Can you please help me with this problem, I am so confused, and I do not know how to set it up, what equation to use, and what do i need to convert.
1 Transportation
Ethanol is being used as an additive to gasoline. The combustion of 1 mol of ethanol releases 1367kJ of energy. How many calories are released.
Thank you
You may be overwhelmed with information that is not needed. You don't care what it's being used for nor even what it is. All you need to do is to convert kJ to calories.
There are 4.184 J in 1 calorie and 1000 J in 1 kJ.
So 1367 kJ x (1000 J/1 kJ) x (4.184 J/cal) = ? calories.
Sure! To solve this problem, first, let's understand the given information:
1 mol of ethanol releases 1367 kJ of energy.
To find out how many calories are released, we need to convert the given value of energy from kilojoules (kJ) to calories (cal).
1 calorie (cal) = 4.184 joules (J)
1 kilojoule (kJ) = 1000 joules (J)
1 calorie (cal) = 0.001 kilojoules (kJ)
Now, let's set up the equation to convert kJ to cal:
1367 kJ × 0.001 kJ/cal = X cal
Simplifying the equation, we have:
1367 × 0.001 = X
X is the number of calories released when 1 mol of ethanol is combusted.
To find X, we need to multiply 1367 by 0.001:
X = 1.367 cal
Therefore, when 1 mol of ethanol is combusted, it releases 1.367 calories of energy.
Sure, I can help you with that problem!
To solve this problem, we need to convert the energy released in kJ (kilojoules) to calories.
First, let's establish the conversion factor between kJ and calories. The conversion factor is that 1 calorie is equal to 4.184 joules, and 1 joule is equal to 0.001 kilojoules.
To convert from kilojoules to calories, we need to multiply the energy value by the conversion factor. In this case, we have:
1 mol ethanol → 1367 kJ
Now, to convert kJ to calories, we multiply 1367 kJ by the conversion factor:
1367 kJ * (4.184 cal / 1 kJ) = 5706.628 cal
Therefore, the combustion of 1 mole of ethanol releases approximately 5707 calories of energy.
I hope this explanation helps! Let me know if you have any further questions.