mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
Answer:
Thecinically it is both because and an un opened bottle of soda is homogeneous and an opened bottle of soda is herterogeneous. I hope this helps please reward branlist.
Answer:
0.25L
Explanation:
Using the dilution formula
C1V1=C2V2
C1=6M
V1?
C2=0.75M
V2=2.0L
V1= C2V2/C1
V1=0.75*2.0/6
V1=0.25L
Answer:
1) synthesis MgI2
2) double replacement CuS + (HCl)2
3) double replacement, not sure ab the formula sorry
English please i cant understand
