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laila [671]
3 years ago
10

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

38 percent. The correct value for the percentage of water in the hydrate is 51 percent. Which of the following is the most likely explanation for this difference.A. Strong initial heating caused some of the hydrate sample to spatter out of the crucibleB. The dehydrated sample absorbed moisture after heatingC. The amount of the hydrate sample used was too smallD. The crucible was not heated to constant mass before useE. Excess heating caused the dehydration sample to decompose
Chemistry
1 answer:
zzz [600]3 years ago
4 0

Answer:

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.

Explanation:

Hi

The percentage of water in the sample is lower than expected.

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:

If part of the sample is splashed from the crucible the mass of water detected will be less.

B. The dehydrated sample absorbed moisture after heating:

If the sample absorbs water after heating the percentage of water would be higher than expected.

C. The amount of the hydrate sample used was too small:

Depending on the sample size, different procedures can be chosen for analysis.

D. The crucible was not heated to constant mass before use:

In many occasions the crucible is heated next to the sample and not in previous form.

E. Excess heating caused the dehydration sample to decompose:

If the sample decomposes during heating, the analysis should be discarded.

success with your homework

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Question:

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a) Number of moles of H:

b) Number of moles of N:

Answer:

a) Number of moles of H: 2

b) Number of moles of N: 2

Explanation:

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Question:

2. Ag.SO.

a) Molar Mass:

b) Percent Composition of Ag:

c) Percent Composition of S:

d) Percent Composition of O:​

Answer:

a) Molar Mass:   155.93 Kg

b) Percent Composition of Ag:   69%

c) Percent Composition of S:  20.5%

d) Percent Composition of O:​ 10.2%

Explanation:

Molar mass =   molar mass of Ag +  molar mass of S +  molar mass of O  

=>107.87+32.06+16

=> 155.93 Kg

Percent Composition of Ag

= \frac{ \text{mass due  to Ag}}{\text {total molar mass}}  \times 100

= \frac{107.87}{155.93} \times 100

= 0.69 \times 100

= 69%

Percent Composition of S:

= \frac{ \text{mass due  to S}}{\text {total molar mass}}  \times 100

=\frac{32.06}{155.93} \times 100

= 0.205 \times 100

= 20.5%

Percent Composition of O:

= \frac{ \text{mass due  to O}}{\text {total molar mass}}  \times 100

= \frac{16.00}{155.93} \times 100

= 0.102 \times 100

= 10.2%

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