Answer:
c) .51835
Explanation:
Let the relative abundance of the lighter of the two isotopes be X we have
Then the relative abundance of the heavier isotope is then (1-X)
Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.
To calculate the relative atomic mass of silver, we have
(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver
106.90509(X) + 108.9047(1-X)
108.9-108.9(x)+106.9(x) = 107.87
-2x-1.03 = 0.517450902926
Closest answer is c
c) .5184
The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.
Answer:39.8375
Explanation:
The mole for the equation is 1:1
Then the molar mass of KCl is 74.5g
Molar mass of k is 39
74.5g of KCl gives 39g of k
76.10g of KCl gives xg of k
X= 76.10×39/74.5
X= 2967.9/7
X= 39.8375
Letter B is the most plausible answer :)
Answer:
a) 7
b) 4
c) 1
d) 4
Explanation:
Significant figures:
All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.
Zero between the non zero digits are consider significant like 104 consist of three significant figures.
The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.
A) 34.50698 g
In this measurement there are 7 significant figures.
B) 30.00 mL
There are 4 significant figures.
c) 0.0008 g
There is only one significant figure which is 8.
d) 125.0 °C
There are 4 significant figures.
Explanation:
We know that relation between and is as follows.
= 14
As it is given that is 8.18. Therefore, calculate the value of as follows.
= 14
= 14
= 14 - 8.18
= 5.82
Similarly, as value of pH is given as 7.18. Therefore, value of pOH will be as follows.
pH + pOH = 14
7.18 + pOH = 14
pOH = 6.82
Let us take that B represents the enzyme. Hence, its reaction with proton will be as follows.
(protonated active enzyme)
Hence, pOH =
6.82 = 5.82 +
= 10
Therefore, percentage of active enzyme = % =
% = 90.9%
Thus, we can conclude that 90.9% is the percentage of the enzyme which is active in a buffer at pH 7.18.