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3 years ago

Why is it advantageous to produce plutonium -239

2 answers:

6 0

By definition, a plutonium 239 is a specific kind of nuclear isotope where it is known to have produced from transmutation of uranium in fuel rods. In addition, among it, main advantages would be that it has a significantly higher chance of being involved in a nuclear fission as compared to uranium.

Sveta_85 [38]3 years ago

6 0

The plutonium is considered to be advantageous. This can be explained considering the following reasons

a) It is an isotope of Plutonium which is radioactive and is being produced from uranium by its transmutation.

b) It is also can be considered as potential and important candidate of production of energy by its nuclear fission in comparison to Uranium

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Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

Calculate the pressure (in kpa) of 1.5 mole of helium gas at 354 k when it occupies a volume of 16.5l.

3241004551 [841]

Answer:

267.57 kPa

Explanation:

Ideal gas law:

PV = n RT R = 8.314462 L-kPa/K-mol

P (16.5) = 1.5 (8.314462)(354) P = 267.57 kPa

8 0

2 years ago

Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz( with units)

kogti [31]

i think it is this

Explanation:

3 0

3 years ago

The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres

Dima020 [189]

Answer:

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as $K_{p}$

$2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)$

Partial pressures at equilibrium:

$p^o_{H_2O}=0.070 atm$

$p^o_{H_2}=0.0035 atm$

$p^o_{O_2}=0.0025 atm$

The equilibrium constant in terms of pressures is given as:

$K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}$

$K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}$

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.

5 0

3 years ago

We have 1700 Tonnes of Amoniac every day. How many tonnes of 63% Nitrogen soda we can get?

V125BC [204]

There are 3 equations involved in manufacturing Nitric Acid from Ammonia.

First the ammonia is oxidized:
4NH3 + 5O2 = 4NO + 6H2O

Then for the absorption of the nitrogen oxides.
2NO + O2 = N2O4

Lastly, the N2O4 is further oxidized into Nitric acid.
3N2O4 + 2H2O = 4HNO3 + 2NO

Then run stoichiometry through these equations.
The first equation produces roughly 271,722,938 grams of NO
The second equation produces roughly 416,606,944 grams of N2O4
The last equation produces roughly 380,412,294 grams of HNO3 (nitric acid)

Convert the exact number back into tons, and your answer is: 419.332775 tons.

Rounded, I'm going to say that's 419.33 tons.
Hope this helps! :)

Also, it seems that commercially, Nitric Acid is commonly made by bubbling NO2 into water, rather than using ammonia.

3 0

3 years ago