An aqueous solution is 4.44 M nitric acid and the density of the solution is 1.42 g/mL. Calculate the mole fraction of this solution.

Answers

Answer 1
Answer:

The mole fraction of HNO3 is  0.225

Explanation:

1.Given data

Density = 1.429 /ml

Mass% = 63.01 g HNO3 / 100g of solution

The mass of 63.01 g is in 100 / 1.142 /ml of solution

Or 63.01 g in 55.7 mL

Molarity = 15.39 moles / L

Mass of water in 100g = 100 - 63.01=36.99 g

So 63.01 grams in 36.99 grams of water

So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703

Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles

Molality = 27.03 molal (mole / Kg)

Mole fraction = Mole of HN03 / Moles of water + mole of HNO3

Mole of water = 62/ 18 = 3.44

Moles of HNO3 = 63.01 / 63.01 = 1.000

Mole fraction = 1.000 / 3.44 + 1.000 = 0.225

The mole fraction of HNO3 is  0.225


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Identify the combustion reaction. identify the combustion reaction. c(s) + o2(g) → co2(g) 2 h2(g) + o2(g) → 2 h2o(g) c3h8(g) + 5 o2(g) → 3 co2(g) + 4 h2o(g) 2 c3h7oh(l) + 9 o2(g) → 6 co2(g) + 8 h2o(g) all of the reactions are examples of combustion reactions.

Answers

Combustion reaction occurs when organic compound reacts with oxygen to form CO₂, H₂O and energy
C(s) + O₂(g) → CO₂(g) is Synthesis reaction not combustion
2 H₂(g) + O₂(g) → 2 H₂O(g) also Synthesis reaction and not combustion
C₃H₈(g) + 5 O₂ → 3 CO₂(g) + 4 H₂O(g) is considered as Combustion reaction
2 C₃H₇OH(l) + 9 O₂(g) → 6 CO₂(g) + 8 H₂O(g) Combustion reaction

Answer:

All of the reactions are examples of combustion reactions.

Explanation:

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.

Answers

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_(solute) = Given mass of solute (CsCl) = 8.00 g

M_(solute) = Molar mass of solute (CsCl) = 168.4  g/mol

W_(solvent) = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2* 0.51^oC/m* (8.00* 1000)/(168.4g/mol* 92)\n\n\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

2 SO2 (g) + O2 (g) 2 SO3 (g)If the TEMPERATURE on the equilibrium system is suddenly increased :The value of Kc A. IncreasesB. DecreasesC. Remains the sameThe value of Qc A. Is greater than KcB. Is equal to KcC. Is less than KcThe reaction must: A. Run in the forward direction to restablish equilibrium.B. Run in the reverse direction to restablish equilibrium.C. Remain the same. Already at equilibrium.The concentration of O2 will: A. Increase.B. Decrease.C. Remain the same.

Answers

Answer:

Part one: B. Kc decreases

Part two: B. Is equal to Kc

Part three: B. Run in the reverse direction to reestablish equilibrium

Part four: A. Increase

Explanation:

Part one: Sulfur dioxide combines with oxygen to form sulphur trioxide in an exothermic reaction. If the temperature is suddenly is increased, while the reaction is at equilibrium, the backward reaction (the endothermic one) is favored to "sweep up the excess heat". An increase in reactants means a decrease in Kc since the denominator(reactants) is becoming bigger while the numerator (products) become smaller.

Part two: Qc is a varying version of Kc. For this set of circumstances, it will be equal to Kc since Kc varies with temperature

Part three: The reaction must run in the reverse to reestablish the equilibrium.

Part four: The concentration of of oxygen will increase as more of the reactants are formed

Final answer:

The increase in temperature for this exothermic reaction will cause the value of Kc to decrease, the value of Qc to be greater than Kc, the reaction to run in the reverse direction, and the concentration of O2 to increase.

Explanation:

The given chemical reaction represents a type of equilibrium reaction, specifically an exothermic reaction, as it produces sulfur trioxide (SO3), which releases heat. According to Le Chatelier's principle, to maintain equilibrium, if a system is disturbed by an external factor, the system will adjust accordingly.

Here are my answers to the specific questions:

  1. When the temperature is increased in an exothermic reaction, the system tries to consume the excess heat by moving in the endothermic direction, which is the reverse reaction in this case. Therefore, the value of Kc decreases (B).
  2. Since the equilibrium has been disturbed, the value of Qc will not be equal to Kc. Considering more products are formed, Qc will be greater than Kc (A).
  3. As a response to the increase in temperature, to re-establish equilibrium, the reaction will run in the reverse direction (B).
  4. As the reaction goes in reverse to establish a new equilibrium, the concentration of reactants increases. Thus, the concentration of O2 will increase (A).

Learn more about Chemical Equilibrium here:

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15. List the substances A-E in order from most dense to least dense based on the facts provided.Please write the letters in the spaces provided below.
Substance A: 8.2 g/cm3

Substance B: 3.5 cm and 30.0g

Substance C: 10.0g and 40mL

Substance D: 0.5 g/cm3

Substance E: 2.0cm by 3.0cm by 1.0cm and 4.0g

Most Dense_ _ _ _ _
Least Dense

Answers

The order of density of substances ranging from most dense to least dense is :substance B>substance A>substance E>substance C>substance D.

What is density?

It is a ratio of substance's mass per unit of volume.Symbol most commonly used for density is р.The SI unit of densityis kilogram per cubic meter .It explains how tightly a material is packed together.

There are2 types of density :1)absolute density 2) relativedensity.Absolute density is the massof any  substance per unit volume and relative density is the ratio of density of a substance to the density of a given reference material.

Reference material used forrelative density is water.The instrument used for measuring density or relative density of liquids is hydrometer. Densityis measured at constant temperature and pressure.

To learn more about density and it's types click here:

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Answer:

So 1st it is B then D then E then a then C

Perform the calculation and report the answer using the proper number of significant figures. Make sure the answer is rounded correctly. 1.012×10^-3 J/(0.015456 g)(298.3682−298.3567)K

Answers

Answer:

=5.694(J)/(g*K)

Explanation:

Hello,

In this case, since the result of the operation between two magnitudes is shown with the same significant figures of the shortest number, we obtain:

1.012x10^(-3) J/[(0.015456 g)(298.3682-298.3567)]K

Next, we proceed as follows:

=0.065476J/[(g)(20.0115K)]\n\n=5.693582(J)/(g*K)

Nevertheless, since 1.012 is the shortest number and has four significant figures, the result is rounded to four significant figures, that is until the three but it rounded due to the fact that the next digit is five:

=5.694(J)/(g*K)

Regards.

If 6.81 mol of an ideal gas has a pressure of 2.99 atm and a volume of 94.35 L, what is the temperature of the sample?

Answers

Answer:

504.57 K.

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 6.81 moles

Pressure (P) = 2.99 atm

Volume (V) = 94.35 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =.?

Using the ideal gas equation, the temperature of the ideal gas can be obtained as follow:

PV = nRT

2.99 × 94.35 = 6.81 × 0.0821 × T

282.1065 = 0.559101 × T

Divide both side by 0.559101

T = 282.1065 / 0.559101

T = 504.57 K.

Thus, the temperature of the ideal gas is 504.57 K.