Extra Credit 16

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Q14.3.9

Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO₃)₂, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO₃ with CuO.

Solution:

Knowing that the chemical formula for nitric acid is HNO₃, and the other reactant is CuO, by intuition, we know this is a single replacement reaction of Cu and H. Then we get the equation to be:

CuO+2HNO₃=Cu(NO₃₎₂+ H₂O

2HNO_3(aq)+ CuO_(s) -> Cu(NO₃)₂_(aq) + H₂O_(l)

Q15.1.X

The calcium ions in human blood serum are necessary for coagulation. Potassium oxalate, K₂C₂O₄, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC₂O₄•H₂O. It is necessary to remove all but 1.0% of the Ca²⁺ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca²⁺ per 100 mL of serum, what mass of K₂C₂O₄ is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K_sp value for CaC₂O₄ in serum is the same as in water.)

Solution:

The pH does not really matter in this scenario. The only thing that does is the number of calcium ions in the solution. Therefore, follow the procedures below and derive an answer from it:

Compute the number of moles of Calcium ions in a standard volume of serum;
Use the number from step 1 to calculate the number of Calcium ions in 5.5mL of serum, the solubility of CaC₂O₄is low, as the Ksp=2*10^-9M²;
Determine the number of moles of Potassium ions needed to replace 1% Calcium in the serum, with the additional 4.5*10^-5 moles of CaC₂O₄ dissolved in the serum;
Note that their ratio should be Ca:K=1:2;
Convert the number of moles of K⁺ into the mass of the substance to add.

0.0022g

Q5.4.26

Using the data in Table T2, calculate the standard enthalpy change for each of the following reactions:

\(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g)\)
\(\ce{Si}(s)+\ce{2Cl2}(g)⟶\ce{SiCl4}(g)\)
\(\ce{Fe2O}(s)+\ce{3H2}(g)⟶\ce{2Fe}(s)+\ce{3H2O}(l)\)
\(\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g)\)

Solution:

The Table 2 is not enough, please refer to Table 1 for more information.

Basic solutions are shown below:

Check out the Enthalpy of formation of non-zero enthalpies for each equation(in kJ/mol) below;
Enthalpy of reaction equals to the total enthalpy of formation of the products minus that of the reactants;
Notice to multiply the enthalpies of formation by the coefficients in front of the chemical quations.

NO: 91.3	SiCl₄ :-657.0	Fe₂O₃: -824.2	H₂O: -285.83
LiOH: -487.5	CO₂ : -393.5	Li₂ CO₃ : -1216.0

90.4 kJ/mol
105 kJ/mol
...there's no neutral molecule Fe₂O, what do I do with this
94.49 kJ/mol

The data in Table T2 only has 3 (water, water vapour and carbon dioxide) out of 8 of the enthalpies of formation required.

Q13.4.2

A reaction is represented by this equation: \(\ce{2W}(aq)⇌\ce{X}(aq)+\ce{2Y}(aq) \hspace{20px} K_c=5×10^{−4}\)

Write the mathematical expression for the equilibrium constant.
Using concentrations of ≤1 M, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.

Solution:

Kc=Mproducts/M reactants= [X][2Y] ² /[2W]² .

b. Using concentrations of ≤1 M, make up two sets of concentrations that describe a mixture of W, X, and Y at equilibrium.

Solution:

Assume 0.1M of X at the beginning, what are the finals?

2W(aq)⇌X(aq)+2Y(aq)

2m 0.1-m 2m

(0.1-m)(2m)²/(2m)²=5*10^(-4)

m=0.0995M

2. Assume 0.2M of W, 0.9M of Y at the beginning?

2W(aq)⇌X(aq)+2Y(aq)

0.2+2m m 0.9-2m

m= 2.47*10_(-5)M

K_c= ([X][Y]^2)/[W]^2 = 5*10^-4
[X]= 0.0125M, [Y] = 0.02M, [W]= 0.1M | [X]= 0.003125M, [Y] = 0.1M, [W]= 0.25M

Q14.3.15

Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.

H₂O or HF
B(OH)₃ or Al(OH)₃
\(\ce{HSO3-}\) or \(\ce{HSO4-}\)
NH₃ or H₂S
H₂O or H₂Te

Solution:

HF would be more acidic, for it is an acid, and water is neutral.

B(OH)₃ or Al(OH)₃

Solution:

B(OH)₃ is stronger, for it can take up three hydroxide groups and Al(OH)₃ only takes one.

\(\ce{HSO3-}\) or \(\ce{HSO4-}\)

Solution:

HSO_4^- is stronger, for its conjugate base is stabler than that of HSO₃ , which is a gas.

NH₃ or H₂S

Solution:

H₂S is more acidic than NH₃ . Because ammonia is very likely to take a hydrogen atom and form NH_{4 ^-} in water, making it a strong base.

H₂O or H₂Te

Solution:

Because Te is larger than O, the Te-H bond is longer and weaker than the OH bond, making H2Te a stronger acid.

HF
B(OH)₃
HSO^3-
H₂S
H₂Te