Answer: 0.374 mol CO2
Explanation:
We can calculate the number of moles of CO, n, required to produce the specified quantity of heat. This is done by dividing the required released heat, q, by the enthalpy of formation per unit mole of the substance,
, based on the chemical equation, such that:
![$$We have the following values for the variables:\displaystyle q = -147\ kJ \\\displaystyle \Delta H = -787\ kJ](https://tex.z-dn.net/?f=%24%24We%20have%20the%20following%20values%20for%20the%20variables%3A%5Cdisplaystyle%20q%20%3D%20-147%5C%20kJ%20%5C%5C%5Cdisplaystyle%20%5CDelta%20H%20%3D%20-787%5C%20kJ)
![$$We proceed with the solution:\begin{align} \displaystyle n &= \frac{q}{ \frac{\Delta H}{2\ mol\ CO}}\\ &= \frac{-147\ kJ}{\frac{-787\ kJ}{2\ mol\ CO}}\\ &\approx \boxed{\mathbf{ 0.374\ mol}} \end{align}](https://tex.z-dn.net/?f=%24%24We%20proceed%20with%20the%20solution%3A%5Cbegin%7Balign%7D%20%5Cdisplaystyle%20n%20%26%3D%20%5Cfrac%7Bq%7D%7B%20%5Cfrac%7B%5CDelta%20H%7D%7B2%5C%20mol%5C%20CO%7D%7D%5C%5C%20%26%3D%20%5Cfrac%7B-147%5C%20kJ%7D%7B%5Cfrac%7B-787%5C%20kJ%7D%7B2%5C%20mol%5C%20CO%7D%7D%5C%5C%20%26%5Capprox%20%5Cboxed%7B%5Cmathbf%7B%200.374%5C%20mol%7D%7D%20%5Cend%7Balign%7D)
Therefore, 0.374 mol of CO2 must be reacted in order to produce 147 kJ of energy?
Answer:
Right
Explanation:
The given parameters are;
The soluble sulfate formed by the metal, M = M₂SO₄
In the galvanic cell formed by the metal M, we have;
The concentration of M₂SO₄ in the left half cell = 50.0 mM = 0.05 M
The concentration of M₂SO₄ in the right half cell = 5.00 M
In the galvanic cell, the metal 'M' will be dissolved into the solution with lower concentration as M²⁺ which is the left half cell, making the cell negative and the solution more concentrated
In the right half cell, the metal 'M²⁺' in the solution will be plated unto the electrode making the solution less concentrated and the electrode in the right half cell will be the positive electrode
Therefore;
The electrode which will be positive is the electrode in the right half cell.
Answer:
Cr₂O₇⁻²(aq) and ClO₃⁻(aq)
Explanation:
At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.
The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:
RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).
Ni⁺²(aq) E° = -0.257 V
I₂(s) E° = +0.535 V
Cr₂O₇⁻²(aq) E° = +1.33 V
ClO₃⁻(aq) E° = +0.890 V
Pb²⁺(aq) E° = -0.125 V
So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).