The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M
1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3
2) Only single bonds => alkane => sufix ane
3) no substitutions
4) 5 carbons = > prefix penta.
Therefore, the name is pentane.
Alkali metals have the lowest electronegativities, while halogens have the highest.
Answer:
The formal charge on nitrogen in
is +1.
Explanation:
The structure of
is as follows.
(In attachment)

From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.
![Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1](https://tex.z-dn.net/?f=Formal%5C%2Ccharge%5C%2Con%5C%2Cnitrogen%20%3D%205-%5B0%2B%20%5Cfrac%7B8%7D%7B2%7D%5D%3D%20%2B1)
Therefore, The formal charge on nitrogen in
is +1.