Answer:
The most important resonance structure is 4 (attached picture). Its bon order is 
or 
.
Explanation:
A picture with 4 forms of the perchlorate structure is attached. The first structure has simple bonds. The second structure contains a double bond, the third structure has two double bonds and the fourth structure has three double bonds.
Formal charge = group number of the periodic table - number of bonds (number of bonding electrons / 2) - number of non-shared electrons (lone pairs)
The formal charges in the first structure is +3 in chlorine and -1 in oxygen.
The formal charges in the second structure is +2 in chlorine, -1 in oxygen and 0 in the double bond oxygen.
The formal charges in the third structure is +1 in chlorine, -1 in the single bond oxygens and 0 in the double bond oxygens.
The formal charges in the fourth structure is 0 in chlorine, -1 in the single bond oxygen and 0 in the double bond oxygens.
The most important resonance structure is given by:
- Most atoms have 0 formal charge.
- Lowest magnitude of formal charges.
- If there is a negative formal charge, it's on the most electronegative atom.
Hence, the fourth structure is the mosr important.
The bond order of the structure is:
Total number of bonds: 7
Total number of bond groups: 4
Bond order= 
<span>There are pros and cons as to whether CCA-treated (pressure-treated) wood should be removed from existing structures, and both sides are subjective.
Some of the arguments for leaving it include:
*When burned, the wood can release dangerous, and sometimes, lethal fumes.
*If buried in a landfill, the chemicals can soak into the ground and eventually contaminate ground water.
*Removing it can expose people to arsenic
*It is costly to remove an existing infrastructure that may or may not be harming people
*Studies conducted within the past decade have determined structures containing CCA-treated wood pose no hazard
*Studies also concluded that children who played on CCA-treated playgrounds were exposed to arsenic levels lower than those that naturally occur in drinking water
Some of the arguments for removing it include:
*The EPA determined that some children could face higher cancer risks from exposure to CCA-treated wood
*If removed, it will need to be disposed of and, as discussed above, that creates another set of problems that could affect a community's health.
A possible solution is to leave existing CCA-treated wood in place but seek viable, safe alternatives for future structures.</span>
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)
Answer:
the positive electrode is an anode, and the negative electrode is a cathode.
Explanation:
Answer:
Atomic neutron mass electron number
Explanation:
