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satela [25.4K]
3 years ago
6

A saturated solution of copper (II) hydroxide is prepared in a beaker by adding the solid to water and stirring. When the soluti

on is
prepared, crystals of copper (ll) hydroxide remain on the bottom of the beaker.

Which change will cause the molar solubility of copper (II) hydroxide to increase?


A)addition of solid copper chloride


B)

addition of solid sodium hydroxide


C)

addition of aqueous hydrochloric acid


D)

dilution of the solution by addition of water
Chemistry
1 answer:
12345 [234]3 years ago
8 0

Answer:

addition of aqueous hydrochloric acid

Explanation:

The solubility of copper (ll) hydroxide in water is negligible. This implies that the substance is only slightly soluble in water.

However, the addition of aqueous hydrochloric acid leads to the following reaction;

Cu(OH)2 + 2HCl ---------> CuCl2 + 2H2O

Copper II chloride is soluble in water hence the crystals dissolve when stirred. and the molar solubility increases.

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seropon [69]
No hay suficiente informacion.
8 0
4 years ago
The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 375 kJ/mol and the change in enthalpy for the reaction
HACTEHA [7]
Answer: 625 kj/mol

Explanation:

As shown below this expression gives the activation energy of the reverse reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

1) The activation energy, EA is the difference between the potential energies of the reactants and the transition state:

EA = energy of the transition state - energy of the reactants.

2) The activation energy of the forward reaction given is:

EA = energy of the transition state - energy of  [ NO2(g) + CO(g) ] = 75 kj/mol

3) The negative enthalpy change - 250 kj / mol for the forward reaction means that the products are below in the potential energy diagram, and that the potential energy of the products, [NO(g) + CO2(g) ] is equal to 375 kj / mol - 250 kj / mol = 125 kj/mol

4) For the reverse reaction the reactants are [NO(g) + CO2(g)], and the transition state is the same than that for the forward reaction.

5) The difference of energy between the transition state and the potential energy of [NO(g) + CO2(g) ] will be the absolute value of the change of enthalpy plus the activation energy for the forward reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

EA reverse reaction = 375 kj / mol + |-250 kj/mol | = 375 kj/mol + 250 kj/mol = 625 kj/mol.

And that is the answer, 625 kj/mol
4 0
3 years ago
A substance is considered to have a smaller surface area when...
Andrews [41]
Answer: C you have a small amount of substance
5 0
3 years ago
PLEASE HELP ME I BEG YOU
VladimirAG [237]

The average kinetic energy of 1 mole of a gas at -32 degrees Celsius is:
 
3.80 x 103 J

The relationship between volume and temperature of a gas, when pressure and moles of a gas are held constant, is: V*T = k.

FALSE

The relationship between moles and volume, when pressure and temperature of a gas are held constant, is: V/n = k. We could say then, that:
If the moles of gas are tripled, the volume must also triple.


 If the temperature and volume of a gas are held constant, an increase in pressure would most likely be caused by an increase in the number of moles of gas.

TRUE

If the vapor pressure of a liquid is less than the atmospheric pressure, the liquid will not boil.

TRUE


35 - AB

36 -  BD

33 - true

34 - False

20 - 6

21 - orthohombic


4 0
3 years ago
Watching the assignment without knowing how to solve the questions is the worst thing to experience hope someone can help with t
frez [133]

Answer:

- 622.5kJ

Explanation:

1)      2NH3 + 3N2O ----> 4N2 + 3H2O    ΔH° 1= - 1010kJ

-  3 (N2O  + 3H2     --->  N2H4 + H2O    ΔH° 2 = - 317 kJ)

2)      2NH3 + 3N2O ----> 4N2 + 3H2O        ΔH° 1= - 1010kJ

<u>      - 3N2O   - 9 H2     ----> - 3N2H4 - 3H2O      - 3ΔH° 2 = 3*317 kJ</u>

2NH3  + 3N2H4 --->4N2+9H2,     ΔH° 1 -  3*ΔH° 2

2NH3  + 3N2H4  --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O,     ΔH°3 = -143 kJ)

   -2NH3 - 1/2O2 ---> - N2H4 - H2O,    - ΔH°3 = 143 kJ

 2NH3  + 3N2H4 --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

<u>  -2NH3 - 1/2O2              ---> - N2H4 - H2O,             - ΔH°3 = 143 kJ</u>

 4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

4)

H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ

9*(H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ)

9H2+ 9/2 O2 ---> 9H2O,  9*ΔH° 4 = 9*(- 286) kJ

4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

<u>9H2+ 9/2 O2    ---> 9H2O,                             9*ΔH° 4 = 9*(- 286) kJ</u>

4N2H4 +4O2 --->4N2+8H2O,         ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

5)

1/4*(4N2H4 +4O2 --->4N2+8H2O,    ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O,        1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

6)

N2H4 +O2 --->N2+2,        

1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

   

7 0
3 years ago
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