Answer:
Explanation:
For transitions:
So, 
and 
(As the hydrogen has to ionize)
Thus,
Answer:
The empirical formula is ZnO2
Explanation:
What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
Step 1: Data given
Suppose the compound has a mass of 100.0 grams
A compound contains:
67.1 % Zinc = 67.1 grams
100 - 67.1 = 32.9 % oxygen = 32.9 grams
Molar mass of Zinc = 65.38 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles of Zinc
Suppose the compound is 100 grams
Moles Zn = 67. 10 grams / 65.38 g/mol
Moles Zn = 1.026 moles
Step 3: Calculate moles of O
Moles O = 32.90 grams / 16.00 g/mol
Moles O = 2.056 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
Zn: 1.026/1.026 = 1
O: 2.056/1.026 = 2
The empirical formula is ZnO2
To control this we can calculate the % Zinc for 1 mol
65.38 / (65.38+2*16) = 0.67.1 = 67.2 %
Answer:
b. 6.02 x 1023 molecules
Explanation:
The formula mass of ammonia is 14 + 1 × 3 = 17.
The number of moles in 27.6g ammonia is 27.6 ÷ 17 = 1.62 mol.
A mole is 6.02 × 10²³, so the number of hydrogen atoms in a 1.62 moles of ammonia is 1.62 × 6.02 × 10²³ × 3 = 2.93 × 10² atoms.
