Unsaturation (IHD) 2 hydrogen Needed
IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)
Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2
The degrees of unsaturation in a molecule are additive — a
molecule with one double bond has one degree of unsaturation, a molecule with
two double bonds has two degrees of unsaturation, and so forth.
DSVR because it only makes sense. Try it out and if it works give me a Brainly
Answer:
C. A hydrocarbon molecule containing six carbon atoms and only
single bonds
Explanation:
hope it helps
Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
Answer:
A) 122 atm
Explanation:
PV = nRT
Solve for P --> P = nRT/V
n = 10.0 mol + 5.0 mol = 15.0 mol
R = 0.08206 L atm / mol K
T = 25 + 273 = 298 K
V = 3.0
P = (15.0)(0.08206)(298) / (3.0) = 122 atm
