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 %
The gravitational force between the objects depends on the mass of the objects and the distance between them.
The Equator is where the doldrums are found.
The best answer is "<span>High temperatures increase the activation energy of the reaction."
The Haber process is an exothermic reaction at room temperature. This means that the reaction actually favors the reverse reaction, especially when the temperature is increased. So why increase the reaction temperature?
The reason for this is that nitrogen is a very stable element. Therefore, more energy is needed to overcome the slow rate of reaction. So the reaction temperature must be low enough to favor a forward reaction, but high enough to speed up the reaction.</span>
Answer:
Mass = 2.12 g
Explanation:
Given data:
Volume of KMnO₄ = 255 mL (255/1000 = 0.255 L)
Molarity = 0.0525 M
Mass in gram = ?
Solution:
First of all we will calculate the number of moles.
<em>Molarity = number of moles of solute / volume in litter</em>
0.0525 M = number of moles of solute / 0.255 L
Number of moles of solute = 0.0525 M ×0.255 L
Number of moles of solute = 0.0134 mol
Mass in gram:
<em>Number of moles = mass/ molar mass</em>
Mass = moles × molar mass
Mass = 0.0134 mol × 158.04 g/mol
Mass = 2.12 g
