Condensation is the opposite of boiling
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
Explanation:
The given data is as follows.
, 
Work produced per kJ of heat extracted from hot reservoir = 0.45 kJ = Efficiency
If the device is Carnot cycle then its efficiency will be maximum and its value will be equal to ![[1 - (\frac{T_{c}}{T_{h}} )]](https://tex.z-dn.net/?f=%5B1%20-%20%28%5Cfrac%7BT_%7Bc%7D%7D%7BT_%7Bh%7D%7D%20%29%5D)
Using this relation we will calculate the efficiency as follows.
Efficiency =
= 
= 0.928
Hence, it means that this type of device is possible and the claim is also believable.
What items are you separating? Please be more specific with your answer!
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2
