Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO
1 answer:
Answer: There are atoms of hydrogen are present in 40g of urea,
.
Explanation:
Given: Mass of urea = 40 g
Number of moles is the mass of substance divided by its molar mass.
First, moles of urea (molar mass = 60 g/mol) are calculated as follows.
According to the mole concept, 1 mole of every substance contains atoms.
So, the number of atoms present in 0.67 moles are as follows.
In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.
Thus, we can conclude that there are atoms of hydrogen are present in 40g of urea,
.
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Answer:
1. Diastereomers have different physical properties (unlike most aspects of enantiomers) and often different chemical reactivity. ... Many conformational isomers are diastereomers as well. Diastereoselectivity is the preference for the formation of one or more than one diastereomer over the other in an organic reaction.
2. The result is a trans dibromide, as shown in the equation below
Explanation:
Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer.[1] Diasteoreomers are defined as non-mirror image non-identical stereoisomers. Hence, they occur when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocenters and are not mirror images of each other.[2] When two diastereoisomers differ from each other at only one stereocenter they are epimers. Each stereocenter gives rise to two different configurations and thus typically increases the number of stereoisomers by a factor of two.
2. the addition of bromine to the trans and
cis isomers of 1,2-diphenylethene, more commonly known as trans- and cis-stilbene.
H
H
H H
trans-stilbene cis-stilbene
m.p. 122-124°C b.p. 82-84°C
density 0.970 g/mL density 1.011 g/mL
M.W. 180.25 g/mol M.W. 180.25 g/mol
In both cases, the nucleophilic double bond undergoes an electrophilic addition reaction
by the bromine reagent which proceeds via a cyclic bromonium ion. The addition of
bromine begins at one side of the double bond (either side is equally likely, but only one
option is drawn) and is followed by attack of bromide ion on the bromonium ion (again,
attack could occur at either carbon since the ion is symmetric, but only one option is
drawn). The result is a trans dibromide, as shown in the equation below:
Since the cis and trans isomers of stilbene have different geometries, it follows
that upon reaction with bromine they give rise to stereoisomeric bromonium ions and,
eventually, products that differ only by their stereochemistry.
Answer : The correct option is, (2) Cr (Chromium)
Explanation :
The reactivity series of metal are arranged of the reactivity from the highest to the lowest. Reactivity series is used to determine the products of the single displacement reactions. In the single displacement reaction, the most reactive metal displaces the least reactive metal.
From the given reactivity series we conclude that there are two metal (Mg and Cr) are more reactive metal than the Ni and there are two metal (Pb and Cr) are less reactive metal than the Zn. So, the Cr (Chromium) is the metal which is more active than Ni and less active than Zn.
Hence, the correct option is, (2) Cr
