Use the equation d=m/v
your mass or "m" is 78 g
your volume or "v" is 60mL
if you plug those values into the equation it will look like this:
d=78/60
d=1.3g/mL should be what you come up with
Answer:
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Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
Given:
Be - Beryllium - 9,3227
C - Carbon - 11,2603
O - Oxygen - 13,6181
Ne - Neon - 21,5645
B - Boron - 8,298
Li - Lithium - 5,3917
F - Fluorine - 17,4228
N - Nitrogen - 14,5341
Arranged from highest ionization energy to lowest ionization energy.
Ne ; F ; N ; O ; C ; Be ; B ; Li