When, it donates electons.
as for example take
NaCl ( sodium chloride)
it's an ionic compund,
that means it is formed by donating or gaining electrons
Na is writen first than, it must be electropositive i.e it has donated electons which made it positive and the clorine gains electron so it's electronegative.
Na is positive because
as we know it's atomic number is 11 that means it has 11 protons and 11 electrons
now, when it donate electon it has, greater number of protons whose change is +ve so the atom becomes overall positively charged ion or cation.
and something same happens in clorine and because it gains one electron and the number of electrons increase in it by 1 whise charge is -ve so, the atom becomes negatively charged ion or anion which has a -1 charge.
Let us check each statement one by one
a) Sb has a lower ionization energy but a higher electronegativity than I. : As per values given : Definitely Sb has lower ionization energy however the electronegativity of Sb is lower than that of iodine
b) Sb has a higher ionization energy but a lower electronegativity than I. FAlse:
Sb has lower ionization energy than I
c) Sb has a lower ionization energy and a lower electronegativity than I. True
d) Sb has a higher ionization energy and a higher electronegativity than I. False
4 moles of oxygen (6.0zzx10
Answer:
Empirical CHO
molecular C4H4O4
Explanation:
From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.
This means : O = 100 - 41.39 - 3.47 = 55.14%
Next is to divide the percentage compositions by their atomic masses.
C = 41.39/12 = 3.45
O = 55.14/16 = 3.45
H = 3.47/1 = 3.47
Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.
Hence the empirical formula of Maleic acid is CHO
Now we go on to deduce the molecular formula.
To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.
Now we can see that 0.378mole = 43.8g
Then 1 mole = xg
x = (43.8*1)/0.378 = 115.87 = apprx 116
[CHO]n = 116
(12 + 1 + 16]n = 116
29n = 116
n = 116/29 = 4
The molecular formula is thus C4H4O4
