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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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ziro4ka [17]

Answer:

L=7.0125*10^-5

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LET  

 Length of polyethene chain= L

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if I add water to 170 mL of a 0.40 M KOH solution until the final volume is 375 mL, what will the molarity M of the diluted solu
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4.7 M It may be wrong, but I hope it helps!

Explanation:

When you're diluting a solution, you're essentially keeping the number of moles of solute constant while changing the total volume of the solution.

Now, let's assume that you don't know the equation for dilution calculations.

In this case, you can use the molarity and volume of the concentrated solution to determine how many moles of hydrochloric acid you start with.

c

=

n

V

⇒

n

=

c

⋅

V

n

HCl

=

18 M

⋅

190

⋅

10

−

3

L

=

3.42 moles HCl

You then add water to get the total volume of the solution from  

190 mL

to  

730 mL

.

The number of moles of hydrochloric acid remains unchanged, which means that the molarity of the diluted solution will be

c

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3.42 moles

730

⋅

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