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ElenaW [278]
2 years ago
8

On a graph , which type of line shows a direct proportion?

Chemistry
2 answers:
SashulF [63]2 years ago
8 0

Answer:

Straight line.

Explanation:

On a graph a straight line shows a direct proportion relation.

The graph equation for a straight line is

y = mx + c

Where

m = slope

c= y-intercept

So here if we change "x" y will change linearly with the change.

B) a jagged line: it is a random relation between two properties

C) downward curved line: It is inverse relation with some log factor or square or other root.

D) upwardly curved line: it is direct relation with some log factor or square or other root.

svetoff [14.1K]2 years ago
5 0
C - straight line
.....................................................................................
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For which of the following equilibria does `"K"_("eq") = ["O"_2]`? A. O2(l) O2(g) B. 2O3(g) 3O2(g) C. 2H2O(l) 2H2(g) + O2(g) D.
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The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

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                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

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3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

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4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

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