What happens to the ions when the drain is turned off? please help :] ❤️
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Answer:
74mL
Explanation:
Given parameters:
Molar mass of citric acid = 192g/mol
Molar mass of baking soda = 84g/mol
Concentration of citric acid = 0.8M
Mass of baking powder = 15g
Unknown parameters:
Volume of citric acid = ?
Solution
Equation of the reaction:
C₆H₈O₇ + 3NaHCO₃ → Na₃C₆H₅O₇ + 3H₂O + 3CO₂
Procedure:
- We work from the known parameters to the unknown. From the statement of the problem, we can approach the solution from the parameters of the baking powder.
- From the baking powder, we can establish a molar relationship between the two reactants. We employ the mole concept in this regard.
- We find the number of moles of the baking powder that went into the reaction using the expression below:
Number of moles =
Number of moles = = 0.179mole
- From the equation of the reaction, we can find the number of moles of the citric acid:
3 moles of baking powder reacted with 1 mole of citric acid
0.179 moles of baking powder would react with :
This yields 0.059mole of citric acid
- To find the volume of the citric acid, we use the mole expression below:
Volume of citric acid =
Volume of citric acid = = 0.074L
Expressing in mL gives 74mL
Answer : The actual cell potential of the cell is 0.47 V
Explanation:
Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
The given redox reaction is :
The balanced two-half reactions will be,
Oxidation half reaction :
Reduction half reaction :
The expression for reaction quotient will be :
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
The value of the reaction quotient, Q, for the cell is, 13.6
Now we have to calculate the actual cell potential of the cell.
Using Nernst equation :
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 316 K
n = number of electrons in oxidation-reduction reaction = 2 mole
= standard electrode potential of the cell = 0.51 V
= actual cell potential of the cell = ?
Q = reaction quotient = 13.6
Now put all the given values in the above equation, we get:
Therefore, the actual cell potential of the cell is 0.47 V