Answer:
5.4 M.
Explanation:
- At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.
<em>(MV)acid = (MV)NaOH</em>
M of acid = ??? M, V of acid = 35.0 mL.
M of NaOH = 3.0 M, V of NaOH = 63.0 mL.
∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.
Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
Answer: ribosomes and membranes.
Explanation:
2 NH3 -> 1 N2 + 3 H2
Explanation:
That would be the answer to this
