Answer: options B,D and F
Explanation:
Since redox reactions are those which involves both oxidation and reduction
In B , Cu is oxidized and S gets reduced
D, Na gets oxidized and hydrogen gets reduced
F, carbon gets oxidized and Oxygen gets reduced
In g, there is no change in oxidation no of s in both product and Reactants is same +4
Similarly in the case of Ag and Mg.
Answer:
Conjugate base: Propionate
Explanation:
- Conjugate base contains one less proton as compared to it's parent acid. Deprotonation occurs from most acidic region.
- Propionic acid deprotonates to produce it's strong conjugate base propionate.
- In propionic acid, -OH group in -COOH functional moiety dissociates to produce
. Because the O-H bond electrons remains highly polarized towards oxygen atom due to electron withdrawing inductive effect as well as resonating effect of -COOH moiety.
- Structure of conjugate base has been shown below.
Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
the titration is the most important part in determining whether
Answer:
The solution is: ZnS + O^2 => ZnSO
