the formula of the compound that is formed between chlorine and calcium is CaCl2
explanation
chlorine (non metal) react with Calcium ( a metal) to form an ionic compound CaCl2
Ionic compound is formed when metal loses electrons to form cation ( a positively charged ion) while non metal gains electrons to form anion ( a negatively charged ion)
An ionic compound is formed when calcium metal(Ca) loses two electron to form Ca2+ cation , while 2 chlorine atom gain one electron each to form cl- anions.
- when writing down the formula of ionic compound cation symbol is written first followed by anion symbol.
Therefore the formula of ionic compound formed between chlorine and calcium is CaCl2
Answer:
Mass of KNO3 in the original mix is 146.954 g
Explanation:
mass of 
in original 254.5 mixture.
moles of 
moles of
= 0.2926 mol of BaSO4
Therefore,
0.2926 mol of BaCl2,
mass of 
= 60.92 g
the AgCl moles 
= 1.3891 mol of AgCl
note that, the Cl- derive from both, 
so
mole of Cl- f NaCl 
mol of Cl-
mol of NaCl = 0.8039 moles
then
KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3
Mass of KNO3 in the original mix is 146.954 g
Answer:
Option A. FeCl3
Explanation:
The following data were obtained from the question:
Mass of iron (Fe) = 6.25g
Mass of the compound formed = 18g
From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow
Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.
Mass of chlorine (Cl) = 18 – 6.25
Mass of chlorine (Cl) = 11.75g
The compound therefore contains:
Iron (Fe) = 6.25g
Chlorine (Cl) = 11.75g
The empirical formula for the compound can be obtained by doing the following:
Step 1:
Divide by their molar mass
Fe = 6.25/56 = 0.112
Cl = 11.75/35.5 = 0.331
Step 2:
Divide by the smallest
Fe = 0.112/0.112 = 1
Cl = 0.331/0.112 = 3
The empirical formula for the compound is FeCl3
<span>26.833 liters
Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).
Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.
Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.
Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815
Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,
1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
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