The given
ketone when reacted with base gave
enolate, the enolate formed due to loss of
methylene proton next to carbonyl group. Enolate when treated with
methyle Bromide gave
alpha substituted product.
Strong absorption around 1713 cm⁻¹ in IR spectrum confirms the presence of
Carbonyl group.
The product along with
¹H-NMR values is given below,
HCl is a polar molecule with the hydrogen part being partial positive while the chlorine end being partial negative. This is because hydrogen has an electronegativity of 2.1, and chlorine has an electronegativity of 3.0. This means that chlorine attracted most of the electron cloud of molecule hence is the negative dipole, The dipole moment of HCl is 1.08 D (debyes). A Debye is equal to 3.34 x 10-30 coulomb-meters (C-m). The charge of each molecule is o.176+ for H and 0.176- for the Cl
Answer:
I think C
Explanation:
because hydrogen on the right side only have two while on the left side it have 4
Solution here,
Volume(V)=67.4 L
Pressure(P)=1 atm
Temperature(T)=(0+273)K=273K
Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1
No. of moles(n)=?
Now,
PV=nRT
or, 1×67.4=n×0.0821×273
or, 67.4=22.4n
or, n=67.4/22.4
or, n=3
therefore, required no. of mole is 3.
Answer:
a. 59 m/atm
Explanation:
- To solve this problem, we must mention Henry's law.
- <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
- It can be expressed as: C = KP,
C is the concentration of the solution (C = 1.3 M).
P is the partial pressure of the gas above the solution (P = 0.022 atm).
K is the Henry's law constant (K = ??? M/atm),
∵ C = KP.
∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.
