Answer:
a. KCl, c. BaCl2 and e. LiF.
Explanation:
Hello,
In this case, we can identify the ionic compounds by verifying the difference in the electronegativity between the bonding compounds when it is 1.7 or more (otherwise it is covalent) as shown below:
a. KCl: 3.0-0.9=2.1 -> Ionic.
b. C2H4: 2.5-2.1=0.4 -> Covalent.
c. BaCl2: 3.0-0.8=2.2 -> Ionic.
d. SiCl4: 3.0-1.8=1.2 -> Covalent.
e. LiF: 4.0-1.0=3.0 -> Ionic.
Therefore, ionic compounds are a. KCl, c. BaCl2 and e. LiF.
Regards.
Its called freezing, the change from a liquid to a gas is called vaporization
P(total)=P1+P2+P3+...
P(total)=P(N2)+P(others)+P(O2)
100kPa=78kPa+1kPa+P(O2)
P(O2)=100-79=21kPa
Gold fold experiment replaced the plum pudding model
First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
