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Flauer [41]
2 years ago
11

Which mineral(s) might you be examining if you place some hydrochloric (hcl) acid on a sample and it fizzes?

Chemistry
1 answer:
makvit [3.9K]2 years ago
4 0
The minerals that one might be examining if you place some hydrochloric acid on a sample and it fizzes are Calcite and Dolomite.
Calcite and dolomite are very similar minerals. Both have the same hardness, the same rhombohedral cleavage, and are found in identical geologic settings. The best way to tell one from the other is the acid test; A drop of 1 M HCl on calcite produces an instant , obvious Fizz; a drop on dolomite produces slow or no obvious bubbling. 
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When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

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2 years ago
After reading the list of physical properties above, you realize that they are describing a substance that is a
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The answer is B

Hope this helps!

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