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Semmy [17]
3 years ago
6

Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH(aq). Calculate the a

mount of Ga(s) that can be deposited from a Ga(III) solution using a current of 0.760 A that flows for 50.0 min.
Chemistry
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

m_{Ga}=0.550gGa

Explanation:

Hello!

In this case, since the applied current for the 50.0 mins provides the following charge to the system:

q=0.760\frac{C}{s}*50.0min*\frac{60s}{1min}=2,280C

As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:

n_{e^-}=2,280C*\frac{1mole^-}{96,485C}=0.0236mole^-

Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:

m_{Ga}=0.0236mole^-*\frac{1molGa}{3mole^-}*\frac{69.72gGa}{1molGa}  \\\\m_{Ga}=0.550gGa

Best regards!

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Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

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Resonance, leaving group, carbonyl carbon delta+, and steric effect is the most crucial variables that affect the relative reactivity of a functional group containing a carbonyl in an addition or substitution process.

Discussion:

1. Carbonyl Carbon Delta+: The carbonyl group becomes more electrophilic and accelerates nucleophilic assault when the carbonyl carbon delta+ is bigger.

2. Resonance: When the carbonyl is transformed into the tetrahedral adduct, it may be lost. Loss of resonance increases the energy of the transition state for this nucleophilic assault because resonance has the function of stabilizing. Therefore, a carbonyl functional group's resistance to nucleophilic attack increases as resonance in the group increases in importance.

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Learn more about carbonyl here:

brainly.com/question/21440134

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