CHEMISTRY HIGH SCHOOL

A chemist combined chloroform ( CHCl 3 ) and acetone ( C 3 H 6 O ) to create a solution where the mole fraction of chloroform, χ chloroform , is 0.203 . The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL , respectively. Calculate the molarity of the solution.

Answers

Answer 1

Answer:

Answer:

Molarity is 13,4M

Explanation:

As mole fraction of chloroform is 0,203; mole fraction of acetone will be 1-0,203=0,797

That means that per 100 moles you have 20,3 moles of chloroform and 79,7 moles of acetone.

Using molar mass and density it is possible to know the volume these moles occupy, thus:

Chloroform: 20,3 moles× $(119,38g)/(1mol)$ × $(1mL)/(1,48g)$ = 1637mL= 1,637L

Acetone: 79,7 moles× $(58,08g)/(1mol)$ × $(1mL)/(0,791g)$ = 5852mL = 5,852L

That means that total volume is 1,637L + 5,852L = 7,489L

As moles are 100, molarity is:

100mol / 7,489L = 13,4M

I hope it helps!

Answer 2

Answer:

Final answer:

To determine the molarity of the solution, calculate the number of moles of each compound in the solution, and then divide the total moles by the total volume of the solution. Ensure all units are in the appropriate metric system and take into consideration potential effects of molecular interactions on the total volume.

Explanation:

The first step to calculate the molarity of the solution of chloroform (CHCl3) and acetone (C3H6O) is to determine the number of moles of each compound in the solution. The mole fraction is given as 0.203 for chloroform, therefore the mole fraction for acetone will be 0.797 (since the total mole fraction in a solution is equal to 1).

Next, we calculate the mass of each compound with the known densities. Upon performing these calculations, we then utilise the definition of molarity, which is moles of solute per liter of solution. This can be calculated by knowing the total moles and total volume of the solution. Note that we'd convert all units to the appropriate metric system before carrying out these calculations (density in g/mL, molar mass in g/mol, etc.).

It is also important to consider that the mass and volume of each solution component might not be directly additive due to potential molecular interactions between acetone and chloroform.

Learn more about Molarity here:

brainly.com/question/8732513

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125 cm3 of a solution contains 0.25 moles of the solute. What is the concentration of the solution?

Answers

Answer:

Explanation:

125 cm3 -> 0.125 L

M = no. mole / volume (L)

M = 0.25 / 0.125 = 2 M

The alkali metals are

Answers

Answer:

They are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr).

Hope this helps :)

Answer:

Groups one of the periodic table.

Lithium Li

Sodium Na

Potassium K

Rubidium Rb

Cesium Cs

Francium Fr

A substance that speeds up the rate of a chemical reaction is calleda. a catalyst.
b. a lipid.
c. a molecule.
d. an element.

Answers

I'm not going toblie I'm stuck with A and B.But to me my gut says A

A substance that speeds up the rate of a chemical reaction is called a catalyst. As a catalyst speeds up the reaction, as an enzyme does.

An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. This is the equation: CO(g)+H2O(g)--><-- CO2(g) + H2(g). How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answers

Answer : The moles of $CO_2$ added will be 1.12 mole.

Solution : Given,

Moles of $CO$ and $H_2O$ at equilibrium = 0.200 mol

Moles of $CO_2$ and $H_2$ at equilibrium = 0.600 mol

First we have to calculate the concentration of $CO,H_2O,CO_2\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CO=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }H_2O=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }CO_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

$\text{Concentration of }H_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$K_c=((0.600)* (0.600))/((0.200)* (0.200))$

$K_c=9$

Now we have to calculate the moles of $CO_2$ added.

Let the moles of $CO_2$ added is 'x'.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initially 0.200 0.200 0.600 0.600

Added moles 0 0 x 0

Change +0.1 +0.1 -0.1 -0.1

Final 0.3 0.3 (0.5+x) 0.5

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$9=((0.5)* (0.5+x))/((0.3)* (0.3))$

$x=1.12mol$

Therefore, the moles of $CO_2$ added will be 1.12 mole.

$\boxed{0.{\text{3 mol}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ are added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol.

