Consider an electrochemical cell based on the reaction: 2H+ (aq) + Sn (s) → Sn2+ (aq) + H2 (g) Which of the following actions would not change the measured cell potential? Consider an electrochemical cell based on the reaction: 2H+ (aq) + Sn (s) Sn2+ (aq) + H2 (g) Which of the following actions would not change the measured cell potential? addition of more tin metal to the anode compartment increasing the pressure of hydrogen gas in the cathode compartment lowering the pH in the cathode compartment increasing the tin (II) ion concentration in the anode compartment Any of the above will change the measured cell potential. Request Answer
Answers
Answer:
The only thing that will not affect the potential is the adition of solid Sn.
Explanation:
The potencial of a cell is linked to the concentration of the substances involved in the reactions by the equation of Nernst. So a change of one of them would change the cell potential.
The Keq for this reaction is:
Sn is not included because it's in solid state.
As can be seen, changing the concentrations of H2 (increasing the pressure), H+ (lowering the pH) or Sn2+ will affect the potential.
The only thing that will not affect it is the adition of solid Sn.
Based on the equilibrium constant equation, addition of Sn will not affect cell potential.
What is cell potential?
Cell potential refers to the potential difference that exists between two points in an electrochemical cell.
The Nerst equation shows the relationship between the cell potential and the concentration of the substances involved in the reactions.
Any change of one values results in a change in the cell potential.
The equilibrium constant for this reaction is given as follows:
Based on equilibrium constant equation, changing the amount of Sn will not affect cell potential since it is not included in the equilibrium constant equation.
Therefore, addition of Sn will not affect cell potential.
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What is the speed of a basketball that is thrown 18meters in 4 seconds?
how does the presence of a strong electrolyte in solution affect the colligative properties of a solution when compared to the same number of moles of a nonelectrolyte solute?
Answers
Answer:
Option 5 → 7.01 g
Explanation:
Molarity . volume (L) = Moles
This can help us to determine the moles of KOH that are in the solution.
We convert the volume from mL to L → 250 mL . 1L / 1000mL = 0.250 L
0.5 mol /L . 0.250L = 0.125 moles of KOH
Now, we only have to convert the moles to mass, by the molar mass:
Moles . molar mass = mass → 0.125 mol . 56.1 g/mol = 7.01 g
Answer:
We need 7.01 grams of KOH (option 5)
Explanation:
Step 1: Data given
Volume aqueous KOH solution = 250 mL = 0.250 L
Molarity = 0.500 M
Molar mass of KOH = 56.10 g/mol
Step 2: Calculate moles KOH
Moles KOH = molarity * volume
Moles KOH = 0.500 M * 0.250 L
Moles KOH = 0.125 moles
Step 3: Calculate mass of KOH
Mass KOH = moles KOH * molar mass KOH
Mass KOH = 0.125 moles * 56.10 g/mol
Mass KOH = 7.01 grams
We need 7.01 grams of KOH
Answers
Answer:
Lemon
HCI
Blood
Saliva
Bleach
NaOH
Explanation:
Blood 7.35-7.45
Bleach 12.6
Saliva 6.2-7.6
Lemon 2-3
HCI 3.01
NaOH 13
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Answers
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
What is meant by rate of a reaction ?
Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.
Here,
The concentration of A, [A] = 1 M
The concentration of B, [B] = 3 M
The partial order with respect to A, m = 2
The partial order with respect to B, n = 1
The rate constant of the reaction, k = 1.5
The rate of the reaction,
r = k[A]^m [B}^n
r = 1.5 x 1² x 3
r = 4.5 mol L⁻¹s⁻¹
Hence,
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
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Answer:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Explanation:
rate = k[A]”[B]"
liters of the 40% solution and
liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.
(Round to the nearest tenth, if necessary.)
Answers
Answer:
16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.
Explanation:
Let the required volume of solution 1 be represented by x.
The required volume of solution 2 would then be 78-x.
The number of moles of solution 1 that would be required = 0.4x
The number of moles of solution 2 that would be required = 0.25(78-x)
The number of moles of the final mixture = 78 x 0.28 = 21.84
moles of solution 1 + moles of solution 2 = moles of final mixture
0.4x + 0.25(78 - x) = 21.84
0.4x + 19.5 - 0.25x = 21.84
0.4x - 0.25x = 21.84 - 19.5
0.15x = 2.34
x = 15.6 liters
To the nearest tenth = 16 liters
Liters of 40% solution needed = 16 liters
Liters of 25% solution needed = 78 - 16 = 62 liters.
Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.
Answers
3H2 + N2 ----> 2NH3
convert from grams of NH3 ----- moles NH3------moles H2------g H2
11.21 gNH3 / ( 17g/ mol NH3)* (3 moles of H2/ 2 moles of NH3) * (2 g H2/ mol H2)
Answers
Answer: Option (2) is the correct answer.
Explanation:
Atomic number of oxygen atom is 8 and its electronic distribution is 2, 6. So, it contains only 2 orbitals which are closer to the nucleus of the atom.
As a result, the valence electrons are pulled closer by the nucleus of oxygen atom due to which there occurs a decrease in atomic size of the atom.
Whereas atomic number of sulfur is 16 and its electronic distribution is 2, 8, 6. As there are more number of orbitals present in a sulfur atom so, the valence electrons are away from the nucleus of the atom.
Hence, there is less force of attraction between nucleus of sulfur atom and its valence electrons due to which size of sulfur atom is larger than the size of oxygen atom.
Thus, we can conclude that the oxygen atom is smaller than the sulfur atom because the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
Final answer:
The oxygen atom is smaller than the sulfur atom because the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
Explanation:
The correct option is (2) the outer orbitals of oxygen are located closer to the nucleus than those of sulfur.
To understand why the oxygen atom is smaller than the sulfur atom, we need to consider their electron configurations. Oxygen has 8 electrons and sulfur has 16 electrons. Oxygen's electron configuration is 1s²2s²2p⁴, while sulfur's electron configuration is 1s²2s²2p⁶3s²3p⁴.
The outer orbitals of an atom, which are the valence orbitals, are the ones involved in bonding. The electrons in these orbitals determine the size of the atom. In the case of oxygen and sulfur, the outer orbitals of oxygen (2p orbitals) are closer to the nucleus compared to sulfur's outer orbitals (3p orbitals). As a result, the oxygen atom is smaller than the sulfur atom.
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