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vichka [17]
3 years ago
13

Plot your values of ln(Ksp) vs. 1/T and find the slope and y-intercept of the best fit line. Use the equation for the best fit l

ine and the following equation l n ( K ) = − Δ H ∘ R T + Δ S ∘ R to calculate ΔH° and ΔS° for dissolving Borax (0/1pts) What is the slope of your best fit line in the plot? -7E-5 highlight_off (-1 pts) Incorrect. (0/1pts) What is ΔH° (kJ/mol)? .000582

Chemistry
1 answer:
labwork [276]3 years ago
5 0

Answer:

a) The slope of the line of best fit plot = -12629.507

b) ΔH∘ = 105 kJ

c) Intercept of the line of best fit plot = 39.099

d) ΔS∘ = 325.1 J/K

e) Option A is correct.

Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.

f) Option D is correct. All of the options are correct.

Explanation:

The complete question is presented in the first attached image to this question. This complete question has the data readings required to plot the graph.

The second attached image has the plotted graph and the regression analysis to obtain the line of best fit.

The equation of the line of best fit obtained is

y = -12629.507x + 39.099

Comparing the given expression for the question with the equation of a straight line

ln (K) = (−ΔH∘/RT) + (ΔS∘/R)

y = mx + c

y = In K

Slope = m = (−ΔH∘/R)

x = (1/T)

Intercept = c = (ΔS∘/R)

So, to answer the question now

a) The slope of the line of best fit plot = -12629.507

b) Slope = (−ΔH∘/R)

(−ΔH∘/R) = -12629.507

But R = molar gas constant = 8.314 J/mol.K

ΔH∘ = 12629.507 × 8.314 = 105,001.721198 J = 105,002 J = 105 kJ

c) Intercept of the line of best fit plot = 39.099

d) Intercept = (ΔS∘/R)

(ΔS∘/R) = 39.099

ΔS∘ = 39.099 × 8.314 = 325.069086 J/K = 325.1 J/K

e) Do you expect the solubility of Borax to increase or decrease as temperature increases?

Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.

f) Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why.

A. To make sure no more sodium borate would dissolve in solution.

B. To ensure the dissolution process was at equilibrium.

C. To make sure the solution was saturated with sodium and borate ions.

D. All of the above

Hope this Helps!!!

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Answer:

La velocidad inicial es 55 \frac{m}{s}y su aceleración es -10 \frac{m}{s^{2} }

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad  varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

x=x0+v0*t+\frac{1}{2} *a*t^{2}

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

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Reemplazando obtenes:

90=v0*2+\frac{1}{2} *a*2^{2}

90=v0*2+\frac{1}{2} *a*4

90=v0*2+2*a

Y por otro lado tenes:

  • x= 120 m
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Reemplazando obtenes:

120=v0*3+\frac{1}{2} *a*3^{2}

120=v0*3+\frac{1}{2} *a*9

120=v0*3+\frac{9}{2} *a

Por lo que tenes el siguiente sistema de ecuaciones:

\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

v0= \frac{90-2*a}{2}

Reemplazando en la segunda ecuación:

120=\frac{90-2*a}{2} *3+\frac{9}{2} *a

Resolviendo:

120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a

120=135-3*a +\frac{9}{2} *a

120-135=-3*a +\frac{9}{2} *a

-15=\frac{3}{2} *a

\frac{-15}{\frac{3}{2} } =a

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Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

v0= \frac{90-2*(-10)}{2}

Resolviendo:

v0= \frac{90+20}{2}

v0= \frac{110}{2}

v0=55

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