Guide :

At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 755.0 K?

At 600.0 K the rate constant is 6.1× 10–8 s–1. C4 H8 ---> 2C2 H4 What is the value of the rate constant at 755.0 K?

Research, Knowledge and Information :

C4H8(g)--->2C2H4(g) At 600.0 K The Rate Constant I... | Chegg.com

... At 600.0 K the rate constant is 6.1 10 8 s 1. What is the value of the rate constant at 755.0 ... C4H8(g)--->2C2H4(g) At 600.0 K the rate constant is 6.1 10 ...

At 600.0 K The Rate Constant Is 6.1 10 8 S ... | Chegg.com

Answer to At 600.0 K the rate constant is 6.1 10 8 s 1. What is the value of the rate constant at 765.0 K?... Textbook Solutions

At 600.0 K the rate constant is 6.1× 10–8 s–1 ... - Answers

At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 780.0 K? ... ABOUT; FIND THE ANSWERS.

At 600.0 K the rate constant is 6.1× 10–8 s–1. C4 H8 -> 2C2 ...

C4 H8 -> 2C2 H4 What is the value of the rate constant at 755.0 K? ... FIND THE ANSWERS. ... At 600.0 K the rate constant is 6.1× 10–8 s–1.

Solved: Question based on activation energy for the reaction ...

At 600.0 K the rate constant is 6.1 x 10-8 s-1. What is the value of the ... At 600.0 K the rate constant is 6.1 x 10-8 s ... Question based on activation energy for ...

Final Flashcards | Quizlet

Final Learn with flashcards, ... At 600.0 K the rate constant is 6.1 x 10⁻⁸s⁻¹. What is the value of the rate constant at 790.0 K?

Arrhenius Equation - Transtutors

... 2C2H4 (g) is 262 kJxmol-1 At 600.0 K the rate constant is 6.1 x 10 -8 s ... Arrhenius Equation 1 ... is 6.1 x 10-8 s-1. What is the value of the rate ...

Use different P and t, plot delta P/delta t against P n obtain the approximate value for r n k.

Inspection of the derivative function shows that it's a parabola​. When P=0 or k, dP/dt=0. And we can rewrite the function: P'=dP/dt=(rP/k)(k-P)=(r/k)(Pk-P^2)=(r/k)(k^2/4-P^2+Pk-k^2/4)=(r/k)((k/2)^2-(k-P)^2). The vertex is at (k/2,rk/2). So the distance between the point where the curve cuts the P (horizontal) axis gives us a measure for k; and, using this value for k we get r by inspecting the vertex height above the P axis and dividing by k/2. The table shows P against t, but we need P' against P; but we can roughly calculate dt and dP by subtracting pairs of numbers. When P=800 in the table we clearly have two values for t and the gradient P'=0. Therefore, k=800, implying that k/2=400. When P=400 in the table we can see that t is between 140 and 182 and P is between 371 and  534. So the gradient is roughly (534-371)/(182-140)=163/42=3.9. Therefore rk/2=3.9 so r=3.9/400=0.01. dP/P(1-P/k)=rdt. So int(dP/P(1-P/k))=rt+C, where C is to be determined from the table. Int(dP/P)-int(dP/(P-k))=rt+C; ln(P)-ln|P-k|=rt+C; ln(P/|P-k|)=rt+C; P/|P-k|=ae^rt where a replaces constant C (a=e^C). P=|P-k|ae^rt, so P=ake^rt/(1+ae^rt). From the table, when t=0, P=49, so 49=ak/(1+a); 49+49a=ak and a(k-49)=49, a=49/(k-49). Using the approximate value of k=800, a is approximately 49/(800-49)=0.065. dP/dt=rP-rP^2/k. Since r and k are constants the DE can be solved by separation of variables: So putting in our approximate values P=0.065*800*e^0.01t(1+0.065e^0.01t)⇒ P=52e^0.01t(1+0.065e^0.01t). The values in the table can be plotted on a graph of P against t, but to plot dP/dt against P would require us to know r and k, for which we have only the approximate values 0.01 and 800.

Determine the maximum value of dP/dt and interpret the meaning of this maximum value.

The maximum value of dP/dt is found by differentiating the differential with respect to t, but to do this we need to find dP/dt in terms of t, which requires integrating the dP/dt. integral(dP/(rP(1-P/k)))=t+c, where c is a constant. Let A/rP+B/(1-P/k)=1/(rP(1-P/k)), where A and B are constants. Then, A(1-P/k)+BrP=1; A-AP/k+BrP=1. Therefore, A=1 and -A/k+Br=0, from which B=1/rk. So integral becomes: integral(dP/rP)+integral(dP/(r(k-P)))=t+c; (1/r)ln(P)-(1/r)ln|k-P|=(1/r)ln|P/(k-P)|=t+c. ln|P/(k-P)|=rt+rc; P/(k-P)=e^r(t+c). At t=0 population is P0, some initial value, so P0/(k-P0)=e^rc, c=(1/r)ln|P0/(k-P0)|. P/(k-P)=e^rt*e^ln|P0/(k-P0)|=(P0/(k-P0))e^rt. dP/dt=rP(1-P/k)=rkP/(k-P)=rk(P0/(k-P0))e^rt; P"=kr^2(P0/(k-P0))e^rt. This expression can never be zero because e^rt cannot be zero, so dP/dt has no maximum value. However, when dP/dt=0 this is a turning point for P, and dP/dt=0 when P=k, the carrying capacity. The population becomes static at this point. If the initial population P0 is less than the carrying capacity the population increases, and if it's greater than the carrying capacity the population decreases. When it has decreased sufficiently to reach the carrying capacity, it remains static. When it has increased sufficiently to reach the carrying capacity it becomes static.

