Question

In: Chemistry

11.5 mL of a 0.399 M sodium phosphate solution reacts with 19.9 mL of a 0.192...

11.5 mL of a 0.399 M sodium phosphate solution reacts with 19.9 mL of a 0.192 M lead(II) nitrate solution. What mass of precipitate will form?

Solutions

Expert Solution


Related Solutions

11.4 mL of a 0.343 M sodium phosphate solution reacts with 16.9 mL of a 0.174...
11.4 mL of a 0.343 M sodium phosphate solution reacts with 16.9 mL of a 0.174 M lead(II) nitrate solution. What mass of precipitate will form?
Determine the mass of sodium phosphate required to prepare 100.00 mL of a 0.500 M solution....
Determine the mass of sodium phosphate required to prepare 100.00 mL of a 0.500 M solution. Solid sodium phosphate, Na3PO4, will be provided; it has a formula weight of 163.94 g/mol.
What is the molarity of a sodium hydroxide solution if 25.0 ml of this solution reacts...
What is the molarity of a sodium hydroxide solution if 25.0 ml of this solution reacts exactly with 22.30 ml of 0.253 M sulfuric acid? Thanks!
when 23.8ml of 0.0985 M sodium phosphate solution is combined with 18.2 ml of 0.0105M copper...
when 23.8ml of 0.0985 M sodium phosphate solution is combined with 18.2 ml of 0.0105M copper (II) sulfate solution the precipitation reaction results in the formation of insoluble Copper(II) phosphate a.) write a balanced molecular equation, total ionic and net ionic equations and idebtify the soectatir ions b.) identify limiting reagent c. calculate theoretical yield in grams of copper (II) phosphate d.) what is the percent yield of the reaction if the actual amount of solid formed is 0.227g?
A. What is the molarity of sodium ions in a solution prepared by mixing 202.1 ml of 0.63 M sodium phosphate with 239 ml of 1.31 M sodium sulfide. Enter to 2 decimal places.
A. What is the molarity of sodium ions in a solution prepared by mixing 202.1 ml of 0.63 M sodium phosphate with 239 ml of 1.31 M sodium sulfide. Enter to 2 decimal places.B. A 17.3 ml sample of a 1.49 M potassium chloride solution is mixed with 58.6 ml of a 1.34 M lead(II) nitrate solution and a precipitate forms. The solid is collected, dried, and found to have a mass of 0.5 g. What is the percent yield?...
29.7 mL of 4.6 M Sodium Phosphate is mixed with 72.0 mL of 4.0 M Calcium...
29.7 mL of 4.6 M Sodium Phosphate is mixed with 72.0 mL of 4.0 M Calcium Bromide to form a precipitate. What precipitate is formed in this reaction? Calculate the theoretical mass (in grams) of the precipitate using only the volume and molarity of the Sodium Phosphate.Calculate the theoretical mass (in grams) of the precipitate using only the volume and molarity of the Calcium Bromide.What is the theoretical yield (in grams) of the precipitate?What is the concentration (in M) of...
Sodium phosphate is added to a solution that contains 0.0051 M aluminum nitrate and 0.014 M...
Sodium phosphate is added to a solution that contains 0.0051 M aluminum nitrate and 0.014 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate?
Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate and 0.021 M...
Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate and 0.021 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate? Hint: Note that well over 99% of the first ion has precipitated before the second ion starts to precipitate. Correct Answer: 2.613e-7 (I need an "EXPERT" to show me how to...
Sodium phosphate is added to a solution that contains 0.0071 M aluminum nitrate and 0.016 M...
Sodium phosphate is added to a solution that contains 0.0071 M aluminum nitrate and 0.016 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate?
24.3 mL of 0.131 M copper(II) sulfate reacts with 16.9 mL of 0.107 M sodium hydroxide...
24.3 mL of 0.131 M copper(II) sulfate reacts with 16.9 mL of 0.107 M sodium hydroxide to form a precipitate (Cu(OH)_2?2??). What is the concentration of Cu^{2+}?2+?? in the resulting solution?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT