Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
The balanced chemical equation is :
5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)
Here the oxidation number of P changed from 0 to -3 in PH₃ and increases from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.
Thus clearly both reduction and oxidation are taking place.
Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.
To know more about oxidation number here:
brainly.com/question/13182308
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<span>This is false. A carbohydrate is a carbon-based molecule that can be utilized by living organisms in order to produce energy. A calorie is a unit of energy often used to measure the amount of energy within food. Another example of energy unit is the Joule, more commonly used within physics.</span>
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
