Asked by Fahmi
A student followed procedure in lab using a mixture of Cobalt (ii) choride hexahydrate: COCl2 * 6H2O ... and NaBr. The crucible weights (1.136x10^1) grams, the crucible and sample weighted (1.31x10^1) grams, and the final mass of the crucible and residue after the last heating was (1.267x10^1) grams.
How many grams of water were lost?
How many grams of water were lost?
Answers
Answered by
DrBob222
13,1 g = weight xble + CoCl2.6H2O + NaBr
-11,36 f = weight xble
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1,74 = weight sample but note that the 4 is not significant and I would record the weight of the sample as 1.7 g.All others were weighed to 4 places; I wonder why this one was weighed only to three places.
12.67 g = final mass after heating
-11.36 g = weight xble
----------------
1.31 g = weight sample less the water.
mass water = 1.74-1.31 >Look at your data. Since that 4 is not significant the correct way to do it is 1.7 - 1.31 = 0.4 g and not 0.43. Leaving that last number off creats an error of (0.43-0.40) or about 7%. at a minimum.
-11,36 f = weight xble
------------------
1,74 = weight sample but note that the 4 is not significant and I would record the weight of the sample as 1.7 g.All others were weighed to 4 places; I wonder why this one was weighed only to three places.
12.67 g = final mass after heating
-11.36 g = weight xble
----------------
1.31 g = weight sample less the water.
mass water = 1.74-1.31 >Look at your data. Since that 4 is not significant the correct way to do it is 1.7 - 1.31 = 0.4 g and not 0.43. Leaving that last number off creats an error of (0.43-0.40) or about 7%. at a minimum.
Answered by
Fahmi
Thank Dr. Bob.
I will check on that. However. I don't understand how you would set up this now that it's asking for moles.
I'm really having trouble solving these three questions.
B. )The mixture is the same with Cobalt (ii) chloride hexahydrate and NaBr.
Now the crucible weight is 1.10x10^1 grams, the crucible and sample weighed 1.307x10^1 grams, and the final mass of the crucible and residue after the last heating was 1.274x10^1 grams. How many MOLES OF WATER were lost?
I will check on that. However. I don't understand how you would set up this now that it's asking for moles.
I'm really having trouble solving these three questions.
B. )The mixture is the same with Cobalt (ii) chloride hexahydrate and NaBr.
Now the crucible weight is 1.10x10^1 grams, the crucible and sample weighed 1.307x10^1 grams, and the final mass of the crucible and residue after the last heating was 1.274x10^1 grams. How many MOLES OF WATER were lost?
Answered by
DrBob222
After you find grams H2O lost, then mols H2O = grams h2O/molar mass H2O.
The question may actually be asking for mole H2O per 1 mol CoCl2. If so that's just one more step. Find the ratio of mols H2O lost per mol CoCl2 and that will be 6 ols H2O/1 mol CoCl2. But alrady know that because the formula ws given as CoCl2.6H2O
The question may actually be asking for mole H2O per 1 mol CoCl2. If so that's just one more step. Find the ratio of mols H2O lost per mol CoCl2 and that will be 6 ols H2O/1 mol CoCl2. But alrady know that because the formula ws given as CoCl2.6H2O
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