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3 years ago

The boiling point of CH3OCH3 would be________ the boiling point of CH3CH2OH.

Chemistry

2 answers:

PolarNik [594]3 years ago

7 0

The ans is c

Hope it’s help

AlexFokin [52]3 years ago

7 0

Answer:

A. Lower than

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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago

The boiling point of bromine is 59 °C. Which of the following best predicts the boiling point of iodine monochloride, a polar co

soldier1979 [14.2K]

Answer:

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Explanation:

I just took the test and i got it right

6 0

4 years ago

Read 2 more answers

To name the compound written as cucl2 you would write\

MA_775_DIABLO [31]

Copper chloride ...............

5 0

3 years ago

What mass of H₂ is needed to react with 8.75 g of O₂ according to the following equation: O2(g) + H2(g) → H₂O(g)?

FromTheMoon [43]

Explanation:

For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

The given balanced equation has been:

\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O

From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.

The mass of oxygen has been: 8.75 g,

Moles = \rm \dfrac{weight}{molecular\;weight}molecularweightweight

Moles of oxygen = \rm \dfrac{8.75}{32}328.75

Moles of oxygen = 0.27 mol

Since,

1 mole Oxygen = 2 mole hydrogen

0.21 mol oxygen = 0.54 mol hydrogen

Mass of hydrogen = moles \times× molecular weight

Mass of hydrogen = 0.54 \times× 2

Mass of hydrogen = 1.08 grams.

Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

6 0

2 years ago

WILL GIVE BRAINLIEST!

Oliga [24]

Answer:

Explanation:

Increasing the the temperature would favour the endothermic reaction which is the forward direction however increasing the pressure would make the reaction try to counteract this change by favouring the reaction that would create more products so the equilibrium will shift left instead of right.

Hope this helps.

4 0

2 years ago