The question is incomplete, the complete question is;
AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —
aluminum bromide
monoaluminum tribromide
aluminide bromine
aluminum tribromide
Answer:
aluminum bromide
Explanation:
Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.
We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.
Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.
Answer:
4feet
Explanation:
Because when he moves he only goes 4 miles every time I
Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
If the grade of the ore is 37.3% nickel, then the unknown quantity to get 10 grams of nickel is 0.373 x = 10 grams or x = 10/0.373=26.8 grams or 0.0268 kg needed to dig up to recover the 10 grams of nickel. At this grade of ore, 1 kilogram would yield 373 grams of nickel.
