Average atomic mass of an element is a sum of the product of the isotope mass and its relative abundance.
For example: Chlorine has 2 isotopes with the following abundances
Cl(35): Atomic mass = 34.9688 amu; Abundance = 75.78%
Cl(37): Atomic mass = 36.9659 amu; Abundance = 24.22 %
Average atomic mass of Cl = 34.9688(0.7578) + 36.9659(0.2422) =
= 26.4993 + 8.9531 = 35.4524 amu
Thus, the term “ average atomic mass “ is a <u>weighted</u> average so it is calculated differently from a normal average
D = M/V = 76g / 22ml = 3.4g/ml
Half ~ D = 38g / 11ml = 3.4g/ml
Even if the object you had was cut in half, it’s density would remain the same.
Are there any choices? Because from what the question is it seems like we need choices to help
Answer:
Explanation:
The formula of the reaction:
KClO₂ → KCl + O₂
To assign oxidation numbers, we have to obey some rules:
- Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
- The charge on simple ions signifies their oxidation number.
- The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.
The oxidation number of K in KClO₂:
K + (-1) + 2(-2) = 0
K-5 = 0
K = +5
The oxidation number of K in KCl:
K + (-1) = 0
K = +1
The oxidation number Cl in KClO₂ is -1
For Cl in KCl, the oxidation number is -1
For O in KClO₂, the oxidation number is (2 x -2) = -4
For O in O₂, the oxidation number is 0
K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.
O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.
By breaking the hydrogen bonds that cause surface tension
