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Komok [63]
3 years ago
7

Describe how you would prepare 350 ml of 0.100 m c12h22o11 starting with 3.00l of 1.50 m c12h22o11

Chemistry
1 answer:
Leona [35]3 years ago
7 0

To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

 

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

 

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

 

 

Steps:

1. Take 23.33 mL of 1.50 M solution

<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M solution</span>

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Biotic factors are living factors in an ecosystem while abiotic factors aren't living

7 0
3 years ago
When ethylbenzene is treated with NBS and irradiated with UV light, two stereoisomeric compounds are obtained in equal amounts.
Stells [14]

Answer: 1-Phenyl ethyl radical is formed as an intermediate in the reaction and since Phenyl ethyl radical has  a trigonal planar geometry so it is a planar molecule having two faces.  So Br radical radical can recombine  with the  two faces with equal probability  leading to a racemic mixture in 50:50ratio of products.Hence two products are formed which are known as enatiomers.

Explanation:

When we irradiate the ethylbenzene with UV light , it leads to homolytic cleavage and 1- Phenyl ethyl free radical is generated.

Phenyl ethyl free radical is generated because it is very stable as it is on a  secondary carbon center as well as on a benzylic position so it can be stabilized by the resonance as well as inductive effect at the secondary carbon center.

NBS(N-bromosuccinimide) is a source of bromine radical and provides bromine free radical.

Once the  1- Phenyl ethyl free radical is generated then bromine free radical can recombine with benzyl free radical leading to product formation.

Since  1- Phenyl ethyl free radical  has a trigonal planar geometry so it is a planar molecule which has two faces  and hence the radical recombination with bromine free radical can occur with either of the two faces available.

Kindly refer the attachments for structure as well as the mechanism of the reaction.

So two isomers which are enantiomers are produced  are obtained.

5 0
3 years ago
Iodine is 80% 127I, 17% 126I, and 3% 128I. Calculate the average atomic mass of Iodine.
trasher [3.6K]
<h3>The average atomic mass of Iodine : 126.86 amu</h3><h3>Further explanation</h3>

Given

80% 127I, 17% 126I, and 3% 128I.

Required

The average atomic mass

Solution

The elements in nature have several types of isotopes

Atomic mass is the average atomic mass of all its isotopes

Mass atom X = mass isotope 1 . % + mass isotope 2.% + ... mass isotope n.%

Atomic mass of Iodine = 0.8 x 127 + 0.17 x 126 + 0.03 x 128

Atomic mass of Iodine = 101.6 + 21.42 + 3.84

Atomic mass of Iodine = 126.86 amu

6 0
3 years ago
A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

8 0
3 years ago
Elements are distinguished from each other by the number of protons present in their nuclei??
Kisachek [45]

False....................

7 0
3 years ago
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