A compound composed of only nitrogen and iodine is found to contain 3.55% nitrogen by mass. Its empirical and molecular formulas are the same. What is the hybridization of the nitrogen atom in this compound?

The nitrogen atom is sp³ hybridized

This is trigonal pyrymid shape

Explanation:

Step 1: Data given

A compound contains only nitroen (N) and Iodine (I)

The procent of nitrogen by mass = 3.55%

Atomic mass of N = 14.0 g/mol

Atomic mass of I = 126.9 g/mol

Step 2: Calculate mass % of iodine

Mass % iodine = 100 % - 3.55%

Mass % iodine = 96.45 %

Step 3: Calculate number of moles

Let's suppose the total mass = 100 grams

Mass of N = 3.55 grams

Mass of I = 96.45 grams

Moles N = 3.55 grams / 14.0 g/mol = 0.254 moles

Moles I = 96.45 grams / 126.9 g/mol = 0.760 moles

Step 4: Calculate empirical formula

For 0.254 moles N we have 0.760 moles I this is 1:3

This means that for 1 mol N we have 3 moles I

The empirical ( and also the molecular) formula is I3N

The nitrogen atom is sp³ hybridized.

This is trigonal pyrymid shape

3

explanation:

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"