coffee cup calorimeter lab
coffee cup calorimeter lab
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coffee cup calorimeter lab Frequently Asked Questions
Relevant but Long, Hard Lab Questions Heat of Acid-Base Neutralization (part 2)?
4) The combustion of 0.1584 grams of benzoic acid increases the temperature of a calorimeter by 2.54 degrees Celsius. Calculate the heat capacity of this calorimeter. The engery released by the combustion of benzoic acid is 26.42 kj/gram.
7) Why Cu^2+ is more active than Cu^1+?
2) The heat of combustion of gasoline can be measured in a way similar to the method you used to measure the
heat of neutralization; a bomb calorimeter is used instead of a coffee cup. The reaction is
C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O
When 1.02 g of gasoline is burned in a bomb calorimeter it raises the temperature of 1 kg of water from 22 °C
to 35.3 °C. The atomic weight of carbon is 12 g/mol and of hydrogen 1 g/mol. The specific heat of water is
4.18 J °C/g. Show your work to answer the following questions.How many moles of gasoline are burned in this experiment? What is the heat of combustion of gasoline?
4)
Heat supplied by the benzoic acid = 26.42*0.1584 = 4.1849 kJ
Heat capacity = Heat supplied/Increase in temperature = 4.1849/2.54=1.6476 kJ/degree celcius
7)
The Cu+2 ion has more charge per mass ratio as compared to the Cu+1 ion. This makes it a better oxidising agent. The relatively less reactive nature of copper is responsible for this fact. So, the Cu+2 ion can accept electrons form a donating species and thus revert back to Cu.
2)
The molecular mass of gasoline = 8(12) + 18(1) = 96 + 18 = 114 grams/mole
The number of moles of gasoline present in 1.02 grams = 1.02/114=0.0089 moles
Thus 0.089 moles of gasoline have been burnt in the experiment.
The heat required for the raising the temoerature of 1 kg water = m * (delta T) * s = 1000 * (35.3-22) * 4.18 = 55594 Joules
The heat of combustion of 0.0089 moles of gasoline = -55594 Joules
So, the molar heat of combution = -55594/0.0089=6.2465 *10^6 Joule/mole = -624.65 kJ/mole
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