What is Mike's speed given the information below

In the first part of the problem, Andrew, traveling at a speed of v1, travels 4 miles, while Mike, traveling at speed v2, travels (d - 4) miles. They leave from their respective starting points at the same time, so the time it takes for them to meet and pass is the same for both. t = d / s 1. t1 = 4/v1 = (d - 4) / v2 Multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 2. (4 / v1) * v1 * v2 = ((d - 4) / v2) * v1 * v2 3. 4v2 = (d - 4) v1 Divide both sides by four to get the value of v2 4. v2 = ((d - 4)v1) / 4 In the second part of the problem, Andrew has reached Simburgh (d) and turned around, travelling another 2 miles, or (d + 2), while Mike has reached Kirkton and turned around, travelling another (d - 2) miles, for a total of d + (d - 2) = (2d - 2) miles. Again, their times are equal when they meet and pass. 5. t2 = (d + 2) / v1 = (2d - 2) / v2 As in the first part, multiply both sides by v1 to eliminate the denominator on the left side, and multiply both sides by v2 to eliminate the denominator on the right side. 6. ((d + 2) / v1) * v1 * v2 = ((2d - 2) / v2) * v1 * v2 7. (d + 2)v2 = (2d - 2)v1 Divide both sides by (d + 2) go get the value of v2 8. v2 = ((2d - 2)v1) / (d + 2) We have two equations for v2, equation 4 and equation 8. The problem states that v2 remains the same throughout the journey. Therefore: ((d - 4)v1) / 4 = ((2d - 2)v1) / (d + 2) Once again, multiply both sides by both denominators. (((d - 4)v1) / 4) * 4 * (d + 2) = (((2d - 2)v1) / (d + 2)) * 4 * (d + 2) v1 * (d - 4) * (d + 2) = v1 * (2d - 2) * 4 Divide both sides by v1, eliminating speed from this equation. (d - 4) * (d + 2) = (2d - 2) * 4 d^2 - 4d + 2d - 8 = 8d - 8 d^2 - 2d - 8 = 8d - 8 Subtract 8d from both sides and add 8 to both sides. (d^2 - 2d - 8) - 8d + 8 = (8d - 8) + 8d + 8 = 0 d^2 - 10d = 0 Factor out a d on the left side. d * (d - 10) = 0 One of those factors is equal to 0 (to give a zero answer). d = 0 doesn't work; we already know the distance is more than 4 miles. d - 10 = 0 d = 10    <<<<<   That's the answer to the first question, how far is it? We'll substitute that into equation 4 to find v2 in relation to v1. v2 = ((d - 4)v1) / 4 = ((10 - 4)v1) / 4 v2 = 6v1 / 4 = (6/4)v1 v2  = 1.5 * v1   <<<<<< That's the answer to the second question No matter what speed you choose for Andrew (v1), Mike's speed is one-and-a-half times faster. Let's set Andrew's speed to 6mph and solve equation 1. t1 = 4/v1 = (d - 4) / v2 t1 = 4mi / 6mph = (10 - 4) / (1.5 * 6mph) 4/6 hr = 6/9 hr 2/3 hr = 2/3 hr With Andrew travelling at 4 mph, and Mike travelling at 6 mph, it took both of them 2/3 of an hour to reach a point 4 miles from Kirkton.

(1, -20), (7, -20), (2, -40), (6, -40), (5, -50), (8, 10), (9, 46), (0, 7), (-1, 47), (-2, 89)

First put all the points (x,y) in order of the x value: -2 89 -1 47 0 7 1 -20 2 -40 5 -50 6 -40 7 -20 8 10 9 46   When x=0 y=7, so the constant term in the function is 7 (y intercept). There's a minimum (turning point) near x=5 because the values of y for x=2 and 6 (-40) are more positive than -50. The gradient (dy/dx) gets positively steeper as x increases after x=5, and negatively steeper between x=-2 and +2. The degree of the polynomial is even because the function is positive for larger magnitude positive and negative values of x. The zeroes are between x=0 and 1 and between x=7 and 8. The function can be split into two parts: (1) y for -2 Read More: ...

I need help, I really need help pleeease.

