Answer:
B
Explanation:
Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation
PV = nRT
n = PV/RT
The parameters have the following values according to the question:
P = 780mmHg, we convert this to pascal.
760mHG = 101325pa
780mmHg = xpa
x = (780 * 101325)/760 = 103,991 Pa
V= 400ml = 0.4L
T = 135C = 135 + 273.15 = 408.15K
n = ?
R = 8314.463LPa/K.mol
Substituting these values into the equation yields the following:
n = (103991 * 0.4)/(8314.463 * 408.15)
= 0.012 moles
Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules
Answer:

Explanation:
Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is
.
Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is
.
Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.
Ca Cl
2 1
Cross multiplying the valency, We get, 
Thus, the formula of calcium chloride is
.
Answer:
ZcH's electrons = 5-2 = +3.
ZcH for inner electrons = -5 - 0 = +5.
Explanation:
Boron 1s² 2s² 2p
As, 2s ⇒ 2.58.
2p ⇒ 2.42.
Electrons = +3.
Greater than 2 cm = 0.35
Slightly greater than +3 .
The earth wouldn’t have an atmosphere
Answer:
The value of dissociation constant of the monoprotic acid is
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)

Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)

![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)


The value of dissociation constant of the monoprotic acid is
.