The maximum number of D orbitals in a principal energy level is 5.
The answer is B. The complete equation is C6H12O6 + 6O2 -->6H2O + 6CO2 + energy. So we can know that A and C and D is right. For B, the reaction release energy so it is exothermic reaction.
So what you’re going to do is basically the + and - in each top hand corner is the charge of compound, so for example Li has a charge of +1 while Br has a charge of -1 , to write the formula you need to get the charges to cancel out ( equal zero) so luckily this was easy because -1 +1 =0 ! So it would be LiBr. Though for another example Al has a charge of 3+ while br has a charge of -1 and these do not equal zero, so as a result you have to add more br making the Formula AlBr3! Hope this helps!
Answer:
1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3
Explanation:
Benzene is a stable aromatic compound hence it undergoes substitution rather than addition reaction.
When benzene undergoes substitution reaction, the substituent introduced into the ring determines the position of the incoming electrophile.
If I want to synthesize m-nitropropylbenzene, I will first carry out the nitration of benzene using HNO3/H2SO4 since the -nitro group is a meta director. This is now followed by Friedel Craft's alkykation using CH3CH2CH2Cl/AlCl3.
