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Thepotemich [5.8K]
3 years ago
11

. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in

the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction Cl2 (g) 3F2 (g) ⟶ 2ClF3 (g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3.
Chemistry
1 answer:
valentina_108 [34]3 years ago
5 0

Answer : The the rate expression will be:

Rate=-\frac{d[Cl_2]}{dt}=-\frac{1}{3}\frac{d[F_2]}{dt}=+\frac{1}{2}\frac{d[ClF_3]}{dt}

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.

Now we have to determine the rate expressions for this given reaction.

The given balanced equations is:

Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)

\text{Rate of disappearance of }Cl_2=-\frac{d[Cl_2]}{dt}

\text{Rate of disappearance of }F_2=-\frac{1}{3}\frac{d[F_2]}{dt}

\text{Rate of formation of }ClF_3=+\frac{1}{2}\frac{d[ClF_3]}{dt}

Thus, the rate expression will be:

Rate=-\frac{d[Cl_2]}{dt}=-\frac{1}{3}\frac{d[F_2]}{dt}=+\frac{1}{2}\frac{d[ClF_3]}{dt}

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Answer:

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The shorthand notation is given as;

Cd: [Kr] 4d¹⁰5s²

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Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

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                Option-D, "their elements have very similar properties and exhibit a clear trend" is the correct answer.

<h3>Explanation:</h3>

                         In periodic table the elements are arranged in tabular form with respect to their atomic masses, atomic numbers, electronic configurations and chemical properties. It is called periodic because the properties of elements repeats periodically.

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Explain why the equation ( x - 4 ) ^2 - 28 =8 has two solutions. Then solve the equation to find the solutions. Show your work.
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Taking into account the discriminant and quadratic formula:

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Zeros or solutions of a function

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

That is, the zeros represent the roots of the polynomial equation that is obtained by making f(x)=0.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

<h3>Discriminant</h3>

The function f(x) = ax²  + bx + c

with a, b, c real numbers and a ≠ 0, is a function  quadratic expressed in its polynomial form (It is so called because the function is expressed by a polynomial).

The following expression is called discriminant:

Δ= b²- 4×a×c

The discriminant determines the amount of roots or solutions of the function.

Then:

  • If Δ <0 the function has no real roots and its graph does not intersect the x-axis.
  • If Δ> 0 the function has two real roots and its graph intersects the x-axis at two points.
  • If Δ = 0 the function has a real root and its graph intersects the x-axis at a single point that coincides with its vertex. In this case the function is said to have a double root.
<h3 /><h3>Amount of solutions of function (x - 4)² - 28 =8</h3>

The function (x - 4)² - 28 =8 can be expressed as:

(x - 4)² - 28 -8= 0

(x - 4)² - 36= 0

x²- 8x + 16 - 36= 0

x²- 8x + 16 - 36= 0

x²- 8x - 20= 0

Being:

  • a= 1
  • b= -8
  • c= -20

the amount of solutions are calculated as:

Δ= (-8)²- 4×1×(-20)

Δ= 144

As Δ> 0 the function has two real roots or solutions and its graph intersects the x-axis at two points.

<h3>Solutions of a cuadratic function</h3>

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or solutions are calculated by:

x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

<h3>Solutions of function (x - 4)² - 28 =8</h3>

The function (x - 4)² - 28 =8 can be expressed as x²- 8x - 20= 0

Being:

  • a= 1
  • b= -8
  • c= -20

the solutions of the function are calculated as:

x1=\frac{-(-8)+\sqrt{(-8)^{2}-4x1x(-20) } }{2x1}

x1=\frac{-(-8)+\sqrt{144 } }{2x1}

x1=\frac{8+\sqrt{144 } }{2}

x1=\frac{8+12 }{2}

x1=\frac{20}{2}

x1= 10

and

x2=\frac{-(-8)-\sqrt{(-8)^{2}-4x1x(-20) } }{2x1}

x2=\frac{-(-8)-\sqrt{144 } }{2x1}

x2=\frac{8-\sqrt{144 } }{2}

x2=\frac{8-12}{2}

x2=\frac{-4}{2}

x1= -2

Finally, the quadratic function (x - 4)² - 28 =8 has two solutions and the solutions are x1= 10 and x2= -2.

Learn more about the zeros of a quadratic function:

<u>brainly.com/question/14477557</u>

<u>brainly.com/question/842305</u>

<u>brainly.com/question/14477557</u>

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