Answer : The correct answer is option 2 : 0.10 M NaCl and 0.10 M NaClO₄
Explanation :
Solutions are classified into 3 categories.
1) Strong electrolytes : These are the solutions that dissociate completely forming ions . They are good conductors of electricity.
Example : All strong acids, bases and the salts made by strong acid/base are strong electrolytes .
2) Weak electrolytes : These substances do not dissociate completely thereby forming fewer ions. They are weak conductors of electricity.
3) Non electrolytes : These are the substances that do not dissociate at all. They do not form ions in aqueous medium. They are bad conductors of electricity.
Let us take a look at the given options and find out what type of solution do we have .
Option 1 : NH₃ is a weak electrolyte whereas NH₄Cl is a strong electrolyte bcause NH₄Cl is made by combination of NH₃ and HCl ( HCl is a strong acid)
Therefore NH₃ would carry electricity less efficiently than NH₄Cl.
Option 2 : Both NaCl and NaClO₄ are strong electrolytes. Therefore they will conduct electrical current equally well
Option 3 : NaNO₃ is a strong electrolyte but HNO₂ is a weak electrolyte. Therefore they will not carry the current equally
Therefore the correct option is option 2
I believe the correct answer from the choices listed above is option B. The stair-step line between the pink squares and the yellow squares separates the metals <span>from the nonmetals. Hope this answers the question. Have a nice day.</span>
Explanation:
The reaction equation will be as follows.
Hence, moles of Na = moles of electron used
Therefore, calculate the number of moles of sodium as follows.
No. of moles = 
= 
(as 1 kg = 1000 g)
= 195.65 mol
As, Q = 
where F = Faraday's constant
= 
= 
mol C
Relation between electrical energy and Q is as follows.
E = 
Hence, putting the given values into the above formula and then calculate the value of electricity as follows.
E =
= 
= 
As 1 J = 
kWh
Hence, 
kWh
= 3.39 kWh
Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.
Answer:
17.57kg of 
and its percentage yield is 81.0%
Explanation:
Through the reaction you can get the theoretical amount of that must be produced.
If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage
=81.0%
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