Further Explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}} + {\text{Q(g)}} \rightleftharpoons {\text{R(g)}} + {\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}}=\frac{{\left[ {\text{R}}\right]\left[ {\text{S}}\right]}}{{\left[{\text{P}} \right]\left[ {\text{Q}} \right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

${\text{CO}}\left(g\right) + {{\text{H}}_2}{\text{O}}\left( g \right)\rightleftharpoons {\text{C}}{{\text{O}}_2}\left( g \right) + {{\text{H}}_2}\left( g \right)$

The expression for the equilibrium constant for the given reaction is as follows:

${\text{K = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}}\right]}}$ ......(1)

Here,

K is the equilibrium constant.

$\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ is the concentration of carbon dioxide.

$\left[{{{\text{H}}_{\text{2}}}} \right]$ is the concentration of hydrogen.

$\left[ {{\text{CO}}}\right]$ is the concentration of carbon monoxide.

$\left[ {{{\text{H}}_2}{\text{O}}}\right]$ is the concentration of water.

Substitute 0.600 mol/L for $\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ , 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}} \right]$ , 0.200 mol/L for $\left[ {{\text{CO}}}\right]$ and 0.200 mol/L for $\left[ {{{\text{H}}_2}{\text{O}}} \right]$ in equation (1).

$\begin{aligned}{\text{K }}&=\frac{{\left( {{\text{0}}{\text{.600 mol/L}}}\right)\left( {{\text{0}}{\text{.600 mol/L}}}\right)}}{{\left( {{\text{0}}{\text{.200 mol/L}}}\right)\left( {{\text{0}}{\text{.200 mol/L}}}\right)}}\n&= 9\n\end{aligned}$

The value of equilibrium constant comes out to be 9 and it remains the same for the given reaction.

Rearrange equation (1) to calculate .

$\left[{{\text{C}}{{\text{O}}_{\text{2}}}} \right]=\frac{{{\text{K}}\left( {\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}} \right]} \right)}}{{\left[ {{{\text{H}}_{\text{2}}}} \right]}}$ ......(2)

Substitute 9 for K, 0.300 mol/L for $\left[{{\text{CO}}}\right]$ , 0.200 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ and 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}}\right]$ in equation (2).

$\begin{aligned}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]&=\frac{{{\text{9}}\left( {{\text{0}}{\text{.300 mol/L}}} \right)\left( {{\text{0}}{\text{.200 mol/L}}} \right)}}{{{\text{0}}{\text{.600 mol/L}}}}\n&= 0.{\text{9 mol/L}}\n\end{aligned}$

Initially, 0.6 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ were present in a 1-L container. But now 0.9 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ are present in it. So the extra amount of ${\text{C}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of C}}{{\text{O}}_{\text{2}}}{\text{ added}} &= 0.{\text{9 mol}} - 0.{\text{6 mol}}\n&= 0.{\text{3 mol}}\n\end{aligned}$

Therefore 0.3 moles of carbon dioxide are added in a 1-L container.

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: CO, H2, CO2, H2O, 0.9 mol/L, 0.2 mol/L, 0.6 mol/L, 0.3 mol/L, K, carbon dioxide, water, hydrogen, carbon monoxide.

An atom with an equal number of electrons and protons has no _________.A. reactivity
B. weight
C. charge

Answers

The answer would be C. charge because the protons, which are positively charged particles, and the electrons, negatively charged particles, balance each other out and the atom becomes of a neutral charge (or no charge)

.Element Z is a neutral element and has 39 protons. What element is it and how many electrons does it have? (Use the periodic table)

Platinum – 78

Yttrium – 39

Argon – 18

There is not enough information given to determine.

Answers

Answer:

Yttrium – 39

Explanation:

Since the element is neutral and has 39 protons s the atomic number of the element must be 39. Yttrium is the only element in the periodic table with atomic number 39.

Atomic number of an element is equal to the number of protons which is equal to the number of electrons. Since the element is neutral, the number of electrons in Yttrium is also 39.

If the element have 39 protons, that means that his atomic number is 39, being Yttrium the element. And, the number of protons is the same of the number of electrons, so, 39 electrons.

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