The question is incomplete, the complete question is;
AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —
aluminum bromide
monoaluminum tribromide
aluminide bromine
aluminum tribromide
Answer:
aluminum bromide
Explanation:
Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.
We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.
Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.
Answer:
16.46 g.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).</em>
- It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).
- We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.
<u><em>Using cross multiplication:</em></u>
2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.
0.503 mole of Cu produces → ??? mole of Zn.
- The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.
∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.
Answer:
It is true answer. Ozone = O3
Answer:
There are 0.5 mole in 20g of argon.
Explanation:
40 g of argon = 1mole
Then 20g of argon is,
→ 1/40 × 20
→ 0.5 mole
9 amino acids (alanine, cysteine, glycine, isoleucine, leucine, methionine, phenylalanine, proline, valine) have no hydrogen donor or acceptor atoms in their side chains.
