Answer:
A3+ and B-
Explanation:
Elements in group 13 have outermost electron configuration, ns2np1 hence they form trivalent positive ions.
Elements in group 17 have outermost electron configuration ns2np5 hence they form univalent negative ions.
This implies that, if element A is in Group 13 and element B is in Group 17, the ions formed are A3+ and B-.
Part 1: Potassium, and Rubidium.
Part 2: Calcium has 20 protons and 20 electrons because the atomic number for calcium is 20 and that determines how many protons there are and in an atom, the number of protons is the same number of electrons. Calcium has about 20 neutrons. I got the number of Neutrons by subtracting the mass number(40.078) and the atomic number(20), I got 20.078. Round to the nearest whole number because you cannot have half or partial neutron. So, Calcium has 20 protons, 20 electrons, and 20 neutrons,
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Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
