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3 years ago

The following questions are about ethyl bromide,n-propyl bromide, isopropyl bromide,t-butyl bromide, & neopentyl bromide

Chemistry

1 answer:

mrs_skeptik [129]3 years ago

7 0

Answer:

a) Watch the attaccment

b) Ethyl bromide is more reactive than n-propyl bromid, and this more than neopentyl bromide. Ethyl bromide has less steric hindrance than the others, to SN2 reactions.

c) t-butyl bromide is more reactive than isopropyl bromide, and this more than ethyl bromide. t-butyl bromide structure stabilize the carbocation, better than the others.

Explanation:

Speed of SN2 reactions depends on steric hindrance, the less hindrance, the most reaction speed, meaning more reactivity. Then, those linear structures are more reactive to SN2 reactions.

In the other hand, speed of SN1 reactions depends on the stability of the carbocation formed. Structure with ramifications can stabilize better the carbocation, these structures are more reactive to SN1 reactions.

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The solid product from reaction of sulfuric acid with sucrose is?

Mama L [17]

$\huge{\mathbb{\tt {QUESTION:}}}$

The solid product from reaction of sulfuric acid with sucrose is?

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Concentrated <u>sulfuric acid</u> is added to sucrose forming carbon, steam and <u>sulfur</u> dioxide.

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$\huge{\mathbb{\tt {What \: is \: solid?:}}}$

<u>Solid is one of the four fundamental states of matter</u>. The molecules in a solid are closely packed together and contain the least amount of kinetic energy.
<u>A solid is characterized by structural rigidity and resistance to a force applied to the surface</u>.

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<u>Sucrose is common sugar. It is a disaccharide</u>, a molecule composed of two monosaccharides: glucose and fructose. Sucrose is produced naturally in plants, from which table sugar is refined. It has the molecular formula C₁₂H₂₂O₁₁.

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3 years ago

If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield

Zigmanuir [339]

<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

$ActualYield=37.6g\\TheoreticalYield=112.8g$

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

$PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)$

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0

4 years ago

Read 2 more answers

Need help with this question

Liula [17]

Answer:

I think it's c

Explanation:

I hope this helps :)

3 0

3 years ago

Read 2 more answers

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

<u>Answer:</u> The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

<u>For $_{17}^{35}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

<u>For $_{17}^{37}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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3 years ago

If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n)?

vekshin1

If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) keto pentose.

These consist of glycogen, cellulose, as well as starch. Benedict's reagent can be used as a test to see if there are lots of simple carbohydrates present. When it interacts with lowering sugars, it changes from turquoise to yellow or orange. These contain unbound aldehyde but rather ketone groups in simple carbohydrates.

Sugars and starches are examples of carbohydrates. They contain carbon, hydrogen, and oxygen, which appear in the ratio 1:2:1. Size-based categories for carbohydrates include monosaccharides, disaccharides, or polysaccharides. Carbohydrates act as sources of power as their main purpose.

Therefore, If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) keto pentose.

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1 year ago