1. If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO,) is mixed with...

Question

Question

1. If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO,) is mixed with 25.0 mL of 0.10 M barium nitrate (Ba(NO,)), what i

Answer 1

Answer #1

Equation : NaNO, (aq) — — Nat(aq) + NO3(aq) No of moles of NO; = 75.0 mL * 1L 0.20 mol NaNO, 1 mol NO; 1000 mL 1L1 mol NaNO,

Answer 2

Similar Homework Help Questions

5. If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO3) is mixed with 25.0 mL of 0.10 M barium nitrate (Ba(NO3)2),...

5. If 75.0 mL of a 0.20 M solution of sodium nitrate (NaNO3) is mixed with 25.0 mL of 0.10 M barium nitrate (Ba(NO3)2), what is the molar concentration of nitrate in the resulting solution? A. 0.10 M D. 0.15 M B. 0.20 M E. 0.18 M C. 0.30 M
Solid lead nitrate is slowly added to 75.0 mL of a 0.0669 M sodium phosphate solution....

Solid lead nitrate is slowly added to 75.0 mL of a 0.0669 M sodium phosphate solution. The concentration of lead ion required to just initiate precipitation is M.
Suppose 2.11 g of barium nitrate is dissolved in 50. mL of a 0.20 M aqueous solution of sodium chromate.

Suppose 2.11 g of barium nitrate is dissolved in 50. mL of a 0.20 M aqueous solution of sodium chromate. Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the barium nitrate is dissolved in it. Round your answer to 2 significant digits.

A 185.0 mL solution of 2.714 M strontium nitrate is mixed with 215.0 mL of a...

A 185.0 mL solution of 2.714 M strontium nitrate is mixed with 215.0 mL of a 3.460 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [Na+]= M [NO−3]= M [Sr2+]= M [F−]= M
A 150.0 mL solution of 3.241 M strontium nitrate is mixed with 205.0 mL of a...

A 150.0 mL solution of 3.241 M strontium nitrate is mixed with 205.0 mL of a 3.569 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [Na+]= M [NO−3]= M [Sr2+]= M [F−]=
A 165.0 mL solution of 2.777 M strontium nitrate is mixed with 210.0 mL of a...

A 165.0 mL solution of 2.777 M strontium nitrate is mixed with 210.0 mL of a 3.278 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate Number Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration. Number Number 2+ Na Number Number

10 mL of a 0.30 M sodium phosphate solution reacts with 20 mL of 0.20 M...

10 mL of a 0.30 M sodium phosphate solution reacts with 20 mL of 0.20 M lead(II) nitrate solution What mass of precipitate will form? What is the concentration of nitrate ions left in solution after the reaction is complete? What is the concentration of phosphate ions left in solution after the reaction is complete?
1. Solid barium sulfide is slowly added to 75.0 mL of a 0.178 M sodium sulfite...

1. Solid barium sulfide is slowly added to 75.0 mL of a 0.178 M sodium sulfite solution until the concentration of barium ion is 0.0288 M. The percent of sulfite ion remaining in solution is %. 2.Solid silver nitrite is slowly added to 125 mL of a potassium carbonate solution until the concentration of silver ion is 0.0134 M. The maximum amount of carbonate remaining in solution is M. 3.Solid sodium chromate is slowly added to 75.0 mL of a silver acetate...
A chemist prepares a solution of sodium nitrate (NaNO,) by measuring out 542 mol of sodium...

A chemist prepares a solution of sodium nitrate (NaNO,) by measuring out 542 mol of sodium nitrate into a 100 ml. volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol/L of the chemist's sodium nitrate solution. Round your answer to 3 significant digits.

1.Solid sodium sulfide is slowly added to 125 mL of a 0.0450 M silver nitrate solution....

1.Solid sodium sulfide is slowly added to 125 mL of a 0.0450 M silver nitrate solution. The concentration of sulfide ion required to just initiate precipitation is M. 2.Solid barium acetate is slowly added to 50.0 mL of a 0.0522 M ammonium sulfite solution. The concentration of barium ion required to just initiate precipitation is M. 3.Solid potassium hydroxide is slowly added to 150 mL of a 0.0329 M iron(III) nitrate solution. The concentration of hydroxide ion required to just initiate precipitation...