Question

a dilute solution is prepared by transferring 40.00mL of a .3433 M stock solution to a...

a dilute solution is prepared by transferring 40.00mL of a .3433 M stock solution to a 750mL volumetric flask and diluting to mark. What is the molarity of this dilute solution
0.01838876 M
0.03661 M
0.06436 M
0.004577 M
0.1716 M
0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
a dilute solution is prepared by transferring 40.00mL of a .3433 M stock solution to a...
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
  • A stock solution of Cu2+(aq) was prepared by placing 0.9157 g of solid Cu(NO3)2∙2.5 H2O in...

    A stock solution of Cu2+(aq) was prepared by placing 0.9157 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and diluting to the mark with water. A standard solution was then prepared by transferring 2.00 mL of the stock solution to a second 25.00-mL volumetric flask and diluting to the mark. What is the concentration (in M) of Cu2+(aq)in the stock solution? What is the concentration (in M) of Cu2+(aq)in the standard solution?

  • 1. A series of dilute NaCl solutions are prepared starting with an initial stock solution of...

    1. A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume....

  • A stock solution was prepared by dissolving 0.593 g of pure ASA (molar mass = 180.2...

    A stock solution was prepared by dissolving 0.593 g of pure ASA (molar mass = 180.2 g/mol) with NaOH solution and diluting with water in a 1000 mL volumetric flask. This was followed by pipetting 4.00 mL of the stock solution into flask A and 5.0 mL of the stock into flask B. Flasks A and B were each diluted to 20 mL mark of the volumetric flask. What are the concentrations, in molarity, of ASA in flasks A and...

  • please answer all please Test No 73 M NasPO42 N 2.(8) a. How many moles of...

    please answer all please Test No 73 M NasPO42 N 2.(8) a. How many moles of Na b. What is the molarity of Na'? NAME: POs are the 0250 L solution of 0.873 M 3. (7) What is the molarity of a NaOH solution th solution? (NaOH Molar Mass: 40.00 g/mol). In that contains 4.27 g of NaOH in 125.0 mL of 4. (8) Na2SO4 + a) Balance the above reaction. b) If 63.8 mol of Na2SO4 is used, how...

  • A Na_2CO_3 standard solution is prepared by transferring 2.4817 g of primary standard grade sodium carbonate...

    A Na_2CO_3 standard solution is prepared by transferring 2.4817 g of primary standard grade sodium carbonate into a 250.0 mL volumetric flask, dissolving the sample in ~ 100 mL of distilled deionized water, and diluting to the mark. A 25.00 mL aliquot is taken and titrated with 42.65 mL of HCI solution. Calculate the concentration of the HCI solution. (Molar masses are: Na_2CO_3 = 105.989 g/mole; HCI = 36.461 g/mole; H_2O = 18.015 g/mole.) Circle the best choice. (Showing calculations...

  • Start with a 1230.0 ppm Stock Zn solution. Prepare control 1 by diluting 2.00 mL of Stock Zn solution in a 200.0 mL...

    Start with a 1230.0 ppm Stock Zn solution. Prepare control 1 by diluting 2.00 mL of Stock Zn solution in a 200.0 mL volumetric flask and dilute to the mark. Prepare control 2 by diluting 2.00 mL of control 1 into a 100.0 mL volumetric flask. What is the Zn concentration (ppm Zn) in control 2? Report your answer to 4 decimal place.

  • c. Calculate the molarity of the solution made by pipetting 5 mL of 2.00 + 0.02...

    c. Calculate the molarity of the solution made by pipetting 5 mL of 2.00 + 0.02 M NaOH with a volumetric pipet and transferring this to a 100 mL volumetric flask and diluting to the mark. Show the equation used to calculate error

  • A stock solution was prepared by dissolving 0.593 g of pure ASA (molar mass = 180.2...

    A stock solution was prepared by dissolving 0.593 g of pure ASA (molar mass = 180.2 g/mol) with NaOH solution and diluting with water in a 1000 mL volumetric flask. This was followed by pipetting 4.00 mL of the stock solution into flask A and 5.0 mL of the stock into flask B. Flasks A and B were each diluted to 20 mL mark of the volumetric flask. What are the concentrations, in molarity, of ASA in flasks A and...

  • What is the final concentration of a solution prepared by pipetting 10.00 mL of a 0.1284...

    What is the final concentration of a solution prepared by pipetting 10.00 mL of a 0.1284 M NaOH into a 25.00 mL volumetric flask and diluting with water to the mark?

  • 3. (5 pts) A solution was prepared by transferring 0.661 g of KSO to a 250.0-ml...

    3. (5 pts) A solution was prepared by transferring 0.661 g of KSO to a 250.0-ml volumetric flask (container 1) and adding water to the mark. A sample of this solution of volume 1.000 ml was transferred to a 500.0-ml volumetric flask (container 2) and diluted to the mark with water. Then 10.0 mL of the diluted solution was transferred to a 250.0-ml flask (container 3) and diluted to the mark with water. What is the final concentration of K2SO4...

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