Answer:
0.15 l of 4.0 m stock KCl solution should betaken
Explanation:
N1V1=N2V2
6*0.1=V2*4
V2=0.15L
D. or C. I think it is more D. than anything else
Moles are the division of the mass and the molar mass. The moles of mercury (ii) oxide in the decomposition reaction needed to produce oxygen are 0.781 moles.
<h3>What is a decomposition reaction?</h3>
A decomposition reaction is a breakdown of the reactant into simpler products. The decomposition of mercury (ii) oxide can be shown as:
2HgO(s) → 2Hg(l) + O₂(g)
From the reaction, it can be said that 2 moles of mercury (ii) oxide decomposes to produce 1 mole of oxygen.
The moles of oxygen that needs to be produced are calculated as:
Moles = mass ÷ molar mass
= 12.5 gm ÷ 32 gm/mol
= 0.39 moles
0.39 moles of oxygen are needed to be produced.
From the stoichiometric coefficient of the reaction, the moles of HgO is calculated as: 2 × 0.39 = 0.781 moles
Therefore, 0.781 moles of HgO are required in the reaction.
Learn more about moles here:
brainly.com/question/3801333
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I think B
STEP-by STEP EXPLANATION
Answer:
Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below
The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.
Explanation:
The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.
In the Left Beaker (Left half cell), their is less concentration
Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side
In the Right Beaker (right half cell), their is more concentration
Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side
Electrons will flow from Left to Right direction.