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stiv31 [10]
2 years ago
8

What are the properties of seismic waves?

Chemistry
1 answer:
natima [27]2 years ago
6 0

Answer: incompressibility, rigidity, and density

Explanation:

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A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you
Anna71 [15]

Answer:

<h2>1.89 atm</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{1.23 \times 23}{15}  =  \frac{28.29}{15}  \\  = 1.886

We have the final answer as

<h3>1.89 atm</h3>

Hope this helps you

7 0
3 years ago
Help!!!<br> (I'll only give brainliest if it's right)
vivado [14]

Answer:

flavio because you need a source of energy in order for it to have energy

Explanation:

can i have brainliest so i can rank up

5 0
3 years ago
Read 2 more answers
Which one of the following species is diamagnetic: a. F b. Zn2+ C. Cu2+ d. Ca+​
diamong [38]

Answer:

The correct answer to this question is C

Explanation:

3 0
3 years ago
Why didn't you just clean the spill with water?
Allisa [31]

Answer:

if you did it would probably make it bigger...

Explanation:

:)

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3 years ago
What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
Levart [38]

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.

7 0
3 years ago
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