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3 years ago

7

Calcium Oxide is used to remove pollutant SO2 from smokestack gases. The ΔG° of the overall reaction is -418.6 kJ. What is PSO2

in equilibrium with air (PO2 = 0.21 atm) and solid CaO?
CaO(s)+SO2 (g)+1⁄2O2 (g)⇔CaSO4 (s)

1 answer:

Gwar [14]3 years ago

5 0

From the change in free energy, we can calculate the equilibrium constant by the equation <span>ΔG°=-RT ln(K) where K is the equilibrium constant.

-418.6x10^2 = -(8.314)(298) ln (K)
K = 2.38x10^73

K is expressed as:

K = 1 / Pso2 (Po2)^0.5 = 1/ </span>Pso2 (0.21)^0.5
2.38x10^73 = 1/ Pso2 (0.21)^0.5
Pso2 = 9.17x10^-74 atm

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A zinc coating is added to the surface of a tin component to provide corrosion protection to the tin. Assume that the coating ad

Norma-Jean [14]

Answer:

In the presence of salt water and oxygen the coating will not corrode. As long as zinc coating is present and remains intact corrosion is not possible.

Explanation:

Here it is given that a tin is present so firstly tin is made of a chemical element

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It is a silvery,soft, white metal with a bluish tinge.

Now the covering which is been done on the tin is Zinc.

so, zinc is known to be served as a sacrificial coater.

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Their is a process which is known as a galvanic corrosion which refers to that "ZINC" defers to the metal to which it is protecting.

It is even more electrochemically active than iron itself.

Here, it is mentioned that zinc coating gets chipped but the coating remains intact. So, if the zinc is not removed from the tin's surface it will not get corroded when it is exposed to salt water and oxygen.

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After a rain a puddle of water remains on a sidewalk after a day o sunshine the puddle is gone which process is most responsible

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Evaporation or boiling

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Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh

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3 years ago

Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t

Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of $HC_2H_3O_2 \ (M_1)$ = 0.105 M

Volume of $HC_2H_3O_2 \ (V_1)$ = 20.0 mL

Concentration of $NaOH (M_2)$ = 0.125 M

The chemical reaction can be expressed as:

$HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}$

Using the ICE Table to determine the equilibrium concentrations.

$HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}$

I 0.105 0 0

C -x +x +x

E 0.105 - x x x

$K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}$

$K_a = \dfrac{(x)(x)}{(0.105-x)}$

Recall that the ka for $HC_2H_3O_2= 1.8 \times 10^{-5}$

Then;

$1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}$

$1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}$

By solving the above mathematical expression;

x = 0.00137 M

$H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )$

pH = 2.86

Hence, the initial pH = 2.86

b) To determine the volume of the added base needed to reach the equivalence point by using the formula:

$M_1 V_1 = M_2 V_2$

$V_2= \dfrac{M_1V_1}{M_2}$

$V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}$

$V_2 = 16.8 mL$

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of $HC_2H_3O_2 =$ molarity × volume

$= 0.105 \ M \times 20.0 \times 10^{-3} \ L$

$= 2.1 \times 10^{-3}$

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = $0.625 \times 10^{-3}$

After reacting with 5.0 mL NaOH, the number of moles is as follows:

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Initial moles $2.1*10^{-3}$ $0.625 * 10^{-3}$ 0 0

F(moles) $(2.1*10^{-3} - 0.625 \times 10^{-3})$ 0 $0.625 \times 10^{-3}$ $0.625 \times 10^{-3}$

The pH of the solution is then calculated as follows:

$pH = pKa + log \dfrac{[base]} {[acid]}$

Recall that:

pKa for $HC_2H_3O_2=4.74$

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

$pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}$

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the $(C_2H_3O^-_2)$ with $H_2O$

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of $(C_2H_3O^-_2)$ = moles/volume

= $\dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}$

= 0.0571 M

Now, using the ICE table to determine the concentration of $H_3O^+$ ;

$C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}$

I 0.0571 0 0

C -x +x +x

E 0.0571 - x x x

Recall that the Ka for $HC_2H_3O_2$ = $1.8 \times 10^{-5}$

$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} } \\ \\ K_b = 5.6 \times 10^{-10}$

$k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}$

$5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}$

$x = [OH^-] = 5.6 \times 10^{-6} \ M$

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }$

$[H_3O^+] =1.77 \times 10^{-9}$

$pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }$

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Before 0.0021 0.002725 0

After 0 0.000625 0.0021

$[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}$

$[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}$

$[OH^-] = 0.0149 \ M$

From above; we can determine the concentration of $H_3O^+$ by using the following method:

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}$

$[H_3O^+] = 6.7 \times 10^{-13}$

$pH = - log [H_3O^+]$

$pH = -log (6.7 \times 10^{-13} )$

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

3 0

3 years ago

What mass of hydrogen contains the same number of atoms as 7.00 g of nitrogen

Leto [7]

Molecular mass of nitrogen, N2 = 2*14 = 28
molecular mass of hydrogen, H2=2*1 = 2

Molecules of both elements contain 2 atoms, so the ratio of molecules is the same as the ratio of atoms.

From the molecular masses above,
ratio of number of molecules = ratio of molecular masses, therefore
7g N2 : x g H2 = 28:2
cross multiply:
x=7*2/28=0.5 g of Hydrogen has the same number of atoms as 7 g of nitrogen (at room temperatures)

7 0

3 years ago