Match the words in the below to the appropriate blanks in the sentences. 1. When comparing the two elements A s and S n , the more metallic element is based on periodic trends alone.
2. When comparing the two elements G e and S b , the more metallic element is based on periodic trends alone.
A. Ge
B. Pb
C. Sb
D. impossible to determine
E. K

Sn and Ge

Explanation:

To obtain the more metallic element, we compare the group in which both elements are. Generally the element with the lower ionzation energy is he more metallic one.

Ionization energy increases fro left to right across a period. Ionization energy decreases down the group.

1. When comparing the two elements A s and S n , the more metallic element is based on periodic trends alone.

Sn has a lower ionization energy so it is the more metallic one.

2. When comparing the two elements G e and S b , the more metallic element is based on periodic trends alone.

Ge has a lower ionizaiton energy compared to Sb. So it is more metallic element than Sb.

The balanced equation

3H₂CO₃ + 2Fe → Fe₂(CO₃)₃ + 3H₂

Given

Reaction

H₂CO₃ + Fe → Fe₂(CO₃)₃ + H₂

Required

Coefficient of the reaction

Solution

aH₂CO₃ + bFe → Fe₂(CO₃)₃ + cH₂

Prioritize balancing the elements that form the most complex compounds

Fe, left = b, right = 2⇒b=2

C, left = a, right = 3⇒a=3

H, left = 2a, right = 2c⇒2a=2c⇒2.3=2c⇒6=2c⇒c=3

The equation becomes :

3H₂CO₃ + 2Fe → Fe₂(CO₃)₃ + 3H₂