A student weighed out a 2.055 g sample of a cobalt chloride hydrate, ConClmpH2O, where n, m, and p are integer values to be determined. The student heated the sample to drive off the waters of hydration, reweighed the anhydride sample and found 1.121 g anhydrate

ANSWER

empirical formula is the simplest ratio of whole numbers of components making up a compound

molecular formula is the actual composition of components making up a compound

to analyse each statement

A.They are always different for the same compound. -

the empirical formula in some instances can be the same as the molecular formula. therefore this statement is incorrect.

B. Subscripts of empirical formulas can be reduced

empirical formula gives the simplest ratios and the subscripts cannnot be further simplified therefore this is incorrect

C. Molecular formulas can be determined from empirical formulas

using the empirical formula, the molecular formula can be determined. therefore its correct.

D. They both show the actual ratio of elements for a compound

only the molecular formula gives the actual ratio of elements whereas the empirical formula gives the simplest ratio of elements. this statement is incorrect

The correct statement is C.

Answer:

Molecular formulas can be determined from empirical formulas.

Explanation:

Answer:

1) 6(1/2)

2) 8(4/5)

Explanation:

1) 7 - 1/2 = 7/1 - 1/2 = 14/2 - 1/2 = 13/2 = 6(1/2)

2) 11(2/5) - 2(3/5) = 57/5 - 13/5 = 44/5 = 8(4/5)

Answer:

1) The dilution scheme will result in a 200μM solution.

2) The dilution scheme will not result in a 200μM solution.

3) The dilution scheme will not result in a 200μM solution.

4) The dilution scheme will result in a 200μM solution.

5) The dilution scheme will result in a 200μM solution.

Explanation:

Convert the given original molarity to molar as follows.

Consider the following serial dilutions.

1)

Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.

Molarity of 500 mL solution:

10 mL of this solution is diluted to 250 ml

Convert μM :

Therefore, The dilution scheme will result in a 200μM solution.

2)

Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.

Molarity of 100 mL solution:

10 mL of this solution is diluted to 1000 ml

Convert μM :

Therefore, The dilution scheme will not result in a 200μM solution.

3)

Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.

Molarity of 100 mL solution:

5 mL of this solution is diluted to 1000 ml

Convert μM :

Therefore, The dilution scheme will not result in a 200μM solution.

4)

Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.

Molarity of 250 mL solution:

10 mL of this solution is diluted to 500 ml

Convert μM :

Therefore, The dilution scheme will result in a 200μM solution.

5)

Dilute 10  mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.

Molarity of 250 mL solution:

10 mL of this solution is diluted to 1000 ml

Convert μM :

Therefore, The dilution scheme will result in a 200μM solution.