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:
Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment, in which he demonstrated that the atom has a tiny, high- mass nucleus. In his experiment, Rutherford observed that many alpha particles were deflected at small angles while others were reflected back to the alpha source.
Answer:
The answer is B. Van der Waals forces are weaker than ionic and covalent bonds.
Explanation:
In general, if we arrange these molecular forces from the strongest to weakest, it would be like this:
Covalent bonds > Ionic bonds > Hydrogen bonds > Dipole-Dipole Interactions > Van der Waals forces
Covalent bonds are known to have the strongest and most stable bonds since they go deep and into the inter-molecular state. A diamond is an example of a compound with this characteristic bond.
Ionic bonds are the next strongest molecular bond following covalent bonds. This is due to the protons and electrons causing an electro-static force which results to the strong bonds. An example would be Sodium Chloride (NaCl), which when separated is Na⁺ and Cl⁻.
Van der Waals forces, also known as Dispersion forces, are the weakest type of molecular bonds. They are only formed through residual molecular attractions when molecules pass by each other. It doesn't even last long due to the uneven electron dispersion. It can be made stronger by adding more electrons in the molecule. This kind of molecular bonds appear in non-polar molecules such as carbon dioxide.
HOPE THIS HELPS!!!!!!!!!!!!!!
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Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
A metalloid is a metal and a nonmetal
