Using the Fenske equation, calculate the number of theoretical plates for a fractional distillation set up used to separate Ethyl acetate (the more volatile component) from hexane (less volatile component) in a mixture with the following experimental data:
n=log(X/Xb) -log(Y a/Yb)/ log α Fenske Equation
Experimental data: l
The following are the data optained from injection of a 1-microliter sample of the equimolar stock solution used in the distillation experiment into a GČ. The percent of the area under the appropriate peak is idicated.
a = 1.6
GC results of the stock mixture used in the experiment
Component Rt (retention time) Percent Area
Ethyl acetate 1.09 53 82
Hexane 1.58 47 18
GC results of a 1-microliter sample after 3 mL had been collected:
Component Rt (retention time) Percent Area
Ethyl acetate 1.09 82
Hexane 1.58 18
a. 3.9
b. 7.2
c. 7.0
d. 3.0

Answer:

It will take 15 days for the same crew to clear the entire plot of 2 1/2 acres.

Step-by-step explanation:

Given that a crew clears brush from 1/3 acre of land in 2 days, to determine how long will it take the same crew to clear the entire plot of 2 1/2 acres, the following calculation must be performed:

1/3 = 0.333

1/2 = 0.5

0.333 = 2

2.5 = X

2.5 x 2 / 0.333 = X

15 = X

Therefore, it will take 15 days for the same crew to clear the entire plot of 2 1/2 acres.

Answer:

3 units to the right and 2 units up

Step-by-step explanation:

9514 1404 393

Answer:

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

Step-by-step explanation:

This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

  AB² = (4-1)² +(-3-0)² +(0-(-1)) = 16 +9 +1 = 26

  BC² = (1-4)² +(2-(-3))³ +(3-0)² = 9 +25 +9 = 43

  CA² = (1-1)² +(0-2)² +(-1-3)² = 4 +16 = 20

From the law of cosines, ...

  ∠A = arccos((AB² +CA² -BC²)/(2·AB·CA)) = arccos((26 +20 -43)/(2√(26·20)))

  ∠A = arccos(3/(4√130)) ≈ 86°

  ∠B = arccos((AB² +BC² -AC²)/(2·AB·BC)) = arccos((26 +43 -20)/(2√(26·43)))

  ∠B = arccos(49/(2√1118)) ≈ 43°

  ∠C = arccos((BC² +CA² -AB²)/(2·BC·CA)) = arccos((43 +20 -26)/(2√(43·20)))

  ∠C = arccos(37/(4√215)) ≈ 51°

The three angles are ...

  ∠CAB = 86°

  ∠ABC = 43°

  ∠BCA = 51°

_____

Additional comment

This sort of repetitive arithmetic is nicely done by a spreadsheet.

Answer:

1,108 in²

Step-by-step explanation:

SA = (12×20) + (2×20×5 + 2×12×5) + (2×½×12×9)

+ (2×20×11)

= 240+320+108+440

= 1,108 in²

Answer:

Step-by-step explanation:

                          Statements                                 Reasons

1). CD is an altitude of ΔABC                1). Given

2). ΔACD and ΔBCD are right              2). Definition of right triangles.

    triangles.

3). a² = (c - x)² + h²                                 3). Pythagoras theorem

4). a² = c² + x² - 2cx + h²                       4). Square the binomial.

5). b² = x² + h²                                       5). Pythagoras theorem.

6). cos(x) = [tex]\frac{x}{a}[/tex]                                           6). definition of cosine ratio for an angle

7). bcos(A) = x                                        7). Multiplication property of equality.

8). a² = c² - 2c(bcosA) + b²                    8). Substitution property

9). a² = b² + c² - 2bc(cosA)                    9). Commutative properties of

                                                                    addition and multiplication.

Answer

y = 3x - 2

Step-by-step explanation

The slope-intercept form of the equation for this line is,

⇒ y  = x + 2

The equation of line in point-slope form passing through the points

(x₁ , y₁) and (x₂, y₂) with slope m is defined as;

⇒ y - y₁ = m (x - x₁)

Where, m = (y₂ - y₁) / (x₂ - x₁)

Given that;

The point-slope form of the equation of the line that passes through points (-9, -2) and (1, 3) is,

⇒ y - 3 = (x - 1)

Now, We can simplify as;

⇒ y - 3 = (x - 1)

⇒ y - 3 + 3 = (x - 1) + 3

⇒ y  = x + 2

Thus, The slope-intercept form of the equation for this line is,

⇒ y  = x + 2

Learn more about the equation of line visit:

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Answer:

1384.5

Step-by-step explanation

Diviseion is just the oppisite of multiplication so you can think to yourself that what times two equals 2769. If you multiply 1384.5 on a calculater you get 2769.

Answer:

1024

Step-by-step explanation:

Given :-

Value of -2x²y³

Answer:

254

Step-by-step explanation:

^ <- this is the square sign

-2x^y^3

x=2

y=4

put x values in to x place and y value in to y place.

