Explanation:
Let relative ratio of one isotope (62.94 u) be X
Then, relative ratio of other isotope (64.93) will be (1 - X)
Now,
(62.94)x + (64.93)(1 - x) = 63.55
1.99x = 1.38
X = 0.69
<u>Relative Abundance</u> :
(62.94 u) isotope = 69 %
(64.93 u) isotope = 31 %
This is an example of Charles’ Law problems, the basic equation is: V1/T1 = V2/T2. One vital thing to recall for all gas law problems is that the temperature must be in Kelvin (not Celsius).
So our given is 10.0 C = 283 K. So
V1/T1 = V2/T2
733/283 = 950/T2
T2 = 367 K
The answer is each indicator has a narrow range. We need many different indicators to span the entire ph spectrum because each indicator has a narrow range.
Soap solution with 9 acidic
