The golden rule is to head for cover.
Electrophiles are reagents attracted to electrons.
Electrophiles tend to be electron-deficient and carry partial positive charges. They are attracted to species with lone pairs of electrons. For example, protons
have no electrons and tend to share ones with other species, hence behaving as electrophiles in aqueous reactions. In the reaction between
and ammonia
, protons would be attracted to lone electron pairs on nitrogen atoms in ammonia molecules, which carry partial positive charges.
The Lewis Acid-base theory define Acids as species that accept electron pairs in a particular acid-base reaction. Electrophiles, by definition, tend to accept electrons. Lewis acids thus behaves as electrophiles in acid-base reactions. In the previous example,
demonstrates acidic behavior and can be inferred as an electrophile.
<span>1 mole of calcium carbonate reacts with 1 mole of sulfuric acid and produces 1 mole of calcium sulfate.
3.1660 g of CaCO3 is how many moles of calcium carbonate? 3.1660 / 100.0869 = 0.031633 moles.
3.2900 g of H2S04 is how many moles of sulfuric acid? 3.2900 / 98.079 = 0.033544 moles.
</span><span>The lesser of the two is 0.031633 moles.
Therefore, 0.031633 moles of calcium carbonate will combine with 0.031633 moles of sulfuric acid to produce 0.031633 moles of calcium sulfate.
Molecular weight of calcium sulfate is 136.14 g/mol.
Therefore, 0.031633 moles of calcium sulfate will weight 0.031633 x 136.14 g/mol = 4.3065 grams.</span>
It would be carbon dioxide because carbon dioxide makes up the vast majority of greenhouse gas .
Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.