Answer:
C. Xenon
Explanation:
The question only requires you to use your visual skills to identify the unknown gas. Look at the spectra of unknown gas and xenon, are they not identical? SInce they are same, the unknown sample must be xenon.
Answer:
C. 75%
Explanation:
From a heterozygous cross (Tt x Tt) the produced offspring would consist of:
- 25% TT -it would be tall-.
- 50% Tt -it would be tall as the gene T is dominant-.
- 25% tt -it would not be tall-.
Thus the produced offspring that would be tall is (50% + 25%) 75%. The answer is option C.
Answer:
At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.
When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.
Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.
Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.
Explanation:
Hope This Helps Amigo!
Answer:
13: if thats one of the options
Explanation:
B/c for pH to be that high you have to have alot of OH ions and almost non H+. pH won't be 0 or 2 cuz that implies ALOT of H ions. pH wont be 7 cuz that implies we have 1*10^-7 H ions.
Answer:
SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.
Explanation:
Oxidation reaction:
3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻
Reduction reaction:
Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)
Now, adding the oxidation and the reduction reactions we get the full net reaction:
Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)
Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.
<u>Therefore, SO₃²⁻ is the reducing agent. </u>
And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.
<u>Therefore, Cr₂O₇²⁻ is the oxidizing agent.</u>