What type of fuel is burned in a nuclear power plant?

What type of organic compound contains the following functional group?

Shtirlitz [24]

That's a ketone's functional group :)

4 0

2 years ago

Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO

zavuch27 [327]

Answer:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

$2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons 4NOBr ( g )$

The suitable equilibrium constant turns out:

$K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}$

Or in terms of the initial equilibrium constant:

$K_2=K_1^2$

Since the second reaction is a doubled version of the first one.

Best regards.

5 0

2 years ago

Due to radioactive decay over many millions of years, it is possible for the element uranium to decay in

ruslelena [56]

The uranium within these items is radioactive and should be treated with care. Uranium's most stable isotope, uranium-238, has a half-life of about 4,468,000,000 years. It decays into thorium-234 through alpha decay or decays through spontaneous fission.

3 0

2 years ago

If 60. liters of hydrogen gas at 546 K is cooled to 273 K at constant pressure, the new volume of the gas would be

Korvikt [17]

Answer:30 L

Explanation:

Initial Volume

=

V

1

=

60

l

i

t

e

r

Initial Temperature

=

T

1

=

546

K

Final Temperature

=

T

2

=

273

K

Final Vloume

=

V

2

=

?

Sol:-

Since the pressure is constant and the question is asking about temperature and volume, i.e,

V

1

T

1

=

V

2

T

2

⇒

V

2

=

V

1

⋅

T

2

T

1

=

60

⋅

273

546

=

60

2

=

30

l

i

t

e

r

⇒

V

2

=

30

l

i

t

e

r

Hence the new volume of the gas is

30

l

i

t

e

r

6 0

2 years ago

What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks

Firlakuza [10]

Answer:

$\large \boxed{1.64\times 10^{-5}\text{ mol/L }}$

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

2x 0.007 50 + x

$K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}$

3 0

2 years ago