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denis23 [38]
3 years ago
5

How much energy is needed when 1.27 moles of sulfur reacts with excess

Chemistry
1 answer:
pav-90 [236]3 years ago
6 0

Answer:

well what do you think it is

Explanation:

huh?..

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A certain half-reaction has a standard reduction potential +0.80 V . An engineer proposes using this half-reaction at the anode
Strike441 [17]

Answer:

a. Minimum 1.70 V

b. There is no maximum.

Explanation:

We can solve this question by remembering that the cell potential is given by the formula

ε⁰ cell = ε⁰ reduction -  ε⁰  oxidation

Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the  oxidized species  0.80 V, thus

ε⁰ reduction -  ε⁰  oxidation ≥  ε⁰ cell

Since ε⁰  oxidation is by definition the negative of ε⁰ reduction , we have

ε⁰ reduction - ( 0.80 V )  ≥  0.90 V

⇒ ε⁰ reduction  ≥ 1.70 V

Therefore,

(a) The minimum standard reduction potential is 1.70 V

(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V

8 0
3 years ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
3 years ago
What is bias in an experiment?
N76 [4]

○ c. ○

◘

because a Research bias, also called experimenter bias, is a process where the scientists performing the research influence the results, in order to portray a certain outcome.

◘

hope that helped ♥

3 0
3 years ago
Calculate the mass of 3.5 mol C6H6
Lilit [14]
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
4 0
3 years ago
A hot gas flowing through a pipeline can be considered as a:________
k0ka [10]

Answer:

B) irreversible process

Explanation:

The process given here is irreversible.

7 0
3 years ago
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