Question

Part A- After 50.0 mL of a 0.250 M solution of calcium nitrate is combined with...

Part A- After 50.0 mL of a 0.250 M solution of calcium nitrate is combined with 100.0 mL of a 0.835 M solution of calcium nitrate, what is the molar concentration of Ca(NO3)2(aq) in the combined solution?

Part B- Once in solution, the calcium nitrate exists not as intact calcium nitrate but rather as calcium ions and nitrate ions. What are the molar concentrations of Ca2+(aq) in the combined solution.

Part C- What are the molar concentrations of NO3−(aq) in the combined solution?

Thank you in advance!!

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Calculate the final concentration of calcium nitrate when a solution is made by dissolving 25.0 mL...
Calculate the final concentration of calcium nitrate when a solution is made by dissolving 25.0 mL of a 0.50 M solution of Ca(NO3)2 to a total volume of 125.0 mL.
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution...
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) Part A What mass of silver chloride can be produced from 1.94 L of a 0.126 M solution of silver nitrate? Part B The reaction described in Part A required 3.49 L of calcium chloride. What is the concentration of this calcium chloride solution?
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution...
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) 1. What mass of silver chloride can be produced from 1.97 L of a 0.285 M solution of silver nitrate? 2.The reaction described in Part A required 3.62 L of calcium chloride. What is the concentration of this calcium chloride solution?
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution...
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) Part A What mass of silver chloride can be produced from 1.28 L of a 0.267 M solution of silver nitrate? Express your answer with the appropriate units. Part B The reaction described in Part A required 3.60 L of calcium chloride. What is the concentration of this calcium chloride solution? Express your answer with the appropriate units
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution...
When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) Part A What mass of silver chloride can be produced from 1.09 L of a 0.281 M solution of silver nitrate? Express your answer with the appropriate units. Part B The reaction described in Part A required 3.16 L of calcium chloride. What is the concentration of this calcium chloride solution? Express your answer with the appropriate units.
A solution is prepared by adding 50.0 mL of 0.040 M HClO4 and 50.0 mL of...
A solution is prepared by adding 50.0 mL of 0.040 M HClO4 and 50.0 mL of 0.060 M HCl.(a) Calculate the concentrations of all ions in solution.(b) Calculate the pH of this solution.
A 100.0 ml portion of 0.250 M copper (ii) chloride is combined with 100.0 ml of...
A 100.0 ml portion of 0.250 M copper (ii) chloride is combined with 100.0 ml of 0.250 M potassium phosphate, resulting in the formation of a plate blue precipitate. Part (A): what is the mass of the precipitate formed? Part (B): how many moles of the excess reagent remain after the precipitation? I need help with this question please.
When 50.0 mL of a 1.00 M solution of Fe(NO3)3 are mixed with 50.0mL of a...
When 50.0 mL of a 1.00 M solution of Fe(NO3)3 are mixed with 50.0mL of a 1.00 M solution of NaOH, a precipitate forms. What Ions remain after the reaction is complete? Fe(NO3)3 (aq) + 3 NaOH (aq) -> Fe(OH)3 (s) + 3 NaNO3 (aq) A. Fe3+, OH-, Na+, and NO3- B. Fe3+ and OH- C. Na+ and NO3- D. Fe3+, Na+, and OH-
Calculate the volume (mL) of 1.0 M metal nitrate solution required to prepare 3.0 g of...
Calculate the volume (mL) of 1.0 M metal nitrate solution required to prepare 3.0 g of CaCO3. Again, using the balanced chemical equation you should be able to relate the moles of pigment to the moles of metal nitrate and from that calculate the volume of 1.0 M metal nitrate solution required. Balanced equation: Ca(NO3)2 (aq) + K2CO3 (aq) ↔ CaCO3 (s) + 2KNO3 (aq)
A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3(aq) with 50.0 mL...
A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3(aq) with 50.0 mL of 0.300 M NH4Cl(aq). The pKb of NH3  is 4.74. 7.50 mL of 0.125 M NaOH is added to the 100.0 mL of the original buffer solution prepared in Question 1 (no HCl added). Calculate the new NH3 concentration for the buffer solution. Calculate the new NH4Cl concentration for the buffer solution. Calculate the new pH of the solution.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT