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Elena L [17]
3 years ago
10

How many Sodium (NA) Atoms are in Methylene?​

Chemistry
1 answer:
postnew [5]3 years ago
4 0

Answer:

No Sodium(Na) in methylene

Explanation:

Methylene is an organic compound

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The decomposition of dinitrogen pentaoxide has an activation energy of 102 kJ/mol and ΔH°rxn = + 55 kJ/mol. What is the activati
ahrayia [7]

<u>Answer:</u> The activation energy of the reverse reaction is 47 kJ/mol

<u>Explanation:</u>

The chemical equation for the decomposition of dinitrogen pentaoxide follows:

N_2O_5\rightleftharpoons 2NO_2+\frac{1}{2}O_2

We are given:

Activation energy of the above reaction (forward reaction) = 102 kJ/mol

Enthalpy of the reaction = +55 kJ/mol

As, the enthalpy of the reaction is positive, the reaction is said to be endothermic in nature.

To calculate the activation energy for the reverse reaction, we use the equation:

E_a_{\text{(forward)}}=E_a_{\text{(backward)}}+|\Delta H|

where,

E_a_{\text{(forward)}} = Activation energy of the forward reaction = 102 kJ/mol

E_a_{\text{(backward)}} = Activation energy of the backward reaction = ?

\Delta H = Enthalpy of the reaction = +55 kJ/mol

Putting values in above equation, we get:

102=E_a_{\text{(backward)}}+55\\\\E_a_{\text{(backward)}}=(102-55)=+47kJ/mol

Hence, the activation energy of the reverse reaction is 47 kJ/mol

3 0
3 years ago
At 25 ∘C , the equilibrium partial pressures for the reaction were found to be PA=5.16 bar, PB=5.04 bar, PC=4.11 bar, and PD=4.8
erastova [34]

Answer: 5.85kJ/Kmol.

Explanation:

The balanced equilibrium reaction is

A(g)+2B(g)\rightleftharpoons 4C(g)+D(g)

The expression for equilibrium reaction will be,

K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}

Now put all the given values in this expression, we get the concentration of methane.

K_p=\frac{(4.85)\times [(4.11)^4}{(5.04)^2\times (5.16)}

K_p=10.6

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

\Delta G^o=-2.303\times RT\times \log K_c

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = 25^0C=(25+273)K=298 K

K_c = equilibrium constant = 10.6

\Delta G^o=-2.303\times 8.314\times 298\times \log (10.6)

\Delta G^o=5850.23J/Kmol

\Delta G^o=5.85kJ/Kmol

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.

3 0
3 years ago
Oxidation-reduction is often the most confusing and abstract part of chemistry for first-time chemistry students. Is it really w
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Which of the following is the atomic number of an alkali metal?​
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