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3 years ago

The highly reactive elements in group 7a are known for forming salts.what are they called

1 answer:

7 0

Group 7A are halogens

If you look at a periodic table, these elements include F, Cl, Br, I, and At. Some well known salts are KCl and NaCl (better known as table salt!)

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3 years ago

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3 years ago

Which element could not form in the sun and would have to be formed in a supernova?.

sasho [114]

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2 years ago

When potassium metal is added to water, a violent reaction occurs that releases a tremendous amount of light and thermal energy.

Thepotemich [5.8K]

Answer:

An Exothermic reaction.

Explanation:

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5 0

2 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

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,

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

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

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

8 0

2 years ago