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2 years ago

7

A student wants to prepare a 1.0-liter solution of a specific Molarity. The student determines that the mass of the solute needs

to be 30. grams. What is the proper procedure to follow?
(1) Add 30. g of solute to 1.0 L of solvent.
(2) Add 30. g of solute to 970. mL of solvent to make 1.0 L of solution.
(3) Add 1000. g of solvent to 30. g of solute.
(4) Add enough solvent to 30. g of solute to make 1.0 L of solution.

1 answer:

Mars2501 [29]2 years ago

4 0

The answer is (4) Add enough solvent to 30.0 g of solute to make 1.0 L solution. The molarity is calculated using volume of the solution. When solute dissolving, the total volume will change. So the final volume of solution need to be 1.0 L.

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12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

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1 year ago

Marie mixed 5g of carbon with 5g of lead oxide. she heated the mixture strongly for 15 minutes in a fume cupboard.

olga2289 [7]

The problem talks about two questions and these are:

1. Metals are very good conductors of electricity and heat. Directing heat is easier. So let Marie heat the beads and also have heat another substance, for instance, water. If the beads heat quicker, then they are metals. Another test to conduct is called flame test. This test should give you a colored flame (blue/white for lead) the metal is lead if the reaction is: 2PbO+C ==> 2Pb +CO2

2. The beads are possibly to be lead since Ferrous(lead) oxide + carbon = carbon dioxide + lead

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1 year ago

In a covalently bonded molecule, the number of electrons that an atom shares with others is usually equal to the number of elect

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In the last shell (valence shell)

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2 years ago

An example of a physical property of an element is the element’s ability to(1) react with an acid(2) react with oxygen(3) form a

ivanzaharov [21]

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2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

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

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

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2 years ago