Please show all work 1 a bomb calorimeter has a heat...

A bomb calorimeter has a heat capacity of 2.47 kJ/K. When a 0.103-g sample of a certain hydrocarbon was burned in this calorimeter, the temperature increased by 2.14 K. Calculate the energy of combustion for 1 g of the hydrocarbon. q_reaction = -q_calorimeter Answer: ΔH°_rxn ______________________________ kJ/g hydrocarbon 2. When a 7.68-g sample of sulfur is burned in a bomb calorimeter with excess oxygen, the temperature increases from 20.34 °C to 36.75 °C. The calorimeter contains 815 g of water, and the bomb’s heat capacity is 923 J/°C. Calculate ΔU for the reaction in kJ/mol. Answer: ΔU = q_v = _________________________________________ kJ/mol

Transcript

00:01 Hi guys, so in this question we have been given the heat capacity that is equal to 2 .47 kilojouze per kelvin.

00:11 And we have been given mass that is equal to 0 .103 grams.

00:20 So we have the temperature also that is equal to temperature increases 2 .14 kelvin.

00:29 So we can write it delta t.

00:31 So we have to calculate the energy of combustion of this hydrocarbon.

00:39 So now, so we have q of reaction is equal to minus of q of calorie meter.

00:55 Now, q of calorie meter is equal to heat capacity of calorie meter multiplied by delta t.

01:03 So c multiplied by delta t.

01:06 So now we have the values that is 2 .5 .5.

01:10 47 kilojoules per kelvin multiplied by 2 .14 kelvin and from here we will get the value of calorie meter that is equal to 5 .2 -8 858 kilojoules so q reaction will be equal to minus of 5 .2858 kilojoules and now if we divided with 0 .103 grams so this will be equal to minus of 51 .3185 kilojoules per gram so this is the final answer.

02:06 Now for the next part of the question we have been given the mass of sulfur that is equal to 7 .68 grams and we have been given the temperature that is that increases from 20 .34 degrees celsius to 36 .75 degrees celsius and we haven't given the mass of water that is 815 grams also the heat capacity bombs is given as c equal to 9 to 3 joules per degree celsius so we have to calculate the delta u for the reaction so so now heat absorbed by water is equal to q w is equal to m multiplied by delta t multiplied by c so from here we can find the value of q w so mass is given that is 815 multiplied by change in temperature so it will be 16 .41 degrees celsius and heat capacity of water is 4184...

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