Most of an atom consists of empty space. Its mass is concentrated in its center, which is called the

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

6 0

3 years ago

In the titration of HCl with NaOH, the equivalence point is determined

kondaur [170]

Answer:

In the titration of HCl with NaOH, the equivalence point is determined from the point where the phenolphthalein turns pink and then remains pink on swirling.

Explanation:

The equivalence point is the point at which exactly enough titrant (NaOH) has been added to react with all of the analyte (HCl). Up to the equivalence point, the solution will be acidic because excess HCl remains in the flask.

Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. Phenolphthalein is naturally colorless but turns pink in alkaline solutions. It remains colorless throughout the range of acidic pH levels, but it begins to turn pink at a pH level of 8.3 and continues to a bright purple in stronger alkalines.

It will appear pink in basic solutions and clear in acidic solutions.

The more NaOH added, the more pink it will be. (Until pH≈ 10)

In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above 13.0 pH

a. from the point where the pink phenolphthalein turns colorless and then remains colorless on swirling.

⇒ the more colorless it turns, the more acid the solution. (More HCl than NaOH)

b. from the point where the phenolphthalein turns pink and then remains pink on swirling.

The equivalence point is the point where phenolphtalein turns pink and remains pink ( Between ph 8.3 and 10). (

Although, when there is hydrogen ions are in excess, the solution remains colorless. This begins slowely after ph= 10 and can be noticed around ph = 12-13

c. from the point where the pink phenolphthalein first turns colorless and then the pink reappears on swirling.

Phenolphthalein is colorless in acid solutions (HCl), and will only turn pink when adding a base like NaOH

d. from the point where the colorless phenolphthalein first turns pink and then disappears on swirling

Phenolphthalein is colorless in acid or neutral solutions. Once adding NaOH, the solution will turn pink. The point where the solution turns pink, and stays pink after swirling is called the equivalence point. When the pink color disappears on swirling, it means it's close to the equivalence point but not yet.

3 0

3 years ago

Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume

DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

4 0

3 years ago

The reaction between carbon tetrachloride, CCl4, and water, H2O, to form carbon dioxide, CO2, and hydrogen chloride, HCl, has a

Ber [7]

Answer:

1) a) It is a slow reaction kinetically.

2) a) It is a catalyst.

3) c) It lowered the activation energy of the reaction.

Explanation:

1) The reaction between carbon tetrachloride, CCl₄, and water, H₂O, to form carbon dioxide, CO₂, and hydrogen chloride, HCl, has a ΔG° value of -232 kJ/mole, and so is thermodynamically favored. But when you mix carbon tetrachloride with water, no change is observed. What is a possible explanation for this?

a) It is a slow reaction kinetically. TRUE. This is the most likely explanation of why no change is observed.

b) The activation energy of the reaction is too small. FALSE. Slow reactions have a large activation energy.

c) The reaction is not spontaneous. FALSE. ΔG° < 0 so the reaction is spontaneous.

d)The reaction is not favored thermodynamically. FALSE. ΔG° < 0 so the reaction is favored thermodynamically.

2) What word best describes the role that the palladium plays in the reaction between propene and hydrogen above?

It is a catalyst because it increases the rate of the reaction without being altered by the end of the reaction.

3) What did the palladium do to increase the rate of the reaction between the propene and hydrogen?

a)It changed the thermodynamic equilibrium. FALSE. Catalysts don't affect equilibrium

b)It made the Gibbs free energy of the reaction more positive. FALSE. Catalysts don't affect Gibbs free energy.

c)It lowered the activation energy of the reaction. TRUE. According to the Arrhenius equation, the lower the activation energy, the higher the rate constant, and the higher the rate of reaction. This is how catalysts act.

d)It increased the activation energy of the reaction. FALSE.

7 0

3 years ago

The phrase “the danger is in the dose” is sometimes used when describing poisonous (toxic) chemicals. What does this mean?

Lostsunrise [7]

It means it depends on how much of the chemical you have. For ex. if 1 drop of snake venom might not kill you, but 20 drop could definitely kill you. Hope this helps!

6 0

3 years ago