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Fudgin [204]
3 years ago
15

Create a graph of y=2x−6. Construct a graph corresponding to the linear equation y=2x−6.

Chemistry
1 answer:
yan [13]3 years ago
7 0
The graph is a straight line with this characteristics:

slope: the coefficient that accompanies the x variable, so it is 2.

x-axis intercept: find from y = 0 => 2x - 6 = 0 => x = 3

y - axis intercept: find from x = 0 => y = 2(0) - 6 = 6.

With that information you can draw you graph.

Just in case you need more information, you can build a table with several values of x and y:

x        y = 2x - 6

-5      2(-5) - 6 = -16

-4      2(-4) - 6 = -14

-3      2(-3) - 6 = -12

-2      2(-2) - 6 = -10

-1      2(-1) - 6 = -8

0        2(0) - 6 = -6

1        2(1) - 6 = -4

2        -2
 
3         0

4         2

5         4

Now you have plenty information to draw the graph.

6         6
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Why can’t elements be separated into smaller parts using chemical means?
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4 0
3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
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Please help me vote you brainiest
Viktor [21]

Answer:

Explanation:

A equal

B opposite

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Define alkanes and give examples
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<span>the simplest organic molecules, consisting of only carbon and hydrogen and with only single bonds between carbon atoms.
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