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AlladinOne [14]
3 years ago
15

A compound was found to be soluble in water. It was also found that addition of acid to an aqueous solution of this compound res

ulted in the formation of carbon dioxide. Which one of the following cations would form a precipitate when added to an aqueous solution of this compound?
Chemistry
1 answer:
quester [9]3 years ago
3 0

A compound was found to be soluble in water. It was also found that addition of acid to an aqueous solution of this compound resulted in the formation of carbon dioxide. Which one of the following cations would form a precipitate when added to an aqueous solution of this compound?

Options

a. NH4

b. K

c. Cr

d. Rb

e. Na

Answer: c: Cr

Explanation:

The aforementioned reaction are banters for to distract ones focus from the question.

However it should be noted that

Cr can be found in tannery waste water showing that it can be soluble in water.

The carbonate compound when reacting with acid CO_2 evolves.

Focusing on the cation that would form a precipitate when added to an aqueous solution.

Chemical precipitation method is a common method for to remove chromium and to recover it from  tannery wastewater.

Cr would form a precipitate.

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Answer:

Carbon Dioxide

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3 years ago
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4 0
3 years ago
A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express
Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

(c) Molarity = 13792 mol HNO3/m3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\  \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3

(c) HNO3 molarity (mol HNO3/m3)

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are  1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3

We can now calculate the molarity as

Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3}  =13792 \frac{molHNO3}{m3}

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What does dinitogen mean in a name?
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The prefix di- means that there are two nitrogens.

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