-2(2)^2(4)^3

Find the squares and - it with 2

-2(4)(64)

2-256=254

:. the value of -2x^2y^3 =254

That the answer.

Hope this is what you asked.

Answer:

0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question:

1200 female students, out of them, 350 have served a mission. So

[tex]p = \frac{350}{1200} = 0.2917[/tex]

0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.

Answer:

y = 14

Step-by-step explanation:

[tex] \frac{15}{21} = \frac{5}{7} [/tex]

[tex] \frac{10}{x} = \frac{5}{7} [/tex]

[tex]x = 14[/tex]

Now,

10/15 = y/21

15y = 10*21

y = 210/15

y = 14

This is a Right answer...

I hope you understand..

Mark me as brainliest...

Answer:

[tex]\frac{-2x+11}{(x-4)(x+1)}[/tex]

Step-by-step explanation:

I don't think we can factor this so we'll have to multiply to make the denominators the same

[tex]\frac{3(x+1)}{(x^2-3x-4)(x+1)}-\frac{2(x^2-3x-4)}{(x+1)(x^2-3x-4)}\\\\\frac{3x+3-(2x^2-6x-8)}{(x^2-3x-4)(x+1)}=\frac{-2x^2+9x+11}{(x^2-3x-4)(x+1)}\\-2x^2+9x+11=(x+1)(-2x+11)\\\\x^2-3x-4=(x+1)(x-4)\\\frac{(x+1)(-2x+11)}{(x+1)(x-4)(x+1)}=\frac{-2x+11}{(x-4)(x+1)}[/tex]

Answer:

24

Step-by-step explanation:

3x5=15

15+9=24

Answer:

B. (-r, 0)

Step-by-step explanation:

W is on the x axis, so the y coordinate is 0.

It is also r away from the origin, so the x coordinate is -r

Hope this helps!

Hello,

[tex]f(x)=2x^4-x^3-18x+9x\\\\=x(2x^3-x^2-18x+9)\\\\=x(x^2(2x-1)-9(2x-1))\\\\=x(2x-1)(x^2-9)\\=x(2x-1)(x-3)(x+3)\\[/tex]

Zeros are : 0;  1/2; -3; 3.

The zeros of the function are -3, 0, 1/2 and 3.

The given function is [tex]f(x)=2x^{4} -x^{3} -18x^{2} +9x[/tex].

Zeros of a function are the points where the graph of the function meets the X-axis i.e., at the solutions of f(x) = 0.

Now, factorise the given function, that is f(x)=x(2x³-x²-18x+9).

=x[x²(2x-1)-9x(2x-1)]

=x(2x-1)(x²-9)

=x(2x-1)(x+3)(x-3)

= -3, 0, 1/2, 3

Therefore, the zeros of the function are -3, 0, 1/2 and 3.

To learn more about the zeros of the function visit:

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Step-by-step explanation:

Area of a square = s x s

(4x-7)x(4x-7)

Open brackets

4x(4x-7)-7(4x-7)

16x^2-28-28x+49

16x^2-28x+21

Answer:

Cuboid = width*height*length

Cuboid = 24 cm^2

Answer:

2x(4x-4+1)

Step-by-step explanation:

i hope it will help you

Answer:

x=1/2

Step-by-step explanation:

press the calculator

8x²-8x+2=0

x=1/2

min,x=1/2

min,y=0

Consider that choice B is the same as saying 2x^2+1x^0.

The exponents 2 and 0 are both even, which is sufficient to say that the entire polynomial function itself is also even.

Something like choice A expands out and simplifies to x^2-2x+2, and that's equivalent to saying x^2+2x^1+2x^0. The presence of the x^1 term, with its odd exponent, is what makes choice A not even (it's not odd either).

Similarly, choices C and D also have exponents of 1, so they aren't even either.

Answer:

G(x)=2x2+1

Step-by-step explanation:

[tex]\bar{x} = 0[/tex]

[tex]\bar{y} =\dfrac{136}{125}[/tex]

Step-by-step explanation:

Let's define our functions [tex]f(x)\:\text{and}\:g(x)[/tex] as follows:

[tex]f(x) = x^2 + 1[/tex]

[tex]g(x) = 6x^2[/tex]

The two functions intersect when [tex]f(x)=g(x)[/tex] and that occurs at [tex]x = \pm\frac{1}{5}[/tex] so they're going to be the limits of integration. To solve for the coordinates of the centroid [tex]\bar{x}\:\text{and}\:\bar{y}[/tex], we need to solve for the area A first:

[tex]\displaystyle A = \int_a^b [f(x) - g(x)]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}[(x^2 + 1) - 6x^2]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}(1 - 5x^2)dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\left(x - \frac{5}{3}x^3 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]

[tex]\:\:\:\:\:\:\:= \dfrac{28}{75}[/tex]

The x-coordinate of the centroid [tex]\bar{x}[/tex] is given by

[tex]\displaystyle \bar{x} = \dfrac{1}{A}\int_a^b x[f(x) - g(x)]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:= \frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} (x - 5x^3)dx[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{75}{28}\left(\dfrac{1}{2}x^2 -\dfrac{5}{4}x^4 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]

