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sp2606 [1]
3 years ago
10

Read the directions and create the graph described below. You will then click the submit button and upload a picture of your gra

ph or some sort of digital version of your graph. It could be done 1) using a computer program or 2) hand-drawn, photographed, and submitted.
You are to create a graph with horizontal and vertical axis. Label "Pressure" on the horizontal axis from O mb to 760 mb. Label "Volume" on the vertical axis from O to 1 mL. Assign values to axes divisions in such a way that you occupy almost all the space on both axes. Now, locate and label the following points:


(90, 0.9)
(100, 0.8)
(400, 0.2)
(600, 0.15)
(760, 0.1)
Chemistry
1 answer:
Alex73 [517]3 years ago
5 0

The characteristics of the graphs we can find the answer on how to draw the graph

  • Scales: The x scale = 40 mb / cm paper and the y scale  = 0.05 mL / cm paper
  • Graphic attached

The graphical representation is one of the best methods to visualize the relationships between a series of experimental data and to be able to find their functional relationships.

A graphic representation has several parts

  • Find the scales.
  • Mark the values ​​on the x and y axis
  • Plot the experimental points
  • Draw the curve through the experimental points

The scales are found with the relationship between the variation of the data between the amount of paper or graph range, let's look for the scales for each axis.

x-axis

The variation of the data  Δx = Final value - initial value

Paper range, generally it is  x = 20cm (sheet size)

They indicate that the graph must start at zero, so the minimum value must be changed,

             initial value = 0

              Δx = 760 -0 = 760

             x scale  = 760/20

             x scale  = 38 mb / cm of paper

y-axis

Data range   Δy= 0.9 -0

Paper range  y= 20 cm (sheet size)

            y-scale  = 0.9 / 20

            y scale  = 0.045 mL / cm paper

One of the characteristics of the scale is that it must be comfortable for graphing, which is why it approximates the following units

Selected scales  

         x scale  = 40 mb / cm paper

         y scale = 0.05 mL / cm paper

The second part consists of marking values ​​equally spaced and facilitating readable in the two axes, see attached for a graph made in a computer program

Plot each experimental point as best as possible, there are always some approximations by the selected scales

Draw a curve that passes through most of the experimental points and those that remain outside are balanced above and below the curve, in this case the program draws a line between each point, but it is preferable to draw a smooth curve.

In conclusion with the characteristics of the graphs we can find the answer on how to draw the graph

  • x scale  = 40 mb / cm paper
  • y scale = 0.05 mL / cm paper
  • Graphic attached

Learn more about graphical representation here:

brainly.com/question/14119615

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The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 17
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Answer:

a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)

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Step 2: Convert 130.0 °C to Kelvin

We will use the following expression.

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K = 130.0°C + 273.15

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P × V = n × R × T

n = P × V/R × T

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The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.

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Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

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In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

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<h3>What is an open system?</h3>

An open system is an interrelated group pf parts that work together to interchange matter and energy with the surrounding environment, while a closed system does not generate an exchange of matter and energy.

Therefore, with this data, we can see that an open system interchange energy and matter with the environment, while closes systems do it.

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