When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:
(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃
Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Answer:
96%
Explanation:
To find the percent yield, we can use this equation
The actual yield of aspirin is 3.0 and the theoretical is 3.14 in this case, so just plug the numbers in.
Thus the percent yield is 96%
;)
Answer:
oh it's easy
Explanation:
Take the hydrate
N
a
2
S
2
O
3
∙
5
H
2
O
. Are there ionic forces between the
N
a
+
and the
S
2
O
2
−
3
and ion-dipole forces between the cation/anions and the water?
