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schepotkina [342]
3 years ago
10

If the equilibrium constant for a reaction is 0.00010, what does this tell us about the position of equilibrium for that reactio

n?
Chemistry
1 answer:
Lilit [14]3 years ago
4 0
When we know the value<span> of the </span>equilibrium constant, we can make guesses about the extent of the chemical reaction. If K is larger than 1, the mixture contains mostly products. If K is less than 1, the mixture contains mostly reactants. Therefore, <span>if the equilibrium constant for a reaction is 0.00010, then the system contains mostly of the reactants.</span>
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What is the answer to-6 + 1 = -5/
Jlenok [28]
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Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the pr
yuradex [85]

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

• ΔG < 0 , the reaction is Spontaneous

• ΔG > 0 , the reaction is non Spontaneous

• ΔG = 0 , the reaction is at equilibrium

a.

N₂(g) + H₂(g)  → N₂H₄ (g)   ; ΔG⁰f = 159.3 kJ/mol

ΔG > 0 , the reaction is non Spontaneous  , the reaction is not feasible in the forward direction

b.

2Na(s) + O₂(g) → Na₂O₂ (s)   ; ΔG⁰ f = − 447.7kJ/mol

ΔG < 0 , the reaction is Spontaneous  , the reaction is feasible in the forward direction .

c.

C(s) + 2S(s)  →  CS₂ (g)   ; ΔG⁰f = 67.1 kJ/mol

ΔG > 0 , the reaction is non Spontaneous  , the reaction is not feasible in the forward direction

d.

Ca(s) + 12 O₂ (g) → CaO (s) ;  ΔG⁰f = −604.0 kJ/mol

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8 0
3 years ago
The sum of protons and neutrons in an atom is called the.
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6 0
1 year ago
Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2
Zinaida [17]

Explanation:

Expression for the kinetic energy is as follows.

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Now, total kinetic energy will be as follows.

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Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.

Relation between energy and photon is as follows.

   Energy of photon = \frac{hc}{\lambda}

                                = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{121.6 \times 10^{-9}}

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    m \times v^{2} = 1.63 \times 10^{-18}

          v = \sqrt{\frac{1.63 \times 10^{-18}}{1.67 \times 10^{-27}}

             = 3.12 \times 10^{4} m/s

Thus, we can conclude that atoms were moving at a speed of 3.12 \times 10^{4} m/s before the collision.

8 0
3 years ago
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