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Troyanec [42]
3 years ago
6

Which is the correct expression for Keq when the chemical reaction is as follows:

Chemistry
2 answers:
zhuklara [117]3 years ago
4 0
The expression of the Keq is [NH3]2/([N2][H2]3). The common formula of reaction: ax+by=cz is Keq - [z]c/([x]a*[y]b). Remember to use a balanced chemical reaction.
skelet666 [1.2K]3 years ago
3 0

Answer : The expression for equilibrium constant will be,

K_{eq}=\fac{[NH_3]^2}{[N_2][H_2]^3}

Explanation :

Equilibrium constant : It is defined as the number of ratio of the concentration between the products and the reactants at equilibrium.

The equilibrium reaction is,

aX+bY\rightleftharpoons cW+dZ

The general expression for equilibrium constant will be,

K_{eq}=\frac{[X]^a[Y]^b}{[W]^c[Z]^d}

The given balanced chemical reaction will be,

N_2+3H_2\rightleftharpoons 2NH_3

The expression for equilibrium constant will be,

K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}

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Which subctant will not conduct electricity a Aluminium b copper c plastic d steel​
jolli1 [7]

Answer:

plastic

Explanation:

4 0
2 years ago
Read 2 more answers
If a hard water solution is saturated with calcium carbonate, what volume of the solution has to evaporate to deposit 1.00×102 m
olasank [31]
I think the Ksp for Calcium Carbonate is around 5×10⁻⁹
(I don't know if this is the Ksp value that you use because I read somewhere that this value can vary.  You should probably check with your teacher with what Ksp value they want you to use)

the equation for the dissociation CaCO₃ in water is CaCO₃(s)⇄Ca²⁺(aq)+CO₃²⁻(aq) which means that the concentration of Ca²⁺ is equal to the concentration of CO₃²⁻ in solution.  For every molecule of CaCO₃ that dissolves, one atom of Ca²⁺ and one molecule of CO₃²⁻ is put into solution which is why the concentrations are equal in solution. 

Since Ksp=[Ca²⁺][CO₃²⁻] and we know that [Ca²⁺]=[CO₃²⁻] we can rewrite the equation as Ksp=x² since if you say that [Ca²⁺]=[CO₃²⁻] when you multiply them together you get the concentration squared (I am calling the concentration x for right now).

when solving for x:
5×10⁻⁹=x²
x=0.0000707
Therefore [Ca²⁺]=[CO₃²⁻]=0.0000707mol/L which also shows how much calcium carbonate is dissolved per liter of water since the amount of Ca²⁺ and CO₃²⁻ in solution came from the calcium in a 1 to 1 molar ratio as shown in the equation (the value we found for x is the molar solubility of calcium carbonate).

Using the fact that the molar mass of calcium carbonate is 100.09g/mol you can use dimensional analysis as fallows:
(0.0000707mol/L)(100.09g/mol)=0.007077g/L
That means that there is 0.007077g of Calcium carbonate that can precipitate out of 1L of water.

since the question is asking for how much water needs to be evaporated to precipitate 100mg (0.1g) of Calcium you have to do the fallowing calculation:
(0.1g)/(0.007077g/L)=14.13L of water.
14.13L of water needs to evaporate in order to precipitate out 100mg of calcium carbonate

These types of questions can get long and confusing so I bolded parts that were important to try to guide you through it more easily.

I hope this helps.  Let me know if anything is unclear.
6 0
3 years ago
The molar mass of Cr(OH)2 is:
son4ous [18]

The molar mass of Cr(OH)_2 is 86.02 g/mole .

<h3><u> Explanation:</u> </h3>

The molar mass of a chemical compound is represented as the mass of a unit of that compound separated by the number of substances in that unit, measured in moles. The molar mass is a volume, not molecular, the property of a substance.

The molar mass is a percentage of various examples of the compound, which usually change in mass due to the appearance of isotopes.

From the below attached table, the Molar mass of Cr(OH)_2 is 86.0108 g/mol.

3 0
3 years ago
May someone assist, please...? I don't know how to do chemistry work...
EastWind [94]

Answer:

1) 90.0 mL

2) 11.25 M

3) 0.477 M

4) 144 mL

Explanation:

The main formula that will be used for all these calculations is:

                                                     C₁V₁ = C₂V₂

C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.

For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.

1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:

5.00 M · V₁ = 500.0mL · 0.900 M                        - divide by 5.00

C₁ = 90.0 mL

2) This time we're finding the initial concentration:

20.0mL · C₁ = 150.0mL · 1.50 M                          - divide by 20.0mL

C₂ = 11.25 M

3) Now we're finding the final concentration:

12.00mL · 3.50 M = 88.0mL · C₂                         - divide by 88.0mL

C₂ = 0.477 M

4) Finally, we're looking for the final volume:

9.0mL · 8.0 M = 0.50 M · V₂                                - divide by 0.50 M

V₂ = 144mL

6 0
3 years ago
Name two compounds in unpolluted air?
goblinko [34]
Nitrogen and oxygen are in unpolluted air

6 0
3 years ago
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