Answer:
False
Explanation:
It is important to give exact measurements.
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of 
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium 
The expression of 
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
Molar mass :
Cl₂ = 71.0 g/mol Na = 23.0 g/mol
<span>2 Na + Cl</span>₂<span> = 2 NaCl
</span>
2 x 23 g Na -------> 71.0 g Cl₂
96.6 g Na ----------> ?
Mass Cl₂ = ( 96.6 x 71.0 ) / ( 2 x 23 )
Mass Cl₂ = 6858.6 / 46
= 149.1 g of Cl₂
hope this helps!
The equation is:
3 O₂ + 4 Co → 2 Co₂O₃
Oxidation half reaction:
Co → Co³⁺ + 3 e
Reduction half reaction:
O₂ + 4 e → 2 O²⁻
To balance the equation number of electrons lost must be equal to number or electrons gained so we must multiply oxidation half time 4 and reduction half times 3