I assume that (3) is a graph, because it doesn't display on my tablet. Question 1 a) 1 because the daily operating cost is stated specifically. b) 3 because the profit of \$500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204. c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of \$2.5, but 136 tickets would make a loss of \$5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137. d) The rate of change is 7.5 from (2).  e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity. Question 2 a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0. b) For (1) you would need to find out how many tickets at \$7.50 you would need to cover the operating costs of \$1025 plus the profit of \$500. That is, how many tickets make \$1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=\$500 somewhere between t=200 and 250 in the table. c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150. d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, \$7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5. e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=\$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add \$375 to the profit.

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0

Show that the equation

This can be written (x^2-2x+1)+(y^2-4y+4)+(z^2-2xz+2z)=0; (x-1)^2+(y-2)^+(z^2-2xz+2z)=0. Put z=0 and we are left with (x-1)^2+(y-2)^2=0. The only solution to this equation is when x=1 and y=2. This means that if we look along z axis towards the x-y plane we see a point at (1,2). When z=1, we have x^2-4x+4+(y-2)^2=0=(x-2)^2+(y-2)^2, another point at (2,2). When z=2, we have x^2-6x+9+(y-2)^2=0=(x-3)^2+(y-2)^2, point at (3,2). When z=3, we have x^2-8x+16+(y-2)^2=0=(x-4)^2+(y-2)^2, point at (4,2). And so on. When z=-1, we have x^2+(y-2)^2=0, point at (0,2). In fact, the equation can be written (x-(z+1))^2+(y-2)^2=0. From this it follows that, from the viewpoint of the z axis, the shape traces the line y=2. Note that if y doesn't equal 2, the graph doesn't exist. Therefore, the plane y=2 for all x and z is the only place where the graph exists. When y=2, x=z+1 traces a straight line in the x-z plane. The figure cannot be a cone. If the expression in x, y and z equalled non-zero (e.g., 1) then at z=0 there would have been a circle of radius 1 centre (1,2). With increasing values of z, the centre would shift along the line x=z+1, but would keep the same radius, so the shape would be a tube. There would be no vertex. As the constant 1 shrinks to zero the diameter of the tube decreases until it becomes a line. To be a cone the constant would need to be a (quadratic) expression in z. (The standard equation for a circular cone is ((x-h)^2+(y-k)^2=a^2(z-j)^2, vertex at (h,k,j), and the radius of the circular base is a. The complete shape is two cones joined at their common vertex like a diabolo.) The expression's changing values would affect the diameter of the circle. When the expression evaluated to zero, the left-hand side expression would describe a point, the vertex of the cone. The sign of z^2 in the question needs to be negative to qualify as the equation of a cone.

find solution to the recurrence relation

Solving a Recurrence Relation Find the solution to the recurrence relation, a_n = 6a_(n-1) + 11a_(n-2) – 6a_(n-3), given a_0 = 20, a_1 = 5, a_2 = 15 Assuming a solution of the form a_n = c.r^n, then a_(n-1) = c.r^(n-1), a_(n-2) = c.r^(n-2), a_(n-3) = c.r^(n-3) Substituting for the above into the recurrence relation, c.r^n = 6c.r^(n-1) + 11c.r^(n-2) – 6c.r^(n-3) r^n = 6r^(n-1) + 11r^(n-2) – 6r^(n-3) 1 = 6r^(-1) + 11r^(-2) – 6r^(-3) r^3 = 6r^(2) + 11r^(1) – 6 r^3 – 6r^2 – 11r + 6 = 0 Solving the cubic equation above would give three root values for r, and a recurrence relation of the form a_n = c1.r1^n + c2.r2^n + c3.r3^n where the three constants, c1, c2, c3 are determined from the initial values for a_0, a_1 and a_2. The solutions for our cubic equation, r^3 – 6r^2 – 11r + 6 = 0, is r1 = -1.8256153, r2 = 0.4453157. r3 = 7.3802996 Our recurrence relation now becomes, a_n = c1*(-1.8256153)^n + c2*(0.4453157)^n + c3*(7.3802996)^n Using the initial values for a_0, a_1 and a_2, we get   a_0 = 20 = c1*(-1.8256153)^0 + c2*(0.4453157)^0 + c3*(7.3802996)^0 a_1 = 5 = c1*(-1.8256153)^1 + c2*(0.4453157)^1 + c3*(7.3802996)^1 a_2 = 15 = c1*(-1.8256153)^2 + c2*(0.4453157)^2 + c3*(7.3802996)^2   20 = c1 + c2 + c3  5 = c1*(-1.8256153 + c2*(0.4453157) + c3*(7.3802996) 15 = c1*(-1.8256153)^2 + c2*(0.4453157)^2 + c3*(7.3802996)^2   The last three simultaneous equations can now be solved for c1, c2 and c3. The values are: c1 = 1.990012, c2 = 17.9216148, c3 = 0.08837315 Our recurrence relation finally becomes, a_n = 1.990012*(-1.8256153)^n + 17.9216148*(0.4453157)^n + 0.08837315*(7.3802996)^n   Check Substituting for n = 0, 1, 2 in the final expression gives results for the initial values of a_0, a_1, a_2. n = 0:    19.99999995  (a_0 = 20) n = 1:    5.000000410   (a_1 = 5) n = 2:    15.00000017   (a_2 = 15)