In a constant pressure calorimeter experiment a student measures a temperature change of 2.5 degrees c. the heat measured was 2.200 kj. what was the volume of the solution? assume that csol=4.18 j/gc

The volume of the solution was 210 mL.

q = mCₛₒₗΔT

m = q/(CₛₒₗΔT) = 2200 J/(4.18 J·°C⁻¹g⁻¹ × 2.5 °C) = 210 g  

If the solution has the same density as water (1.00 g·mL⁻¹) ,

Volume = 210 g × (1 mL/1.00 g) = 210 mL

Note: The answer can have only two significant figures because that is all you gave for the temperature change.

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.