Answer:
acid
Explanation:
Methyl acetate, also known as MeOAc, acetic acid methyl ester or methyl ethanoate, is a carboxylate ester with the formula CH3COOCH3. It is a flammable liquid with a characteristically pleasant smell reminiscent of some glues and nail polish removers.
The answer is A, between 0 and 7.
In a pH scale from 0 to 14, we can groups these numbers into acidic, neutral, and alkaline. 7 is the neutral pH value, therefore, 0-7 is always acidic, and 7-14 is alkaline.
The smaller the number is, the more acidic the solution will be. This applies same in alkalis, the larger the pH value is, the more alkaline the solution is.
We can measure the pH of solution with many methods, the easiest way include using a pH paper, more advanced and accurate methods includes using a pH meter.
Answer: The sequence from longest wavelength (radio waves) to shortest wavelength (gamma rays) is also a sequence in energy from lowest energy to highest energy. ... The energy carried by a radio wave is low, while the energy carried by a gamma ray is high. Different materials can block different types of light.
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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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