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jolli1 [7]
3 years ago
6

Consider the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Chemistry
1 answer:
arsen [322]3 years ago
6 0

Answer:

K = 361.369

Explanation:

C2H4(g) + H2O(g) → CH3CH2OH(g)

∴ ΔG°f(298.15K) CH3CH2OH(g) = - 174.8 KJ/mol

∴ ΔG°f(298.15) C2H4(g) =  68.4 KJ/mol

∴ ΔG°f(298.15) H2O(g) = - 228.6 KJ/mol

⇒ ΔG°f(298.15) = - 174.8 - ( - 228.6 + 68.4 ) = - 14.6 KJ/mol

  • K = e∧(-ΔG°f / RT)

∴ R = 8.314 E-3 KJ/mol.K

∴ T = 298.15 K

⇒ K = e∧(-(-14.6)/((8.314 E-3)(298.15)))

⇒ K = e∧(5.889)

⇒ K = 361.369

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Give the electron configuration of vanadium (V), atomic number 23
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1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³ or [Ar] 4s²3d³

<h3>Further explanation</h3>

Given

Vanadium, atomic number 23

Required

The electron configuration

Solution

In an atom, there are levels of energy in the shell and subshell  

This energy level is expressed in the form of electron configurations.  

Charging electrons in the subshell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

The electron configuration is based on the atomic number which indicates the number of electrons of the atom

Vanadium has the atomic number 23, so the number of electrons = 23

Electron configuration:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d³

or we can write it using the noble gas notation

[Ar] 4s²3d³

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stiv31 [10]

Answer:

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3 years ago
A grapefruit has a weight on earth of 4.9 newtons. what is grapefruit's mass?
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<span>F = ma </span>
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irga5000 [103]
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8 0
4 years ago
Given the propane molecule C3H8(g)! a) What is the balanced equation for the combustion of propane in O2?(20 pts.) b) What is th
lubasha [3.4K]

Answer:

The answer to your question is below

Explanation:

Propane = C₃H₈

Process

1.- Write the chemical reaction

                     C₃H₈  +  O₂   ⇒   CO₂  +  H₂O

balanced chemical reaction

                      C₃H₈  +  5O₂   ⇒  3CO₂  +  4H₂O

               Reactants      Elements     Products

                      3                    C                  3

                      8                    H                  8

                     10                    O                 10

b) Standard enthalpy

Propane                  -104.7 kJ/mol

Oxygen                         0   kJ/mol

Carbon dioxide      -393.5 kJ/mol

Water                      -241.8 kJ/mol

ΔH° = ∑ΔH° products - ∑H° reactants

ΔH° = 3(-393.5) + 4(-241.8) - [(-104.7) + 0]

-Simplification

ΔH° = -1180.5 kJ/mol - 967.2 kJ/mol + 104.7 kJ/mol

ΔH° = - 2042.5 kJ/mol

This reaction is exothermic because ΔH° is negative.

4 0
3 years ago
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