Chemistry check

First, percent error can be negative, correct? Because in my lab I got -32%.

Here are my questions that I need checked:


1. List possible manipulation errors that might contribute to the following errors in results:
1) obtaining a significantly larger amount of NaCl than was present in the original sample:

Answer) Higher percent error and possibly a bad experiment.

2) obtaining a significantly smaller amount of SiO2 than was present in the original sample.

Answer) Higher percent error and less chance an experiment will work correctly.

2. Use the information in the table to answer the following questions.

-----------soluble in:------------------
-----cold water--hotwater--3MHCl--3MNaOh
benzoic acid--no--yes-------no-----yes--
Mg(OH)2-------no--no--------yes----no---
Na2SO4-------yes--yes-------yes----yes--
Zn(OH)2-------no--no--------yes----yes--

1) Could you separate the components of a mixture of Mg(OH)2 and Zn(OH)2 by using 3M HCl? Briefly explain.

Answer) Yes I could separate the components by using 3M HCl because it is soluble in both Mg(OH)2 and Zn(OH)2.

2) Briefly describe how you could separate the components of a mixture of benzoic acid and Na2SO4 and recover the two separated substances.

Answer) I could separate the components of benzoic acid and Na2SO4 and recover the substances by adding hot water or 3MNaOH. The substances are soluble in both of these.

3) Could you separate the components of a mixture of benzoic acid, Mg(OH)2 and Na2SO4 by using only cold water and 3M HCl? Briefly explain.

Answer) I could not separate the components of this mixture because benzoic acid is not soluble in cold water and 3M HCl, and Mg(OH)2 is not soluble in cold water.

4) A mixture was known to contain three of the four compounds in the table. After the mixture of the three solids was extracted with hot water and filtered, Compound A was obtained by evaporating the filtrate to dryness.
When 3M HCl was added to the solid residue, a clear solution resulted. After adding excess 3M NaOH solution to the acid solution, a precipitate formed. After filtration, the insoluble solid was found to be Compound B. Compound C was recovered by evaporation of the filtrate. Neither Compound A, B, nor C was soluble in all of the solvents, that is water, 3M HCl, and 3M NaOH.
Identify Compounds A, B, and C. Briefly explain.

Answer) Benzoic acid is compound A because it was extracted with hot water, and Na2SO4 isn't possible because it is soluble in all solvents. Compound A, B, and C are not. Compound B is Zn(OH)2 because it was recovered through adding 3M NaOH and 3M HCl, while only 3M HCl is soluble in Mg(OH)2. Compound C is Mg(OH)2 because it was the final one left and can only be recovered through 3M HCl.


First, percent error can be negative, correct? Because in my lab I got -32%.

Yes, percent error may be negative

For the other answers, here is my honest opinion, in bold to separate the answer from the question.

1. List possible manipulation errors that might contribute to the following errors in results:
1) obtaining a significantly larger amount of NaCl than was present in the original sample:

Answer) Higher percent error and possibly a bad experiment.

It sounds as if you don't know the answer but you feel you must write something.

2) obtaining a significantly smaller amount of SiO2 than was present in the original sample.

Answer) Higher percent error and less chance an experiment will work correctly.

same comment as #1.

2. Use the information in the table to answer the following questions.

-----------soluble in:------------------
-----cold water--hotwater--3MHCl--3MNaOh
benzoic acid--no--yes-------no-----yes--
Mg(OH)2-------no--no--------yes----no---
Na2SO4-------yes--yes-------yes----yes--
Zn(OH)2-------no--no--------yes----yes--

1) Could you separate the components of a mixture of Mg(OH)2 and Zn(OH)2 by using 3M HCl? Briefly explain.

Answer) Yes I could separate the components by using 3M HCl because it is soluble in both Mg(OH)2 and Zn(OH)2.

First, your answer is incorrect. You cannot separate these two components with 3 M HCl and precisely because both ar soluble in HCl. Second, you didn't explain HOW you could separate the mixture.

2) Briefly describe how you could separate the components of a mixture of benzoic acid and Na2SO4 and recover the two separated substances.

Answer) I could separate the components of benzoic acid and Na2SO4 and recover the substances by adding hot water or 3MNaOH. The substances are soluble in both of these.

You can NOT separate two substances because both are soluble in the same solvent. The only way to separate two materials (by solubility)is to place them in some solvent that dissolves one of them but doesn't dissolve the other.

3) Could you separate the components of a mixture of benzoic acid, Mg(OH)2 and Na2SO4 by using only cold water and 3M HCl? Briefly explain.

Answer) I could not separate the components of this mixture because benzoic acid is not soluble in cold water and 3M HCl, and Mg(OH)2 is not soluble in cold water.

You CAN separate this mixture by treating with cold water first (which dissolves the Na2SO4), then treating the residue with 3 M HCl which will dissolve the Mg(OH)2 but not the benzoic acid.

4) A mixture was known to contain three of the four compounds in the table. After the mixture of the three solids was extracted with hot water and filtered, Compound A was obtained by evaporating the filtrate to dryness.
When 3M HCl was added to the solid residue, a clear solution resulted. After adding excess 3M NaOH solution to the acid solution, a precipitate formed. After filtration, the insoluble solid was found to be Compound B. Compound C was recovered by evaporation of the filtrate. Neither Compound A, B, nor C was soluble in all of the solvents, that is water, 3M HCl, and 3M NaOH.
Identify Compounds A, B, and C. Briefly explain.

Answer) Benzoic acid is compound A because it was extracted with hot water, and Na2SO4 isn't possible because it is soluble in all solvents. Compound A, B, and C are not. Compound B is Zn(OH)2 because it was recovered through adding 3M NaOH and 3M HCl, while only 3M HCl is soluble in Mg(OH)2. Compound C is Mg(OH)2 because it was the final one left and can only be recovered through 3M HCl.

The question isn't clear, but if I understand it correctly, compound A is, indeed, benzoic acid. You are also correct that the compounds can not be Na2SO4 because it is soluble in all three solvents. However, I think you have compounds B and C reversed.


the equation Mg + O forms what new substance?

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asked by Jay
  1. no

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    posted by nicv

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