Mass = 473.2 g
Explanation:
Given data:
Mass of cobalt(III) nitrate = 206 g
Mass of silver bromide produced = ?
Solution:
Chemical equation:
CoBr₃ + 3AgNO₃ → 3AgBr + Co(NO₃)₃
Number of moles of cobalt(III) nitrate:
Number of moles = mass/ molar mass
Number of moles = 206 g/ 245 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of cobalt(III) nitrate with silver bromide.
Answer:
+6
Explanation:
We are given;
We are required to determine the oxidation number of S in the compound.
We need to know that;
- The total oxidation number of this compound is Zero
- Oxidation number of Ba metal is +2
- Oxidation number of Oxygen atom is -2
- There is 1 atom of Ba, 1 atom of S and 4 atoms of O in the compound.
Therefore; assuming oxidation number of S is x
Then, (1× 2) + (1 × x) + ( 4 ×(-2)) = 0
2 + x - 8 = 0
x = 6 or +6
Thus oxidation number of Sulfur in BaSO₄ is +6.
Answer:
The volumes of 1 mol of CO, and 1 mol of N, gases are the same.
Explanation:
Equal volumes of gases have the same number of moles.
