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3 years ago

How does nuclear fission of Uranium - 234 result in electricity being generated?

Chemistry

1 answer:

nevsk [136]3 years ago

7 0

Answer:

See explanation

Explanation:

The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.

As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.

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Plz help (it’s a picture)

Feliz [49]

Umm I’ll figure it out rn! Will come back in 1 min

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3 years ago

II need this asap !!!!!!!

Xelga [282]

Answer:

1. B.10 mol/l

2. D.

3.A.

Explanation:

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7 0

2 years ago

Which corresponds to the composition of the ion typically formed by magnesium?

nekit [7.7K]

The ion composition of Magnesium is 12,10, 2+.
Magnesium is a chemical element with symbol Mg and an atomic number 12, it has 12 protons, and 12 electrons with a chemical configuration of 2:8:2. It requires to loose two electrons to form a stable configuration forming a cation (positively charged ion) with a charge of +2 and a configuration of 2:8 ( 12 protons and 10 electrons).

3 0

3 years ago

How would i write this balanced equation using the coefficients?

ivanzaharov [21]

Just the number 2 to get the cofficients

7 0

3 years ago

The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O

Colt1911 [192]

Answer:

pH of a 0,245 M ammonia solution is 11,3 and percent ionization is 0,86%

Explanation:

For the equilibrium buffer of NH₃:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻; kb = 1,8x10⁻⁵

kb = [NH₄⁺] [OH⁻] / [NH₃] (1)

When 0,245 M of NH₃ is added, the equilibrium concentrations are:

[NH₃] = 0,245 - x

[NH₄⁺] = x

[OH⁻] = x.

Replacing this values in (1)

$1,8x10^{-5} = \frac{x^2}{0,245 - x}$

x² + 1,8x10⁻⁵x - 4,41x10⁻⁶ = 0

Solving for x:

x = -0,00211 No physical sense. There are not negative concentrations.

x = 0,00211 Real answer

Thus [OH⁻] in equilibrium is 0,00211 M.

As pOH = -log [OH⁻] and 14 = pH + pOH

pH of 0,00211 M is 11,3

It is possible to calculate the percent ionization thus:

Percent ionization = [OH−] equilibrium / [B] initial×100%

Replacing:

0,00211 / 0,245 × 100 = 0,86%

I hope it helps!

6 0

3 years ago