The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =
The empirical formula : MnO₂
The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.
Answer,
For example, silver ion can be precipitated with hydrochloric acid to yield solid silver chloride. Because many cations will not react with hydrochloric acid in this way, this simple reaction can be used to separate ions that form insoluble chlorides from those that do not.
Answer:
I don't know I'm sorry I will tell you another answer asks me to
Answer:
Option B. Cation that is smaller than the original atom.
Explanation:
Magnesium is a divalent element. This implies that magnesium can give up 2 electrons to become an ion (cation) as shown below:
Mg —> Mg²⁺ + 2e¯
Next, we shall write the electronic configuration of magnesium atom (Mg) and magnesium ion (Mg²⁺). This can be written as follow:
Mg (12) = 2, 8, 2
Mg²⁺ (10) = 2, 8
From the above illustration, we can see that the magnesium atom (Mg) has 3 shells while the magnesium ion (Mg²⁺) has 2 shells.
This simply means that the magnesium ion (Mg²⁺) i.e cation is smaller that the original magnesium atom (Mg).
