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2 years ago

8

A certain reaction is endothermic in the forward direction. The reaction has less moles of gas on the product side. Which of the

following stresses would increase the
yield of the products (shift right)?
Increasing the volume
Decreasing the pressure
Increasing the temperature
Decreasing the reactant concentration
Please help

1 answer:

bazaltina [42]2 years ago

6 0

Answer:

decreasing the pressure

Step-by-step explanation:

i just took the test

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Mohamed decided to track the number of leaves on the tree in his backyard each year. The first year, there were 500500500 leaves

Paha777 [63]

Answer:

$f(n) = (1.4)^{(n - 1)} \times 500$

The sequence is a geometric sequence.

Step-by-step explanation:

The number of leaves in the tree in the first year is 500.

And the number of leaves in the tree in the second year increases by 40% of the number in the first year.

So, it will be $500(1 + \frac{40}{100}) = 1.4 \times 500$

Again, the number of leaves in the tree in the third year increases by 40% of the number in the second year and so on.

Therefore, the number of leaves in the tree in the third year will be $1.4 \times 500 \times (1 + \frac{40}{100}) = (1.4)^{2} \times 500$ and so on.

If we want to calculate the number of leaves in the tree in the nth year by f(n) then it will be given by

$f(n) = (1.4)^{(n - 1)} \times 500$

Therefore, the sequence is a geometric sequence. (Answer)

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2 years ago

This background applies to the next several questions. Assume a 15 cm diameter wafer has a cost of 12, contains 84 dies, and has

nalin [4]

Answer:

1) $Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

2) $Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

3) $Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

And for the new yield we need to take in count the increase of 15% for the area and we got this:

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

4) $DR_{old}=\frac{1}{\sqrt{0.92}} -1$ =0.0426 defects/cm^2

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$ =0.0260defects/cm^2

Step-by-step explanation:

Part 1

For this part first we need to find the die areas with the following formula:

$Area= \frac{W area}{Number count}$

$Area_1 = \frac{\pi (7.5cm)^2}{84}=2.104 cm^2$

$Area_2 = \frac{\pi (10cm)^2}{100}=3.1415 cm^2$

Now we can use the yield equation given by:

$Yield=\frac{1}{(1+ DR\frac{Area}{2})^2}$

And replacing we got:

$Yield_1= \frac{1}{(1+ 0.02 \frac{1}{2} 2.104)^2}=0.959$

$Yield_2= \frac{1}{(1+ 0.031 \frac{1}{2} 3.1415)^2}=0.909$

Part 2

For this part we can use the formula for cost per die like this:

$Cost/die = \frac{Cost per day_i}{Number count_i x Yield_i}$

And replacing we got:

$Cost/die_1 = \frac{12}{84 x 0.959}=0.149$

$Cost/die_2 = \frac{15}{100 x 0.909}=0.165$

Part 3

For this case we just need to calculate the new area and the new yield with the same formulas for part a, adn we got:

$Area_1 = \frac{1.1 \pi (7.5cm)^2}{84}=\frac{2.104 cm^2}{1.1}=1.913 cm^2$

$Area_2 = \frac{1.1 \pi (10cm)^2}{100}=\frac{3.1415 cm^2}{1.1}=2.856 cm^2$

And for the new yield we need to take in count the increase of 15% for the area and we got this:

$Yield_1= \frac{1}{(1+(1.15) 0.02 \frac{1}{2} 1.913)^2}=0.957$

$Yield_2= \frac{1}{(1+(1.15) 0.031 \frac{1}{2} 2.856)^2}=0.905$

Part 4

First we can convert the area to cm^2 and we got 2 cm^2 the yield would be on this case given by:

$Yield= \frac{1}{(1+DR\frac{2cm^2}{2})^2}=\frac{1}{1+(DR)^2}$

And if we solve for the Defect rate we got:

$DR= \frac{1}{\sqrt{Yield}}-1$

Now we can find the previous and new defect rate like this:

$DR_{old}=\frac{1}{\sqrt{0.92}} -1$ =0.0426 defects/cm^2

And for the new defect rate we got:

$DR_{new}=\frac{1}{\sqrt{0.95}} -1$ =0.0260defects/cm^2

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2 years ago

A manager records the number of hours, X, each employee works on his or her shift and develops the probability distribution belo

Reika [66]

Answer:

c. 2

Step-by-step explanation:

Given : X = No. of hours worked

No. of people work for the manager = 50

X = 3, 4 , 5 , 6 , 7, 8

P(X) = 0.1 , ? , 0.14 , 0.3 , 0.36 , 0.06

To Find : No. of people work for four hours

Solution : First understand the fact that sum of all probabilities is equal to 1

So, sum of all values of P(X) = 1

⇒ $0.1+?+0.14+0.3+0.36+0.06 = 1$

⇒ $0.96+? = 1$

⇒ $?= 1-0.96$

⇒ $? = 0.04$

So, the probability of no. of people worked for 4 hours is 0.04.

⇒P(4)=0.04

Thus , To calculate no. of people work for four hours :

$0.04\times 50$

⇒ 2 no. of people work for four hours per shift .

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2 years ago

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GuDViN [60]

A straight line (180o) is still an angle. Would you be wondering if the question said "Can 179o be a real angle?" No you would simply draw it. 1 degree more and you have 180o and it is an angle as well.

So you have a line that is both a line and an angle with A in the middle.

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The instructor’s friend also plans to rent an apartment in the same complex. Use the graph to identify the y-intercept and the s

Marta_Voda [28]

Answer:

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Step-by-step explanation:

Y intercept is found where x = 0 or on the y axis, so the only place on this graph where x = 0 and there is a number on the y axis is at (0, 3000), making 3000 your y intercept. The slope is figured from two points, and taking y2-y1/x2-x1. This would give you 3000-2450/2-0 which is -550/2 which is -275. That's your slope. Slope intercept form is y = mx+b, where m is the slope and b is the y intercept. Just plug in, and you get y = -275x+3000

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