Answer:
26.67 mol HCl
Explanation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.
To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:
- 8.89 mol Al(OH)₃ *
= 26.67 mol HCl
Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
Answer:
B
Explanation:
I looked it up and found the answer lol
<span>false - sodium is not a member of the transition elements, however </span><span>copper is a </span><span>member of the transition elements.</span>
