All categories

2 years ago

What ocurrrs when the vapor pressure of a liquid is equal to the external atmospheric pressure

Chemistry

2 answers:

babymother [125]2 years ago

7 0

Answer:

The change from a liquid phase to a gaseous phase.

so the answer would be it changed to a gaseous phase

---------------

This is what occurs when the vapor pressure of the liquid is equal to the atmospheric pressure exerted on the liquid.

ElenaW [278]2 years ago

4 0

Changes to gaseous phase

You might be interested in

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium

lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0

3 years ago

Which has the largest atomic radius calcium bromide barium and astatine

Fantom [35]

The one with the largest atomic radius is barium

Hoped this helped:)

7 0

3 years ago

Read 2 more answers

Question 1: A substance that is soluble in two liquids and makes an emulsion last longer is called what?

Pavlova-9 [17]

<u>Answer:</u>

1. A substance that is soluble in two liquids and makes an emulsion last longer is called "Emulsifier".

2. The process that reduces the size of particles so emulsions will last longer is called "Homogenization".

<u>Explanation:</u>

Emulsifiers are additives which enable two liquids to mix around each other. Water and oil separate in a container, for an instance, but using an emulsifier can make the liquids blend along. It is widely used on various foods and beverages. Egg yolks and mustard are a few examples of emulsifiers.

Homogenization is the physical mechanism by which the fat molecules in milk are broken down because then they stay incorporated instead of segregated as cream. Majority of the milk sold in the United States is homogenized.

3 0

3 years ago

Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0.2

Allushta [10]

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

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse.

The chemical reaction is:

$XY(g)\rightleftharpoons X(g)+Y(g)$

Concentration(M) [XY] [X] [Y]

(M)initial: 0.200 0.300 0.300

change: +x -x -x

equilibrium: 0.200+x 0.300-x 0.300-x

The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a Kc value of 0.140 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answer : The Equilibrium concentrations of XY, X, and Y is, 0.104 M, 0.204 M and 0.204 M respectively.

Explanation :

The chemical reaction is:

$XY(g)\rightleftharpoons X(g)+Y(g)$

initial: 0.200 0.300 0.300

change: +x -x -x

equilibrium: (0.200+x) (0.300-x) (0.300-x)

The equilibrium constant expression will be:

$K_c=\farc{[X][Y]}{[XY]}$

Now put all the given values in this expression, we get:

$0.140=\frac{(0.300-x)\times (0.300-x)}{(0.200+x)}$

By solving the term 'x', we get:

x = 0.0963 and x = 0.644

We are neglecting the value of x = 0.644 because the equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.0963

Equilibrium concentrations of XY = 0.200+x = 0.200+0.0963 = 0.104 M

Equilibrium concentrations of X = 0.300-x = 0.300-0.0963 = 0.204 M

6 0

3 years ago

What three compounds are produced by the combustion of hydrocarbon fuels?

MakcuM [25]

The most common hydrogen carbon fuels are ethanol and diesel and their product of combustion is carbon dioxide, water and heat .

8 0

3 years ago