Clearly volume will INCREASE almost sevenfold, as we would expect if we reduce the pressure. ... Assuming this is an ideal gas, and that the number of molecules *n* of this gas remains constant, and given that the temperature is also ... That is, in these conditions, pressure times volume is constant :
There are zero lone pairs of electrons on the central carbon atom in a Lewis Structure of CHI3. The correct option is b.
Explanation:
Lone pair electrons are those valence electrons of an atom which do not take part in bonding or left after the bonds are formed.
CH
carbon is the central atom in the compound having 4 valence electrons.
Each iodine molecule has 7 valence electrons.
H has 1 valence electron.
from the Lewis structure of CH
, it is seen that all four valence electrons of the carbon is involved in bonding with 3 atoms of iodine and 1 atom of hydrogen.
Thus when all valence electrons are involved in bonding no lone pair of electron is found in CH
that means zero lone pairs of electrons.
The octet of the iodine, carbon and hydrogen gets completed to form the compound.
Answer:
Learn from their experiment and know the possible results from trial and error
Answer:

Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-3.4%7D%3D3.98x10%5E%7B-4%7D)
And the percent ionization is:
![\% \ ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20%5C%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
We compute the concentration of the acid, HA:
![[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\](https://tex.z-dn.net/?f=%5BHA%5D%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5C%25%20%5C%20ionization%7D%2A100%5C%25%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%7D%7B66%5C%25%7D%20%20%2A100%5C%25%5C%5C%5C%5C)
![[HA]=6.03x10^{-4}](https://tex.z-dn.net/?f=%5BHA%5D%3D6.03x10%5E%7B-4%7D)
Thus, the Ka is:
![Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%2A3.98x10%5E%7B-4%7D%7D%7B6.03x10%5E%7B-4%7D%7D%5C%5C%20%20%5C%5CKa%3D2.63x10%5E%7B-4%7D)
So the pKa is:

Regards.
answer: its 7290 joules.
explanations: the first procedure is to convert 1 pound to kilogram. 1 kg = 2.205 hence given 100 lb so we cross multiply. 1 kg * 100 = 2.205 * x
hence x= 45 kg. let's convert 1 mile per hour = 0.45 metre per second we cross multiply by 40 mile per hour. x= 40 * 0.45= 18 m/s.
KE= 1/2 * 45 * (18)^2
= 1/2 * 45 * 14580
= 7290joules