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Vilka [71]
2 years ago
8

6. What is the equilibrium constant for the following reaction? C+02 = CO2

Chemistry
1 answer:
Orlov [11]2 years ago
8 0
<h3>\tt Kc=\dfrac{[CO_2]}{[C][O_2]}</h3><h3>Further explanation</h3>

Given

Reaction

C+02 = CO2

Required

The equilibrium constant

Solution

The equilibrium constant is the ratio of concentration or pressure between the product and the reactant with each reaction coefficient raised  

The equilibrium constant is based on the concentration (Kc) in a reaction  

pA + qB -----> mC + nD  

\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}

So for the reaction :

C+O₂ ⇔ CO₂

\tt Kc=\dfrac{[CO_2]}{[C][O_2]}

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Answer:

0.727 l.

Explanation:

For the new dilute Sulphuric acid,

Number of moles = molar concentration × volume

= 2.5 × 5

= 12.5 mol

Volume of concentrated solution = 12.5/17.2

= 0.727 l of 17.2 M of sulphuric acid to be diluted.

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3 years ago
Word equation of nitrogen + oxygen —--&gt; ammonia
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Answer:

it is ammonia nitro oxide

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If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?
NARA [144]
I have attached a photo of the structure. 
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5 0
3 years ago
When you are converting grams to moles , are you multiplying or dividing ?
Effectus [21]
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4 0
3 years ago
The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
3 years ago
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