The formal charge is = -3 on the central iodine atom
Answer:
Explanation:
Given :
Mass of a bar of lead = 115.2 g
Initial water level 
= 25 mL
Final water level = 35.5 mL
Difference in the water level = 35.5 - 25
= 10.5 mL
= 
We know that when a body is submerged in water, it displaces its own volume of water.
Therefore, the volume of the lead bar = volume of the water displaced = 10.5 mL =
We know that mathematically, density is the ratio of mass of body to its volume.
Density of the lead bar is given by :
=
The mass of the liquid water will be greater because it takes more space and room. will ice cubes arnt liquid and can only take a little space of room.
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31
Answer:
N-ethyl-2-methylpropan-2-amine
Explanation:
In this case, we have to start with the <u>IR info</u>. The signal on 3400 cm^-1 indicates the presence of a <u>hydrogen bonded to the heteroatom</u>. In this case, we have nitrogen in the formula, so we will have the <u>amine group</u>.
On the other hand, we have to analyze the NMR info:
a) We have 2 singlets => This indicates the presence of 2 different hydrogens without neighbors.
b) We have a triplet => This indicates the presence of <u>CH3 bonded to a CH2</u>.
c) We have a quartet => This indicates the presence of <u>CH2 bonded to a CH3</u>.
From b) and c) we can conclude that we have the <u>ethyl group</u> bonded to a nitrogen.
Finally, we have to add 4 more carbons in such a way that we only have a single signal. In this case the <u>ter-butyl group</u>.
In that way, we will have <u>2 singlets</u> (from the CH3 groups in the ter-butyl and the H on the N). Also, we will have the <u>quartet </u>on the CH2 in the ethyl group and the <u>triplet</u> on the CH3 in the ethyl group
