Answer:
What type of work do u have? LOL
Explanation:
Answer: Rate in terms of disappearance of
= ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
Rate in terms of disappearance of
= ![-\frac{1d[Cl_2]}{1dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BCl_2%5D%7D%7B1dt%7D)
Rate in terms of appearance of
= ![\frac{1d[NOCl]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNOCl%5D%7D%7B2dt%7D)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of = ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
Rate in terms of disappearance of = ![-\frac{1d[Cl_2]}{1dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BCl_2%5D%7D%7B1dt%7D)
Rate in terms of appearance of
= ![+\frac{1d[NOCl]}{2dt}](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BNOCl%5D%7D%7B2dt%7D)
Answer:0,25 g/cm3 creo
Explanation:densidad es masa entre volumen
Answer:
Explanation:
SODIUM ATOM;
SODIUM ATOM IS NEUTRAL
SODIUM ION;
IT IS A CHARGED SPECIE WITH A CHARGE OF +1
SODIUM ATOM:
THE NUMBER OF PROTONS AND ELECTRONS ARE SAME ie:11
SODIUM ION:
NUMBER OF PROTONS AND ELECTRONS ARE NOT SAME ie. ELETRON: 10, PROTONS:11
HOPE IT WILL HELP:)
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02