Answer:

Explanation:
When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.
So:
24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O
The next step is to divide each mass by their molar mass to convert your grams to moles.
24.42/40.08 = 0.6092 mol
17.07/14.01 = 1.218 mol
58.85/15.99 = 3.680 mol
Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.
0.6092 mol/0.6092 mol = 1
1.218 mol/0.6092 mol = 2
3.680 mol/0.6092 mol = 6
So the empirical formula is 
Answer:
Explanation:
Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.
Charges on group one.
Hydrogen = +1
Lithium = +1
Sodium = +1
Potassium = +1
Rubidium = +1
Cesium = +1
Francium = +1
Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.
Charges on group two.
Beryllium = +2
Magnesium = +2
Calcium = +2
Strontium = +2
Barium= +2
Radium = +2
Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.
Charges on group 13.
Boron = +3
Aluminium = +3
Gallium = +3
Indium = +3
Thallium= +3
Group 13 elements are also shows +1 charge by losing one valance electron.
Answer:
a) 2-bromopyrrole
Explanation:
Our options for this questions are:
a) 2-bromopyrrole
b) 2,3-dibromopyrrole
c) N-bromopyrrole
d) 3-bromopyrrole
To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.
The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the
and
. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).
I hope it helps!