from:

0,050M×0,250L× = 1,06 g of NaNO₃

The 0,040 M of SO₄⁻ comes from:

0,040M×0,250L× = 1,42 g of Na₂SO₄

I hope it helps!

idk

explanation:

dasdad

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: