HELP ASAP PLEASAE!!!

At what temperature are the liquid and the vapour of Bromine in equilibrium (e.g. boiling point)?

irina [24]

Explanation:

Boiling point is defined as the point at which liquid state and vapor state of a substance are existing in equilibrium.

Equilibrium is defined as the state in which rate of forward and rate of backward reaction are equal to each other.

For example, $Br(l) \rightleftharpoons Br(g)$

So, when we boil bromine which is present in liquid state then at the boiling point its vapors will exist in equilibrium. And unless all the liquid state of bromine will not convert into vapors its temperature will not change.

Therefore, we can conclude that at boiling point the liquid and the vapur of Bromine are in equilibrium.

7 0

3 years ago

What Phase is water at 120 degrees Celsius?

Rina8888 [55]

At atmospheric pressure water boils at 100 degC.
So the water would be gas/vapor/steam.

4 0

3 years ago

How does the sun compare to other stars

Reptile [31]

By Our sun is a bright, hot ball of hydrogen and helium at the center of our solar system.

5 0

3 years ago

Read 2 more answers

Draw the structures of organic compounds a and b omit all by products thf

Brums [2.3K]

The reaction is shown below,

Step 1: Hydration of Alkene:
In first step Ethene is hydrated to Ethanol through Hydroboration Reaction.

Step 2: Oxidation of Primary Alcohol:
In this step partial oxidation of ethanol is carried out using mild Oxidizing agent Pyridinium chlorochromate (PCC) and Acetaldehyde is produced.

Step 3: Reduction of Acetaldehyde followed by Oxidation:
In this step, first acetaldehyde is reduced to secondary alcohol using grignard reagent. After that the sec. alcohol is oxidized to ketone by using oxidizing agent CrO₃.

6 0

3 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$

Mass of 44 moles of water ;

$=44mol\times 18g/mol=792g$

792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

brainly.com/question/7144022?referrer=searchResults

3 0

3 years ago