What is the ratio of moles of sulfur to moles of sulfur trioxide in 25 +302-->2SO3
1 answer:
You might be interested in
<h3>Answer:</h3>
The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.
<h3>Solution and Explanation:</h3>- To find how fast molecules or particles of gases move at a particular temperature, the root mean square speed is calculated.
- Root mean square speed of a gas is calculated by using the formula;
Where R is the molar gas constant, T is the temperature and M is the molar mass of gas in Kg.
<h3>Step 1: Root mean square speed from O₂</h3>Molar mass of Oxygen is 32.0 g/mol or 0.032 kg/mol
Temperature = 65 degrees Celsius or 338 K
Molar gas constant = 8.3145 J/k.mol
The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol
Therefore; the root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.
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.
