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
Iodine 131 is a radioisotope with a very short half-life of 8.02 days, making it highly radioactive. Frequently used in small doses in thyroid cancers therapies, it is also one of the most feared fission products when accidentally released into the environment. Radiotoxicity of iodine 131.
The power used by Alex to drag the log across the yard is determined as 2,656 W.
<h3>Mass of the log</h3>
The mass of the log is calculated as follows;
W = mg
m = W/g
m = (400)/9.8
m = 40.82 kg
<h3>Velocity of the log</h3>
K.E = ¹/₂mv²
v² = 2K.E/m
v² = (2 x 900)/(40.82)
v² = 44.096
v = 6.64 m/s
<h3>Power used by Alex</h3>
P = Fv
P = 400 x 6.64
P = 2,656 W
Learn more about power here: brainly.com/question/13881533
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Answer:
98.8
Explanation:
CsF + XeF6 --> CsXeF7
37.8g ................. ?g
37.8g CsF x (1 mol CsF / 151.9g CsF) x (1 mol CsXeF7 / 1 mol CsF) x (397.2g CsXeF7 / 1 mol CsXeF7) = 98.8g CsXeF7 .......... to three significant digits
A) the independent variable is the variable that doesn’t rely on another variable. For this question the independent would be the days.
b) the dependent variable is the number of bacteria
c) ‘Number of bacteria across a number of days’