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they are dry mixtures that is all stirred up, now i have to find a way to not necessarily separate them, but to find the masses of each of them.

2007-09-27 10:02:25 · 3 answers · asked by Anonymous in Science & Mathematics Chemistry

they are dry mixtures that is all stirred up, now i have to find a way to not necessarily separate them, but to find the masses of each of them. I'm not so worried as the salt part, but more of the sand and zinc part...I also know that a strong acid, like Hydrochloric adic will dissolve the zinc, but i think it will get some of the sand too. I've read somewhere that it will disolve the calcium from the sand.

2007-09-27 10:09:47 · update #1

3 answers

If you add water, then filter the mixture and dry the filtrate, you will have separated the salt from the rest. I don't know how to separate sand and zinc, but zinc'll probably be heavier than sand...

2007-09-27 10:06:29 · answer #1 · answered by Anonymous · 0 0

Weigh the mixture.

Take half the mixture and add EXCESS HCl to react ALL the Zn. This will liberate H2, and note the volume of the gas liberated. From the volume of gas, calculate the moles of gas, thereby moles of zinc hence mass of zinc.

To the other half, add EXCESS water to dissolve ALL the salt. As Sand and Zinc are insoluble filter off. This leaves a solution of salt. Evaporate, dry and weigh.

Beacuse the mass of Zinc and Salt are now known, by subtraction from the original mass, the mass of sand is found.

2007-09-27 10:16:54 · answer #2 · answered by lenpol7 7 · 0 0

You can add strong acid to half the mixture and measure the hydrogen evolved. From this, you find the mass of the zinc. Take the other half and add water. Since salt solubility is 35g/L, you can find an amount of water to use. Remove the water from the mix and dry the solid. This gives you the mass of salt in that half. From the result of the first experiment (salt+sand remains) and the second (zinc + sand remains), you can back-calculate the sand.

2007-09-27 10:11:01 · answer #3 · answered by cattbarf 7 · 0 0

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