Write in short adsorbent chromatography

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

makvit [3.9K]

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.

4 0

3 years ago

A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa

Elina [12.6K]

Answer: The identity of the unknown acid is butanoic acid or ascorbic acid.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.570 M

Volume of solution = 39.55 mL

Putting values in above equation, we get:

$0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol$

The chemical equation for the reaction of NaOH and monoprotic acid follows:

$NaOH+HX\rightarrow NaX+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

So, moles of monoprotic acid = 0.0225 moles

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.0225}{2}=0.01125moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For butanoic acid:

Mass of butanoic acid = 2.002 g

Molar mass of butanoic acid = 88 g/mol

Putting values in above equation, we get:

$\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol$

For L-tartaric acid:

Mass of L-tartaric acid = 2.002 g

Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

$\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol$

For ascorbic acid:

Mass of ascorbic acid = 2.002 g

Molar mass of ascorbic acid = 176 g/mol

Putting values in above equation, we get:

$\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol$

As, the number of moles of butanoic acid and ascorbic acid is equal to the number of moles of acid getting neutralized.

Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

5 0

3 years ago

In the following neutralization reaction, which substance is the acid?

denis-greek [22]

Hydrochloric acid is a colorless, highly pungent solution of hydrogen chloride in water. It is a corrosive

5 0

3 years ago

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Calculate the volume of 0.684mol of carbon dioxide at s.t.p. show working

solong [7]

Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.684

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$ (at STP)

$V=\frac{nRT}{P}$

$V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}$

$V=15.3L$

Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

6 0

3 years ago

HELP!!!!!!! 40 points! I WILL GIVE BRAINLIEST IF RIGHT

Korolek [52]

b) High-frequency waves travel somewhat faster than low-frequency waves

4 0

3 years ago

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