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balandron [24]
3 years ago
14

When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the ca

lorimeter is 69.6 J∘C and it is submerged in 944mL of water, how much heat (in kJ) was produced by the hydrocarbon combustion?
Chemistry
1 answer:
schepotkina [342]3 years ago
7 0

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

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Answer:

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Explanation:

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The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, ass
arlik [135]

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Hydrogen spectrum

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harina [27]

Answer:

86.3 g  of N₂ are in the room

Explanation:

First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.

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Mole fraction N₂ = N₂ pressure / Total pressure

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Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K

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3 years ago
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