Question



e) All three students ar 21. What volume (mL) of a 5.45 M lead nitrate solution must be diluted to 820.7 mL make a 1.41 M sol
0 0
Add a comment Improve this question Transcribed image text
Answer #1

This problem is based on dilution equation which is discussed in detail as follows-

Que. 21 : Given :- 1] Initial concentration of lead nitrate - M:= 5.45 mg 2Volume of 5.45M Lead nitrate need to diluted = V1

Add a comment
Know the answer?
Add Answer to:
e) All three students ar 21. What volume (mL) of a 5.45 M lead nitrate solution...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • L M 21. What volume (ml) of a 5.45 M lead nitrate solution must be a...

    L M 21. What volume (ml) of a 5.45 M lead nitrate solution must be a to 20.7 ML M solution of lead nitrate? VPL NO 3 = ? a) 0.00471 MPENOS = 3.45m b) 6310 c) 3170 d) 212 e) 0.00936 V-2 Mand M=545

  • What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to...

    What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 200.0 mL to make a 0.880 M solution of sodium hydroxide? Question 37 options: A) 50.0 B) 29.3 C) 176 D) 26.4 E) 2.64

  • What volume of 0.100 M sodium chloride must be added to 75.0 mL of 0.200 M...

    What volume of 0.100 M sodium chloride must be added to 75.0 mL of 0.200 M lead(II) nitrate to precipitate all of the lead ions? A 15 mL B 30 mL C 150 mL D 300 mL E 600 mL

  • please number and write the correct answer Ar have a total pressure of 1.6 bar. What...

    please number and write the correct answer Ar have a total pressure of 1.6 bar. What is the partial 17. A mixture of 100 g of Ne and 10.0 Ne? a) 0.40 bar b) 1.1 bar c) 0.54 bar GP 0.80 bar e) 1.3 bar 18. A sample of N, effuses in 255 s. How long will the same size sample of Cl2 take to effuser c12 a) 388 5 b) 1555 c) 247 s d) 406 5 e) 6455...

  • d. 0.750 M 9. What volume of a 3.50 M Na PO, solution should you use...

    d. 0.750 M 9. What volume of a 3.50 M Na PO, solution should you use to make 1.50 L of a 2.55 M Naz PO, solution? a. 0.917 L b. 13.4 L C. 2.06 L 2 Kl(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + Pblz (s) What minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all of the lead in 155.0 mL of a 0.112 M lead(II) nitrate solution? a. 348 ml b. 86.8 ml...

  • suppose 50.00 ml of a 1x10^-7 M solution of lead(II) nitrate is mixed with 50.00 ML...

    suppose 50.00 ml of a 1x10^-7 M solution of lead(II) nitrate is mixed with 50.00 ML of a 1X10^-8 solution of sodium phosphate which of the following statement is true? for lead(II) phosphate, Ksp=1x10^-44 Options: a: no precipitate forms because Qc<Ksp b no precipitate forms because Qc>Ksp c a precipitate forms because Qc<Ksp d a precipitate forms because Qc>Ksp e no precipitate forms because they are Qc=Ksp

  • Suppose 1.32 g of lead(II) nitrate is dissolved in 50. mL of a 0.20 M aqueous solution of ammonium sulfate

    Suppose 1.32 g of lead(II) nitrate is dissolved in 50. mL of a 0.20 M aqueous solution of ammonium sulfate Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) nitrate is dissolved in it. Round your answer to 2 significant digits 

  • 6.80 mL of a 1.30 M solution of lead(II) nitrate was combined with 3.20 mL of...

    6.80 mL of a 1.30 M solution of lead(II) nitrate was combined with 3.20 mL of a 5.0 M solution of ammonium sulfide. 1. Write the chemical equation for this reaction and balance it. 2. What is the total concentration of all soluble ions after these two solutions are mixed and allowed to react completely? 3. A solid product precipitated out of this reaction, and was collected. If 2.01 g of this product were collected, what is the percent yield?

  • 14./What volume of a 6,00 M solution of sodium hydroxide must be diluted to 200.0 mL...

    14./What volume of a 6,00 M solution of sodium hydroxide must be diluted to 200.0 mL to m .00 make a 0.880 M solution of sodium hydroxide?u =1.00 M2.880 M üni make V1=200ml Va=! 6.00 200.880 -

  • Part A When 75.5 mL of a 0.100 M lead (II) nitrate solution is mixed with...

    Part A When 75.5 mL of a 0.100 M lead (II) nitrate solution is mixed with 104.5 mL of a 0.185 M potassium iodide solution, a yellow-orange precipitate of lead (II) iodide is formed What mass (in grams) of lead (II) iodide is formed, assuming the reaction goes to completion? Submit My Answers Give Up Part B What is the molarity of Pb2+ in the resulting solution? Pb2+ molarity = Submit My Answers Give Up Part C What is the...

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