Answer:
Explanation:
1. Please provide the enthalpy info - I will work on it with the info
2.
i) Reaction a should be modified to match the number of S in equation:
2S + 2O2 -> 2SO2 deltaH = -370kJ
ii) Reaction b should be written reversely to match the reactants of SO2:
2SO2 + O2 -> 2SO3 deltaH = 256kJ
iii) Adding the equations together:
2S + 3O2 -> 2SO3
iv) Enthalpy of the combined reaction = -370+256 = -114kJ
It is negative so the reaction is exothermic.
Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.
Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
in digital signals, codes are used while in analogue signals do not use signals.
<h3>What are signals?</h3>
Signals are defined as the instructions which are transferred to another source in other to deliver a message.
There are two types of signals which include the digital and analogue signals.
The digital signals uses codes which is a secret way of sending messages while analogue doesn't use codes.
Learn more about codes here:
brainly.com/question/22654163
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