First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
![\frac{[H^+][HC2O4^-]}{[H2C2O4]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BHC2O4%5E-%5D%7D%7B%5BH2C2O4%5D%7D%20)
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
Answer:
a. 0.119mol Kr
Explanation:
To solve this problem, we must understand that;
Mass = number of moles x molar mass
Molar mass of Kr = 83.3g/mol
Ar = 40g/mol
He = 4g/mol
Ne = 20.18g/mol
a0.119 mol Kr mass = 0.119 x 83.3 = 9.9g
b 0.400 mol Ar mass = 0.4 x 40 = 16g
C 1.25 mol He mass = 1.25 x 4 = 5g
d 2.02 mol Ne mass = 2.02 x 20.18 = 40.8
Krypton is the answer
The experimental control is the standard used as a comparison for the experimental groups.
For example, you may be trying to find out how different types of disinfectants affect bacterial growth. The control group would receive <em>no</em> disinfectant whereas the experimental groups would be the ones on which the disinfectants were tested.
Hope this makes sense!
Atomic weight = 197
symbol = Au
electrons = 79
neutrons = 197 - 79 = 118
<u>answer: E</u>
Answer:
OPTUON C,CHEMICALLY BONDED