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3 years ago

Which of the following statements describes

Chemistry

1 answer:

kozerog [31]3 years ago

5 0

Answer:

Energy is released as heat during the reaction.

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4 years ago

How does the way in which the solar system formed explain the orbits, sizes, and compositions of the planets?

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3 years ago

Which word describes the amount of matter an object contains?

AnnyKZ [126]

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3 0

3 years ago

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Be sure to answer all parts. Plutonium−242 decays to uranium−238 by emission of an α particle with an energy of 4.853 MeV. The 2

zheka24 [161]

Answer:

(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ

(b) 4.898 MeV

Explanation:

In the nuclear reaction, it was stated that plutonium-242 decayed firstly to uranium-238 and alpha, and lastly to a stable uranium-238 by emitting a gamma ray. The balanced equation for the nuclear reactions is shown below:

(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ

(b) The energy emitted by releasing a gamma ray is calculated using:

E = hc/λ

where

h = 4.136*10^-15 eV.s

c = 299792458 m/s

λ = 0.02757 nm = 0.02757*10^-9 m

Therefore:

E = (4.136*10^-15)*(299792458)/0.02757*10^-9 = 44974.31 eV

The total energy if the stable 238U was produced directly would be

4.853*10^6 + 44974.31 = 4.898 MeV

4 0

4 years ago

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150

Effectus [21]

Given:

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

To determine:

The cell potential, Ecell

Explanation:

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

8 0

3 years ago

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