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denpristay [2]
3 years ago
11

How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy

l ether = 26.5 kj/mol?
Chemistry
1 answer:
malfutka [58]3 years ago
5 0

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01  moles of diethyl ether require heat = \frac{26.5}{1}\times 1.01=26.8kJ

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

You might be interested in
determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m
mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) 1s^22s^22p^63s^23p^5

(b) 1s^22s^22p^63s^23p^63d^74s^2

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(d) 1s^22s^22p^63s^23p^63d^4s^1

Answer :

(a) 1s^22s^22p^63s^23p^5   → Halogen

(b) 1s^22s^22p^63s^23p^63d^74s^2    → Transition metal

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6   → Transition metal

(d) 1s^22s^22p^63s^23p^63d^4s^1   → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: ns^2np^6 where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: ns^2np^5 where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: ns^1 where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: ns^2 where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: (n-1)d^{1-10}ns^{0-2} where n is the outermost shell.

(a) 1s^22s^22p^63s^23p^5

The element having this electronic configuration belongs to the halogen family.

(b) 1s^22s^22p^63s^23p^63d^74s^2

The element having this electronic configuration belongs to the transition family.

(c) 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

The element having this electronic configuration belongs to the transition family.

(d) 1s^22s^22p^63s^23p^63d^4s^1

The element having this electronic configuration belongs to the transition family.

4 0
3 years ago
Please help me! Will mark brainliest! Thank you!
Kamila [148]

Answer:

gas

Explanation:

8 0
2 years ago
A student made a copper bracelet by hammering a small copper bar into the desired shape. The bracelet has a mass of 30.1 grams a
Tcecarenko [31]
69. The student's skin was obviously warmer than the initial temperature of the bracelet. According to the second law of thermodynamics, heat transfers from a higher temperature to a lower temperature hence heat transfers from the student's skin to the bracelet warming up the bracelet. 

70. moles of copper=mass of copper/molar mass of copper
=30.1g/(63.55g/mol)= 0.4736 moles Cu

71. Delta H( heat released) = mCp(deltaT) = 30.1g(0.385 J/gK)(33-19)K = 162.23J

72. Copper is chemically less active than iron. 
7 0
3 years ago
How many kilojoules are equivalent to 10 joules?
miss Akunina [59]
1 Kj ------------- 1000 J
?? --------------- 10 J

10 x 1 / 1000 => 0.01 Kj
7 0
3 years ago
You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 2
Katyanochek1 [597]

<u>Answer:</u> The mass of acetic acid used is 0.12 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is CH_3COOH

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL

Putting values in above equation, we get:

1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g

Hence, the mass of acetic acid used is 0.12 grams

5 0
3 years ago
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