Answer:
Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.
Explanation:
The substance having highest positive reduction 
potential will always get reduced and will undergo reduction reaction.
Reduction : cathode
..[1]
Oxidation: anode
..[2]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
The overall reaction will be:
2 × [1] + 3 × [2] :
Electrons on both sides will get cancelled :
Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.
The formula of compound is LiClO4.3H2O
<em><u>calculation</u></em>
- <em><u> </u></em>find the mole of each element
that is moles for Li,Cl,O and that of H2O
- moles = % composition/ molar mass
For Li = 4.330/ 6.94 g/mol= 0.624 moles
Cl=22.10/35.5=0.623 moles
39.89/16 g/mol =2.493 moles
H20= 33.69/18 g/mol= 1.872 moles
- find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)
that is for Li= 0.624/0.623= 1
Cl= 0.623/0.623=1
O = 2.493/0.623 =4
H2O= 1.872/0.623=3
<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />
Answer: <u><em>Option B; It traps light energy and converts it into chemical energy.</em></u>
<h2>
Explanation: This substance is chlorophyll. It is a pigment present in leaves of all plants. It absorbs light energy and provides it to carry out the process of photosynthesis. Light energy is converted into chemical energy, in form of NADPH and ATP, which can be used by plants for photosynthesis.</h2><h2>
</h2><h2>
This pigment is present only in plants, so option A is incorrect.</h2><h2>
</h2><h2>
This pigment only absorbs and transfers energy to other molecules, and is not associated with carbon dioxide directly, so option C and D are also incorrect.</h2>
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
