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3 years ago

PLEASE HELP ASAP Half equation for zinc atoms changing into zinc ions

Chemistry

1 answer:

Fiesta28 [93]3 years ago

3 0

Answer:

$Zn^{0} --->Zn^{2+} +2e^{-}$

Explanation:

$Zn^{0} --->Zn^{2+} +2e^{-}$

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Which set of numbers gives the correct possible values of l for n = 3?

Free_Kalibri [48]

0 1 2 3
0 and 3 is 3
1 and 2 is 3

3 0

3 years ago

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30.0 mL of 0.50 M NaOH to neutralize 15.0 mL of HNO3, what is the concentration (molarity) of the HNO3?

vredina [299]

First find the no. of moles of NaOH :
30/1000 = 0.3 dm3 so no. of moles = 0.3*0.5 = 0.15 moles 

as NaOH reacts with HNO3 in a ratio of one to one, there must have been 0.15 moles of HNO3 too 
moles/volume = concentration 
volume= 15/1000 = 0.15 dm3 

concentration = 1.15/0.15 = 1 mol.dm-3 

The quicker way would be to realize that you used twice as much NaOH so the HNO3 had to be twice as strong

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

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

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

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2 years ago

How is a pure substance different from a mixture? A. pure substances cannot be seperated by physical means.

jeyben [28]

Answer: Pure substances cannot be separated by physical means.

Explanation:

A pure substance is a substance which contains definite composition of only one type of component. Hence, it cannot be separated by physical means.

On the other hand, a mixture is a substance which contains two or more than two types of components and they can be separated by physical means as well.

Thus, option (a) is correct one that is pure substances cannot be separated by physical means.

6 0

3 years ago

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What is the molar mass and identity of a 5.28g sample of a gas that occupies 2.96 L at STP?

Mrrafil [7]

Answer:

40 g/mol, Argon

Explanation:

We can find the number of moles of the gas by using the equation of state for an ideal gas:

$pV=nRT$