Answer : The equilibrium concentration of NO is, 0.0092 M.
Solution :
First we have to calculate the concentration of NO.
The given equilibrium reaction is,
Initially conc. 0 0 0.1576
At eqm. (x) (x) (0.1576-2x)
The expression of 
will be,
![K_c=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
By solving the term, we get:
Neglecting the 0.0839 value of x because it can not be more than initial value.
Thus, the value of 'x' will be, 0.0742 M
Now we have to calculate the equilibrium concentration of NO.
Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M
Therefore, the equilibrium concentration of NO is, 0.0092 M.
0.086mole
Explanation:
The problem is very simple. It involves conversion from mass to number of moles of NaCl
What is mole?
A mole is a unit of measuring chemical substances. It is the amount that contains the avogadro's number of particles.
How is it related and can be converted from mass;
Number of moles of a substance = 
Since we know the mass, we can find the molar mass and put into this equation;
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Number of moles of NaCl = 
0.086mole
learn more:
Number of moles brainly.com/question/1841136
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Just find the energy of the <span>blueviolet light with a wavelength of 434.0 nm using the formula:
E = hc / lambda
E = energy
c= speed of light = 3 x 10^8 m/s
h = planck's constant = 6.6 x 10^{-34} m^2 kg / s
lambda = 434 nm = 434 x 10^{-9} m
Putting these values (with appropriate units) in the above formula :
we get: Energy, E = 4.5 x 10^{-19} J
E = 0.45 x 10^{-18} J
Now, the </span>minimum energy is 2.18×10^-{18} J but our energy is 0.45 x 10^{-18} J which is less.
<span>Means the electron will not be removed
</span>
Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Answer:
i think it will be warm air rises and takes heat with it eventually it cools and sinks
Explanation:
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