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ser-zykov [4K]
3 years ago
8

Calculate the standard enthalpy of formation of ammonium carbonate in Kcal/mol?

Chemistry
1 answer:
Nina [5.8K]3 years ago
6 0

Answer:

gymbct yet ff y hcyxuci jc bbn ruf it ux7zhchchx xyzycyxhc8

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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
  • brainly.com/question/19671384
5 0
3 years ago
Will give brainliest!!!!
Luda [366]
The answer is aldehyde
4 0
3 years ago
Look at the above table.
MariettaO [177]

Answer:

<em> 1</em>. A. 0

<em>2</em>. B. 7

<em>3. </em>C<em>.</em><em> </em>4

Explanation:

1. charge is equal to the number of protons minus the number of electrons!

2. neutrons is equal to mass number minus atomic number!

3. valence electrons equal 4!

Hope this helped you! :)

7 0
3 years ago
How and why are Endothermic and Exothermic reactions chemical changes?
Olin [163]

Answer:

Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic.

Explanation:

8 0
3 years ago
the combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how many moles of water formed
enyata [817]
2C_4H_{10}+13O_2 ⇒ 8CO_2 + 10H_2O

n= \frac{m}{M} =  \frac{m}{2M(H)+M(O)}= \frac{2,46}{2*1,0+16,0}  = 0,14mol

So 0,14mol are formed.


6 0
4 years ago
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