Answer:
a)there would be no reaction
Explanation:
The activity series of metals has many functions. The one applicable to this problem is that it can be used to determine whether a reaction will occur or not. Also, based on the positions of metals in the series, we can know how reactive a metal is compared to another.
In a single displacement reaction, a metal replaces another metal based on their position on the activity series. Metals that are higher in the series are generally more reactive than others below them and so will displace them.
Would aluminum replace magnesium to form a new compound or would there be no reaction?
Magnesium is higher than aluminum in the activity series. Therefore it is more reactive than aluminum. No reaction will occur.
Hybrid Orbitals: are used to describe the orbitals in covalently bonded atoms sp,sp2,sp3
You can find the hybridization by adding the number of bonded atoms and the number of lone pairs.
For example in BF3. The central atom (B) is bonded to three atoms. So the hybridization is sp2
In NH3, the central atom (N) is bonded to three atoms and has one lone pair. The hybridization is sp3
Answer:
1. Dmitri Mendeleev
2. Johann Dobhereiner
3. John Newlands
4. Henry Moseley
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In
Methyl Amine the Nitrogen atom is attached to two Hydrogen Atoms and one Methyl (CH₃-) group. From
Lewis structure it is shown that the Nitrogen atom also contains a
lone pair of electron.
So,
According to
Valence Shell Electron Pair Repulsion (VSEPR)
Theory, those central atoms which have four electron pairs and all of them are bonding electron pairs gives
Tetrahedral Geometry.
Methylamine also contains four electron pairs, but out of four only
three are bonding pair electrons and
one lone pair electron. Now, due to presence of lone pair of electron (which has
greater repulsion effect than bonding electron pair) the bond angle between H-N-H will decrease from
109° to 107° and results in the formation of <span>
Trigonal Pyramidal geometry as shown below,</span>
Answer and Explanation:
The IUPAC substitutive name of the major product when 2-methylbut-1-ene reacts are as follows
a) HCl = 2-chloro-2-methylbutane
b) HBr in the presence of peroxides = 1-bromo-2-methylbutane
c) H2O/ H2SO4 (dilute sulfuric acid) = 2-methylbutane-1,2-diol
d)BH3-THF followed by basic hydrogen peroxide = 2-methylbutan-1-ol
e) Chlorine in water = 2-chloro-2-methylbut -1 -ol
f) Bromine in carbon tetrachloride = 1,2- dibromo-2-methybutane
g) H2 in the presence of a metal catalys = isopentane or 2- methyl butane