?? Is that the whole question?
0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
Answer:
6
Explanation:
The coordination number of a particular crystal structure arrangement is defined as the number of nearest neighbor atoms or ions surrounding an atom or ion.
We must note that in salts of the type MX, the coordination number of each ion must be equal. The structure of LiCl is built by placing each unit cell next to another such that ions residing in the corner, edge or face sites are shared between adjacent unit cells. The LiCl is in FCC arrangement.
Hence in LiCl, each Li+ ion is surrounded by 6 chloride ions.
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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