Answer:
24.32 amu
Explanation:
From the question given above, the following data were obtained:
Isotope A (Mg–24):
Mass of A = 24 amu
Abundance (A%) = 79%
Isotope B (Mg–25):
Mass of B = 25 amu
Abundance (B%) = 10%
Isotope C (Mg–26):
Mass of C = 26 amu
Abundance (C%) = 11%
Average atomic mass of Mg =?
Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
= [(24 × 79)/100] + [(25 × 10)/100] + [(26 × 11)/100]
= 18.96 + 2.5 + 2.86
= 24.32 amu
Thus, the average atomic mass of Mg is 24.32 amu
Answer:
- <u><em>The first statement is false: a.At equilibrium, equal amounts of products and reactants are present. ΔG° is a function of Keq.</em></u>
Explanation:
When one part of a statement is false, the whole statement is false.
At <em>equilibrium,</em> the amounts of products and reactants does not have to be equal.
At equlibrium the rates of the forward reaction and the reverse reaction must be equal.
An equilibrium reaction may be represented by:
That represents two reactions:
- Direct reaction: A + B → C + D (A and B yield C and D)
- Reverse reaction: A + B ← C + D (C and D yield A and B: note that the arrow goes from right to left)
So, it is when the direct and the forward rates are equal that there is not net change in the amounts of all the species and so the reaction is is equilibrium).
As per the other statement, both parts are true:
- When reactants become products, they do so through an intermediate transitrion state: when the reactants approach each other and collide with enough energy and appropiate position, the bonds start to break and the bonds of the products start to form. This is the transition state.
- Most biocatalysts are proteins: enzymes are simply proteins, with specific structures, that may accelerate or even deceralate biochemical reactions.
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
Prokaryotic cells do not contain membrane-closed organelles, (??) and they are not the same size at all. (prokaryotic are way smaller) Just to help narrow down, if I'm right! (: Sorry, I know that's not completely helpful.