Write the balanced half-equation describing the oxidation of mercury to hgo in a basic aqueous solution. Please include the stat
1 answer:
Answer:
Explanation:
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In this case, mercury (II) oxide (HgO) is obtained via the reaction:
Nonetheless, since it is a reaction carried out in basic solution, mercury's half-reaction only, must be:
Thus, it is seen that OH ionis should be added due to the basic aqueous solution considering that 2 electrons are transferred from 0 to 2 in mercury.
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Explanation:
The given data is as follows.
= 85.0 ml,
= 0.9 M
= 85.0 ml,
= 0.9 M
Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.
So, No. of moles = Molarity × Volume
= (as 1 L = 1000 ml so, 85 ml = 0.085 L)
= 0.076 mol
As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.
= 4.271 kJ
or, = 4271 J (as 1 kJ = 1000 J)
Therefore, heat released = - heat of gained by calorimeter
Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.
Hence, mass of HCl = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Similarly, mass of KOH = = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Hence, total mass of the solution = 85 g + 85 g
= 170 g
Also, q =
4271 J =
0.0773 =
=
Thus, we can conclude that final temperature of the mixed solution is .