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3 years ago

What are valence electrons

Chemistry

2 answers:

tatiyna3 years ago

8 0

These are electrons that hang around in the outermost shell, with the purpose of bonding. So A is correct!

stealth61 [152]3 years ago

3 0

Answer: an electron of an atom, located in the outermost shell (valence shell) of the atom, that can be transferred to or shared with another atom.

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Which type of bond will form between these elements? *

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Answer:

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Explanation:

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3 years ago

What are the three main gases that make up the atmosphere?

Nat2105 [25]

Nitrogen — 78 percent.

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3 years ago

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3 years ago

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3 years ago

At 25 ∘ C , the equilibrium partial pressures for the reaction 3 A ( g ) + 4 B ( g ) − ⇀ ↽ − 2 C ( g ) + 3 D ( g ) were found to

VMariaS [17]

<u>Answer:</u> The standard Gibbs free energy of the given reaction is 6.84 kJ

<u>Explanation:</u>

For the given chemical equation:

$3A(g)+4B(g)\rightleftharpoons 2C(g)+3D(g)$

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_C)^2\times (p_D)^3}{(p_A)^3\times (p_B)^4}$

We are given:

$(p_A)_{eq}=5.70atm\\(p_B)_{eq}=4.00atm\\(p_C)_{eq}=4.22atm\\(p_D)_{eq}=5.52atm$

Putting values in above expression, we get:

$K_p=\frac{(4.22)^2\times (5.52)^3}{(5.70)^3\times (4.00)^4}\\\\K_p=0.0632$

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = 8.314 J/K mol

T = temperature = $25^oC=[273+25]K=298K$

$K_{eq}$ = equilibrium constant at 25°C = 0.0632

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.0632)\\\\\Delta G^o=6841.7J=6.84kJ$

Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ

6 0

4 years ago