If the density of water is 1.00 g/mL and the density of mercury is 13.6 g/mL, how high a column of water in meters can be suppor
1 answer:
Answer:
The answer to this is
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
Explanation:
To solve this we first list out the variables thus
Density of the water = 1.00 g/mL =1000 kg/m³
density of mercury = 13.6 g/mL = 13600 kg/m³
Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals
Therefore from the equation for denstity we have
Density = mass/volume
Pressure = Force/Area and for a column of water, pressure = Density × gravity×height
Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa
This value of pressure should be supported by the column of water as follows
Pressure = 101396.16 Pa = kg/m³×9.81 m/s² ×h
∴ = 10.336 meters
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
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Explanation:
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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