Given:
<span>CS2 + 3O2 → CO2 + 2SO2
</span><span>114 grams of CS2 are burned in an excess of O2
</span>
moles CS2 = 114 g/76.143 g/mol → 114g * mol/76.143 g = 1.497 mol
<span>the ratio between CS2 and SO2 is 1 : 2 </span>
moles SO2 formed = 1.497 x 2 = 2.994 moles → 2nd option
This is because copper is a transition metals.
Transition metals don't have a definite charge, you can see this on the periodic table. However, Alkali Earth, Alkali, Metalloids, Non-metals, Halogens, Noble Gases, etc, all have a specified charge.
Although, there are exceptions: Zinc, Gold, and Cadmium.
Aluminum is a poor metal, but it has it's own charge (3+).
Therefore, Copper has roman numerals due to the fact that it's a transition metal and Aluminum is a poor metal.
Hope this helps!
I think it becomes an Ion. If i'm not mistaking.
Radioactivity is produced when unstable nuclei B. Break apart.
The reaction of benzene with an alkyl chloride and aluminum trichloride is called a Friedel-Crafts alkylation. The aluminum trichloride is a lewis acid that reacts with the alkyl chloride, and this converts the alkyl group into a very good electrophile. The overall reaction is an electrophilic aromatic substitution. A pi bond of the benzene ring will attack the electrophile as a nucleophile and form a new carbon-carbon bond. The result is a monosubstituted isopropyl group on the benzene ring.
However, we are told that the conditions are set up to encourage dialkylation. The isopropyl group is a group that donates electron density through induction. Therefore, it is an ortho- and para-directing group. This means that once one isopropyl group is substituted on to the benzene ring, the second substitution will be favoured in the ortho- and para-positions. In this instance, the major product is the 1,4-disubstituted product. The reason being is that the isopropyl groups are very bulky, and the steric hindrance of having two isopropyl groups adjacent leads to a small amount of this product being formed.
There will also be trace amounts of the meta-substituted product, but the 1,4- and 1,2-substituted products are more favoured.