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aivan3 [116]
2 years ago
12

What patterns do you notice in your data?

Chemistry
1 answer:
tino4ka555 [31]2 years ago
7 0
My data is all over and inaccurate
You might be interested in
What are valency and radicals
PolarNik [594]
Valency- it means the combing capacity if an element.
<span> radical- it  is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell.
</span>
6 0
3 years ago
Cual es mi nombre ? me pide en mi examen de kinder :(
erastova [34]

Answer:

dude

Explanation:

3 0
3 years ago
Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for ho
nikdorinn [45]
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr] 

HOBr  = 0.50 M 
KOBr  = 0.30 M = OBr-

<span>     HOBr + H2O <-> H+ + OBr- </span>
<span>I     0.50        -              0          0.30 </span>
<span>C       -x                        x             x 
</span>---------------------------------------------
<span>E(0.50-x)                    x       (0.30+x) </span>

<span>Assuming that the value of  x is small as compared to 0.30 and 0.50 </span>

<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>

<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
3 0
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
On x
Lena [83]

Answer:

2

Explanation:

1+1=2

6 0
2 years ago
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