HomeworkLib
Question

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

science   chemistry   
1 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

✔ Recommended Answer
Answer #1

hlaite a balanud chamical egualuen S& (NO3) taNaF Sa +2 NaND3 calculate mo of mols fer sst2 mats = Valume x Molaity (motj 2.7Calculate theoetcal mas ok SaF 0-439 molFx mol St 2molF 46.2 43 X 195.62-9mel 14 mols pin alution Moles Na 11 07439 mol D.4 LLonc 0.50209 mal Sa+ 0.7439 Tnitially Sol contained moles Sat2 plecipitated 0.37 19 S2 ppt moles a Si semaining in sol- 0.502

0 0
Add a comment
Know the answer?
Add Answer to:
A 185.0 mL solution of 2.714 M strontium nitrate is mixed with 215.0 mL of a...
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

Similar Homework Help Questions

  • 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

  • This is the fifth part of a five part problem: A 185.0 mL solution of 2.049...

    This is the fifth part of a five part problem: A 185.0 mL solution of 2.049 M strontium nitrate is mixed with 210.0 mL of a 2.458 M sodium fluoride solution. Assuming complete precipitation of strontium fluoride, calculate the final concentration of nitrate ions.

  • This is the third part of a five part problem: A 185.0 mL solution of 2.049...

    This is the third part of a five part problem: A 185.0 mL solution of 2.049 M strontium nitrate is mixed with 210.0 mL of a 2.458 M sodium fluoride solution. Assuming complete precipitation, calculate the final concentration of fluoride ions.

  • 375 mL of a 0.150 M aqueous solution of silver (I) nitrate is mixed with 125...

    375 mL of a 0.150 M aqueous solution of silver (I) nitrate is mixed with 125 mL of a 0.125 M aqueous solution of sodium phosphate. Calculate the mass of precipitate that forms and the final concentration of each ion in the mixed solution. Volumes are additive and the precipitation reaction goes to completion. Can you show all work for calculating the final concentration of each ion in the reaction including Ag, NO3^-1, Na and PO4^-3

  • A 829.0-mL aliquot of 0.880 M aqueous potassium hydroxide is mixed with 829.0 mL of 0.880...

    A 829.0-mL aliquot of 0.880 M aqueous potassium hydroxide is mixed with 829.0 mL of 0.880 M aqueous magnesium nitrate. ( just need d.) a.)Write a balanced chemical equation for any reaction that occurs. b.) The precipitate is magnesium hydroxide Mg(OH)2 c.) What mass of precipitate is produced? d.)Calculate the concentration of each ion remaining in solution after precipitation is complete. Assume that the precipitate is completely insoluble The concentration of K+ = ____ M The concentration of OH- =...

  • 100.0 mL of 0.100 M sodium sulfide is mixed with 100.0 mL of 0.100 M chromium...

    100.0 mL of 0.100 M sodium sulfide is mixed with 100.0 mL of 0.100 M chromium (III) nitrate. Calculate the mass of solid that forms and the concentration of the remaining species in solution: assume complete precipitation.

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

    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 is the molar concentration of nitrate in the resulting solution?

  • If a solution containing 30.61 g of lead(II) nitrate is allowed to react completely with a...

    If a solution containing 30.61 g of lead(II) nitrate is allowed to react completely with a solution containing 5.102 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: How many grams of the reactant in excess will remain after the reaction? mass of excess reactant: Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (O) for the number...

  • A 30.0 ml. stock solution of 0.447 M ammonium nitrate, NH.NO3, is mixed into a beaker...

    A 30.0 ml. stock solution of 0.447 M ammonium nitrate, NH.NO3, is mixed into a beaker containing 220.0 mL of water. What is the nitrate ion concentration in the resulting solution? 0708 00610M DO

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT