Chemistry, 08.11.2019 18:31 luv4appleallday

Look back at your data and your regression equation. (sample data is shown to the right, with x= pressure and y=volume.) click all statements that are true.
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Look back at your data and your regression equation. (sample data is shown to the right, with x= pre

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Answer from: moomoofower

Select the block of cells to be included in the scatter plot by clicking and dragging, then from the Insert ribbon under Chart drop down the Scatter or Bubble menu and select Scatter. A chart will appear on the spreadsheet.

To set up a scatter plot in Excel, enter the pairs of data in two columns with each value of a pair on the same row. By default, Excel considers the column on the left to contain the horizontal (X) values and the column on the right to contain the vertical (Y) values.

If you click on the + sign at the upper right of the chart, a list of checkboxes will appear. Check Axes, Axis Titles, and Trendline. Uncheck everything else. You should edit the Axis Titles to include the name of the factor and any units associated with it. Double-click on the Axis numbers to bring up the Format Axis dialog, then click on the bar-graph icon to access Axis Options. Set the bounds and units appropriately and set the tick marks to something sensible.

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

Answer from: Quest

Particle exchange energy though elastic collision , hope this .

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Look back at your data and your regression equation. (sample data is shown to the right, with x= pre...

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