[tex]\:\:\:\:\:\:\:= 0[/tex]

The y-coordinate of the centroid [tex]\bar{y}[/tex] is given by

[tex]\displaystyle \bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}[f^2(x) - g^2(x)]dx[/tex]

[tex]\displaystyle \:\:\:\:\:\:\:=\frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} \frac{1}{2}(-35x^4 + 2x^2 + 1)dx[/tex]

[tex]\:\:\:\:\:\:\:=\frac{75}{56} \left[-7x^5 + \frac{2}{3}x^3 + x \right]_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{136}{125}[/tex]

Answer:

[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Step-by-step explanation:

Solving (a): The frequency distribution

Given that:

[tex]Lowest = 138[/tex] --- i.e. the lowest class value

[tex]Class = 5[/tex] --- Number of classes

From the given dataset is:

[tex]Highest = 459[/tex]

So, the range is:

[tex]Range = Highest - Lowest[/tex]

[tex]Range = 459 - 138[/tex]

[tex]Range = 321[/tex]

Divide by the number of class (5) to get the class width

[tex]Width = 321 \div 5[/tex]

[tex]Width = 64.2[/tex]

Approximate

[tex]Width = 64[/tex]

So, we have a class width of 64 in each class;

The frequency table is as follows:

[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]

Solving (b) The relative frequency histogram

First, we calculate the relative frequency by dividing the frequency of each class by the total frequency

So, we have:

[tex]\begin{array}{ccc}{Class}& {Frequency} & {Relative\ Frequency} & 138 - 202 & 14 & 0.53 & 203 - 267 & 5 & 0.19 & 268 - 332 & 3 & 0.12 & 333 - 397 & 1 & 0.04 & 398 - 462 & 3 & 0.12 \ \end{array}[/tex]

See attachment for histogram

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Answer:

It has a side that is 9 units long.

Step-by-step explanation:

Answer:

B) It has a side that is 9 units long.

Step-by-step explanation:

Since it does not have two angles on the X-axis, a side that lies on the X-axis, or an obtuse angle the reasonable answer would be B as it is true, and all of the others are false.

Answer:

D. Critical value =± 0.707; there is not sufficient evidence to support the claim of a linear correlation between heights of mothers and heights of their daughters.

Step-by-step explanation:

Given that :

Correlation Coefficient, r = 0.693

The sample size, n = 8

The degree of freedom used for linear correlation :

df = n - 2

df = 8 - 2 = 6

Using a critical value calculator for correlation Coefficient at α = 0.05

The critical value obtained is : 0.707

The test statistic :

T = r / √(1 - r²) / (n - 2)

T = 0.693 / √(1 - 0.693²) / (8 - 2)

T = 0.693 / 0.2943215

T = 2.354

Since ;

Test statistic < Critical value ; we fail to reject the null and conclude that there is not sufficient evidence to support the claim of a linear correlation between heights of mothers and heights of their daughters.

Answer:

2250 g of nut mixture

Step-by-step explanation:

In order to use the ratio, we have to divide that 750 g by 3.

So 250 g multiplied by 5 is 1250. That's the amount of peanuts. Then just 1 for the cashews.

So, adding all of that up, the answer is 2250 g of nut mixture.

Answer:

1350g of nut mixture

Step-by-step explanation:

5:3:1 = 5x + 3x + x

5x = 750

x = 150

5x + 3x + x

= 5(150) + 3(150) + (150)

=750 + 450 + 150

=1350g

Hope it helps...

Answer:

sin(x)

Step-by-step explanation:

sec x tanx(1 - sin^2 x)

1 - sin^2 x = cos^2 x

sec(x)tan(x)cos^2(x)

[tex]\frac{1}{cos(x)}[/tex] * [tex]\frac{sin(x)}{cos(x)}[/tex] * cos^2(x)

[tex]\frac{sin(x)cos^2(x)}{cos^2(x)}[/tex]

sin(x)

Answer:

Hope it is helpful and useful

Answer:

The price of 1 exercise book is $122.45.

Step-by-step explanation:

This question is solved using a system of equations.

I am going to say that:

x is the price of one exercise book.

y is the price of one textbook.

Total cost of 10 exercise books and 2 textbooks is $1400.00.

This means that:

[tex]10x + 2y = 1400[/tex]

Since we want x:

[tex]2y = 1400 - 10x[/tex]

[tex]y = 700 - 5x[/tex]

One also finds the total cost of 3 textbooks and 30 exercise books is $3000.

This means that:

[tex]3x + 30y = 3000[/tex]

Since [tex]y = 700 - 5x[/tex]

[tex]3x + 30(700 - 5x) = 3000[/tex]

[tex]3x + 21000 - 150x = 3000[/tex]

[tex]147x = 18000[/tex]

[tex]x = \frac{18000}{147}[/tex]

[tex]x = 122.45[/tex]

The price of 1 exercise book is $122.45.