Answer:
explanations below
Explanation:
The alkaline earth metals are the 6 chemical elements that can be found in group two of the periodic table. These elements have a lot of properties in common, in the sense that they are generally shiny, reactive at standard temperature and pressure and they are also silvery-white. The elements are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra)
These metals have two electrons at their orbits, which can be easily lost to form cations (with charge +2, and an oxidation state of +2)
Below are some of their details
Name: Beryllium
Number of Protons/Electrons: 4
Number of Neutrons: 5
Name: Magnesium
Number of Protons/Electrons: 12
Number of Neutrons: 12
Name: Calcium
Number of Protons/Electrons: 20
Number of Neutrons: 20
Name: Strontium
Number of Protons/Electrons: 38
Number of Neutrons: 50
Name: Barium
Number of Protons/Electrons: 56
Number of Neutrons: 81
Name: Radium
Number of Protons/Electrons: 88
Number of Neutrons: 138
Answer:
96.32 %
Explanation:
Given that:
The solubility of compound in hot water = 4.35 g / 100 mL
The solubility of compound in cold water = 0.16 g / 100 mL
Which means that in 100 mL of hot water, the dissolved compound is 4.35 g and in cold water, the dissolved compound is 0.16 g
Hence, on transition, compound that will catalyze is 4.35 - 0.16 g = 4.19 g
So,
Percent recovery for re-crystallization of this compound from water= 96.32 %
name= calcium
atomic mass= 40.078 atomic mass unit
no of protons= 20
no of electrons=20
Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.
A) Reaction with NaI :
Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .
The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)
NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.
1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane
The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)
B) Reaction with AgNO3 :
Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.
AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )
The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.
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