All categories

1 year ago

how did the boiling point of plain water compare to that of water with salt? Compared to water with sugar?

Chemistry

1 answer:

elixir [45]1 year ago

7 0

The boiling point of plain water is less than the boiling point of both salt and sugar water.

<h3>What is boiling point?</h3>

Boiling point can be defined as the point when the pressure exerted by the surroundings upon a liquid is equal to the pressure exerted by the vapour of the liquid.

The boiling point of plain water is 100°C which increases upon addition of solute substances such as salt and sugar.

But salts are usually made up of ionic bonds while sugar are made up of covalent bonds. This means that more energy would be required to boil salt solution due to its ionic bonds.

Therefore, the boiling point of salt water is highest following sugar water before plain water which is the lowest.

Learn more boiling point here:

brainly.com/question/14008526

#SPJ1

You might be interested in

Which represents the empirical formula for a3B9

Viktor [21]

A3B9 represents a molecular formula. The representation of the empirical formula for this compound is AB3. This is so because the empirical formula is the simplest ratio of the atoms present in the molecule. You get AB3 when you divide the subscripts of A3B9, this is 3 and 9, by the greatest common factor, which is 3. 3/3 = 1 and 9/3 = 3, so the subscripts for the empirical formula are 1 and 3, which is what AB3 represents. <span>Answer: AB3.</span>

3 0

2 years ago

Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra

Anit [1.1K]

Answer:

β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal.

Explanation:

When acetaldehyde is treated with dil.NaOH it undergoes self condensation as it contains alpha-hydrogen atom in its compound forming β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal. This compound upon further heating will eliminate a molecule of water forming aldol condensation product namely Crotonaldehyde Or But-2-en-al. see the diagram attached.

5 0

3 years ago

How much (mL) of a 2.0 M sodium hydroxide solution would it take to neutralize 50 mL of a 6.0 M solution of hydrochloric acid?

Tom [10]

Answer:

150ml

Explanation:

For this question,

NaOH completely dissociates. It is a strong base

HCl also completely dissociates. It is a strong acid

So we have this equation

m1v1 = m2v2 ----> equation 1

M2 = 2m

V1= ??

M2 = 6m

V2 = 50m

When we input these into equation 1, we have:

2m x v1 = 6m x 50ml

V1 = 6m x 50ml/2

V1 = 300/2

V1 = 150ml

Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.

Thank you

8 0

2 years ago

What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa

Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

T1 = 30 Degrees or 273.15 + 30 = 303.15 K

P1 = 91 KPa

V2 =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

$\frac{P1VI}{T1}$ = $\frac{P2V2}{T2}$

putting the values in the equation of Gas Law:

V2 = $\frac{P1V1T2}{P2T1}$

V2 = $\frac{91 X 75 X 273.15}{303.15 X 101.3}$

V2 = 60.7 ml

at STP the volume of carbon dioxide gas is 60.7 ml.

4 0

3 years ago

Quick question

IceJOKER [234]

Answer:Um... I think 5000 i am not really sure

Explanation: I Dont Really Know

4 0

3 years ago