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What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m.

Answers

Answer 1

Answer:

Answer:

freezing point (°C) of the solution = - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3

$K_f$ = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

$molality = (mole \ of \ solute )/(kg \ of \ solvent )$

which can now be re-written as :

$molality = (mole \ of \ Ca(NO_3)_2)/(kg \ of \ water)$

$molality = ((mass \ of \ \ Ca(NO_3)_2)/(molar \ mass of \ Ca(NO_3)_2) )/(kg \ of \ water)$

$molality = ((11.3 \ g )/(164 \ g/mol) )/(0.115 \ kg )$

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

$\Delta T =3 * (1.86 \ ^0C/m) * (0.599 \ m)$

$\Delta T =3.34^0 \ C$

$\Delta T =$ the freezing point of water - freezing point of the solution

3.34° C = 0° C - freezing point of the solution

freezing point (°C) of the solution = 0° C - 3.34° C

freezing point (°C) of the solution = - 3.34° C

Calculating the pH

a) 0 mL

Initially, the pH of the solution is given by the dissociation of NH₃ in water.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)

The constant of the above reaction is:

$Kb = ([NH_(4)^(+)][OH^(-)])/([NH_(3)]) = 1.76\cdot 10^(-5)$ (2)

At the equilibrium, we have:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)

0.030 M - x x x

$1.76\cdot 10^(-5)*(0.030 - x) - x^(2) = 0$

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]

Now, we can calculate the pH of the solution as follows:

$pH = 14 - pOH = 14 + log(7.18\cdot 10^(-4)) = 10.86$

Hence, the initial pH is 10.86.

b) 10 mL

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)

We can calculate the pH of the solution from the equilibrium reaction (3).

$1.76\cdot 10^(-5)(Cb - x) - (Ca + x)*x = 0$ (5)

Finding the number of moles of NH₃ and NH₄⁺

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

$n_(b) = n_(i) - n_(HCl)$ (6)

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^(-4) moles$

$n_(a) = n_(HCl)$ (7)

$n_(a) = 0.025 mol/L*0.010 L = 2.5 \cdot 10^(-4) moles$

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are given by:

$Cb = (6.5\cdot 10^(-4) moles)/((0.030 L + 0.010 L)) = 0.0163 M$ (8)

$Ca = (2.5 \cdot 10^(-4) mole)/((0.030 L + 0.010 L)) = 6.25 \cdot 10^(-3) M$ (9)

Calculating the pH

After entering the values of Ca and Cb into equation (5) and solving for x, we have:

$1.76\cdot 10^(-5)(0.0163 - x) - (6.25 \cdot 10^(-3) + x)*x = 0$

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(4.54\cdot 10^(-5)) = 9.66$

Hence, the pH is 9.66.

c) 20 mL

We can find the pH of the solution from the reaction of equilibrium (3).

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are (eq 8 and 9):

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 8.0\cdot 10^(-3) M$

$Ca = (0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 0.01 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(8.0\cdot 10^(-3) - x) - (0.01 + x)*x = 0$

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(1.40\cdot 10^(-5)) = 9.15$

So, the pH is 9.15.

d) 35 mL

We can find the pH of the solution from reaction (3).

Calculating the concentrations of NH₃ and NH₄⁺

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 3.85\cdot 10^(-4) M$

$Ca = (0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 0.0135 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(3.85\cdot 10^(-4) - x) - (0.0135 + x)*x = 0$

x = 5.013x10⁻⁷ = [OH⁻]

Then, the pH is:

$pH = 14 + log(5.013\cdot 10^(-7)) = 7.70$

So, the pH is 7.70.

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0$

$n_(a) = 0.025 mol/L*0.036 L = 9.0 \cdot 10^(-4) moles$

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

Ca - x x x

$Ka = (x^(2))/(Ca - x)$

$Ka(Ca - x) - x^(2) = 0$ (10)

Calculating the acid constant of NH₄⁺

We can find the acid constant as follows:

$Kw = Ka*Kb$

Where Kw is the constant of water = 10⁻¹⁴

$Ka = (1\cdot 10^(-14))/(1.76 \cdot 10^(-5)) = 5.68 \cdot 10^(-10)$

Calculating the pH

The concentration of NH₄⁺ is:

$Ca = (9.0 \cdot 10^(-4) moles)/((0.030 L + 0.036 L)) = 0.0136 M$

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:

$pH = -log(H_(3)O^(+)) = -log(2.78\cdot 10^(-6)) = 5.56$

Hence, the pH is 5.56.

f) 37 mL

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

Calculating the concentration of HCl

$C_(HCl) = (0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L)/((0.030 L + 0.037 L)) = 3.73 \cdot 10^(-4) M = [H_(3)O^(+)]$

Calculating the pH

$pH = -log(H_(3)O^(+)) = -log(3.73 \cdot 10^(-4)) = 3.43$

Therefore, the pH is 3.43.

Find more about pH here:

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I hope it helps you!

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction: NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:

[NH3] = 0.030M [NH4+] = 0M [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 = x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

Energy levels in an electron configuration correspond to what on the periodic table?

Answers

I believe energy levels correspond to periods on the periodic table.

Which of the following do you need to know to be able to calculate the molarity of a salt solution ? I. the mass of salt added II. the molar mass of the salt III. the volume of water added IV. the total volume of the solution A) II and III only B) I, II, and IV only C) I, II, and III only D) I and III only E) You need all of the information.

Answers

According to molar concentration, for calculating molarity of salt solution mass,molar mass of salt and volume of solution is required , hence option B is correct.

What is molar concentration?

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.

In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.

Learn more about molar concentration,here:

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Explanation:

to find molarity

1the mass of salt added

Find the number of moles of solute dissolved in solution,

2 molar mass of salt

Find the volume of solution in liters,

4 total volume of solution

so B

An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of 55 degrees C is mixed with the water in the insulated container. When thermal equilibrium is reached, the temperature of the mixture is 25 degrees C. Assume that heat flows only between the alloy and the water. What is the specific heat of the alloy?

Answers

Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

Step 1: Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the alloy is 2.324 J/g°C

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 35 cm3. If the combustion of this mixture releases 775 J of energy, to what volume will the gases expand against a constant pressure of 710. torr if all the energy of combustion is converted into work to push back the piston

Answers

Answer:

The gases will expand 8.2 L against the constant pressure of 710 torr.

Explanation:

Given that:

the original volume V₁ = 35 cm³ = 35 × 10⁻⁶ m³

Since the combustion of the mixture releases energy then :

the work W = - 775 J

Pressure = 710 torr

Since 1 torr = 133.322 Pa

710 torr = 94658.62 Pa

We all know that:

W = -PdV

-775 = - 94658.62 Pa ( V₂ - V₁ )

-775 = - 94658.62 ( V₂ - 35 × 10⁻⁶)

-775/ - 94658.62 = V₂ - 35 × 10⁻⁶

0.008187 = V₂ - 35 × 10⁻⁶

V₂ = 0.008187 + 35 × 10⁻⁶

V₂ = 0.008222 m³

The change in volume dV = V₂ - V₁

The change in volume dV = 0.008222 m³ - 35 × 10⁻⁶ m³

The change in volume dV = 0.008187 m³

To litres

The change in volume dV = 8.2 L

Thus, the gases will expand 8.2 L against the constant pressure of 710 torr.

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