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3 years ago

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

2. Add 326.67 mL of water to make 350 mL of 0.100 M solution

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How are biotic factors different from abiotic factors?

Gala2k [10]

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 reaction between HCl and KOH:

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

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

$M=\frac{mol}{L}$

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 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract 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 concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

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