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 %
D. breakdown of rocks through mechanicals or chemicals processes
Opportunity cost
Opportunity cost is the benefit, profit or value that could have been received or obtained but instead given up in favor of another alternative or course of action. Every situation has several alternatives and every resource has several alternative uses as well. Thus, every decision comes with it an opportunity cost. <span>
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Answer and explanation:
The relative rates of free radical halogenation is in the order of,
F₂ (10⁸) > Cl₂ (1) > Br₂ (10⁻¹¹) > I₂ (10⁻²²)
The above order also show a decreasing reactivity from left to right
Hence, reaction of fluorine with alkanes is highly reactive and it is too difficult to control.
The reaction with Cl₂ is moderately fast, while with Br₂ is slow and with I₂ is too slow reaches to equilibrium.
Hence in general we don't prefer radical fluorination of alkanes.
