Answer:
The elements in group 13 and group 15 form a cation with a -3 charge each.
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
Answer:
Option D. equation D
Explanation:
Positron is simply a beta plus decay or superscript 0 e subscription 1 (0 1 e).
Now, let us consider the options given:
Option A is emitting alpha decay
Option B is emitting beta minus decay.
Option C is emitting beta minus decay.
Option D is emitting beta plus decay also known as Positron.
Answer:
The partial pressure of BrCl at equilibrium is 0.08 atm.
Explanation:
The equilibrium constant of the reaction is given by =

initial
0 0 0.500 atm
At equilbrium
p p (0.500-2p)
The equilibrium constant's expression of the reaction is given by ;
![K_p=\frac{[BrCl]^2}{[Br_2][Cl_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BBrCl%5D%5E2%7D%7B%5BBr_2%5D%5BCl_2%5D%7D)

Solving for p:
p = 0.21 atm
The partial pressure of BrCl at equilibrium is:
(0.500-2p) = (0.500 - 2 × 0.21 )atm = 0.08 atm[/tex]