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9966 [12]
3 years ago
7

While the carbonyl stretching frequency for simple aldehydes, ketones, and carboxylic acids is about 1710 cm-1, the carbonyl str

etching frequency for acid chlorides is about ________.
A) 1800 cm-1
B) 1660 cm-1
C) 2200 cm-1
D) 1735 cm-1
E) 1700 cm-1
Chemistry
1 answer:
Virty [35]3 years ago
5 0

Answer:

A

Explanation:

Acid chloride: --C=O-Cl. Stretch range ~1750-1820cm-1. I'd probably pick

(A) 1800 cm-1, as this value is in the middle (-ish) of that range.

Hence the correct answer is 1800 cm^-1

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What is the common name of the following compound
Masja [62]
Hey there!

Here is your answer:

<u><em>The proper answer to this question is option D "sec-butyl bromide".</em></u>

Reason:

<u><em>S</em></u><span><u><em>ec-butyl bromide looks often like this:</em></u>

<em>Therefore the answer is option D!</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
</span>

4 0
3 years ago
Read 2 more answers
Which of the following is a balanced chemical equation?
Artemon [7]
`2H + 02  ----->  2H2O is a Balanced chemical equation. 


WATCH THESE FOR A BETTER UNDERSTANDING :
https://www.youtube.com/watch?v=yA3TZJ2em6g

https://www.youtube.com/watch?v=eNsVaUCzvLA
5 0
3 years ago
Read 2 more answers
What is the pH of a solution with a 4.60 × 10−4 M hydroxide ion concentration?
guapka [62]

Answer:

pH

=

10.66

Explanation:

For pure water at  

25

∘

C

, the concentration of hydronium ions,  

H

3

O

+

, is equal to the concentration of hydroxide ions,  

OH

−

.

More specifically, water undergoes a self-ionization reaction that results in the formation of equal concentrations of hydronium and hydroxide anions.

2

H

2

O

(l]

⇌

H

3

O

+

(aq]

+

OH

−

(aq]

At room temperature, the self-ionization constant of water is equal to

K

W

=

[

H

3

O

+

]

⋅

[

OH

−

]

=

10

−

14

This means that neutral water at this temperature will have

[

H

3

O

+

]

=

[

OH

−

]

=

10

−

7

M

As you know, pH and pOH are defined as

pH

=

−

log

(

[

H

3

O

+

]

)

pOH

=

−

log

(

[

OH

−

]

)

and have the following relationship

pH

+

pOH

=

14

In your case, the concentration of hydroxide ions is bigger than  

10

−

7

M

, which tells you that you're dealing with a basic solution and that you can expect the pH of the water to be higher than  

7

.

A pH equal to  

7

is characteristic of a neutral aqueous solution at room temperature.

So, you can use the given concentration of hydroxide ions to determine the pOH of the solution first

pOH

=

−

log

(

4.62

⋅

10

−

4

)

=

3.34

This means that the solution's pH will be

pH

=

14

−

pOH

pH

=

14

−

3.34

=

10.66

Indeed, the pH is higher than  

7

, which confirms that you're dealing with a basic solution.

4 0
4 years ago
Read 2 more answers
How much 0.02 m kmno4 solution should be needed if the solutions tested have a composition of 3% h2o2 by mass?
Nostrana [21]
When the concentration is expressed in mass percentage, that means there is 3 g of solvent H₂O₂ in 100 grams of the solution. Then, that means the remaining amount of solute is 97 g. We use the value of molarity (moles/liters) to determine the amount of solution in liters, denoted as V. The solution is as follows:

0.02 mol KMnO4/L solution * 158.034 g KMnO4/mol * V = 97 g KMnO4

Solving for V,
V = 30.69 L
6 0
3 years ago
When the u-235 nucleus is struck with a neutron, the ce-144 and sr-90 nuclei are produced, along with some neutrons and beta par
butalik [34]
Answer: 2 (2 neutrons are produced).

Explanation:

1) In the left side of the transmutation equationa appears:
²³⁵U + ¹n →

I am omitting the atomic number (subscript to the leff) because the question does not show them as it is focused on number of neutrons.


2) The right side of the transmutation equation has:

→ ¹⁴⁴Ce + ⁹⁰Sr + ?

3) The total mass number of the left side is 235 + 1 = 236

4) The total mass number of Ce and Sr on the right side is 144 + 90 = 234

5) Then, you are lacking 236 - 234 = 2 unit masses on the right side which are the 2 neutrons that are produced along with the Ce and Sr.

The complete final equation is:

²³⁵U + ¹n → ¹⁴⁴Ce + ⁹⁰Sr + 2 ¹n

Where you have the two neutrons produced.
6 0
4 years ago
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