I really need help on this problem...
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Hydrocarbon Formula Delta H (kj/mol)
1,3-Butadiene C4H6(g) 111.9
a) calculate the fuel value in kJ/g for this compound and determine the percentage of hydrogen by mass.
I've tried it for hours yet I can't figure out how to do this.
Please help thank you@!!
You have kJoules/mol. How many grams of this compound are in one mole? Divide your given number by the grams in a mole to get kJ/g.
Percentage of H?
6(atomic mass H)/molmassC4H6 * 100
Here is the problem-
Calculate the total enery change when warming 5.00g of ice at -15 degrees celcius to liquid water at 25 degrees celcius.
could i get a step by step solution and answer to this problem??? What is the thermal energy change as 175.og of water drops from 75.00 degrees celcius to 15.50 degrees celcius? help?
To calculate the total energy change when warming 5.00g of ice at -15 degrees Celsius to liquid water at 25 degrees Celsius, we need to consider three different steps:
1. Heating the ice from -15 degrees Celsius to 0 degrees Celsius:
- The heat absorbed or released during a phase change can be calculated using the formula: q = m * ΔH, where q is the heat absorbed or released, m is the mass, and ΔH is the enthalpy change for the phase transition.
- For ice to water, ΔH is known as the heat of fusion, which for water is 334 J/g.
- So, the heat absorbed during this step can be calculated as: q1 = 5.00g * 334 J/g = 1670 J.
2. Melting the ice at 0 degrees Celsius:
- The heat absorbed during the phase change from solid to liquid is given again by the formula: q = m * ΔH.
- For ice to water, ΔH is the heat of fusion, which remains the same at 334 J/g.
- Therefore, the heat absorbed during this step is: q2 = 5.00g * 334 J/g = 1670 J.
3. Heating the liquid water from 0 degrees Celsius to 25 degrees Celsius:
- The heat absorbed or released during a temperature change is given by the formula: q = m * c * ΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity of the substance, and ΔT is the temperature change.
- For liquid water, the specific heat capacity is 4.18 J/g·°C.
- The temperature change is ΔT = 25°C - 0°C = 25°C.
- Therefore, the heat absorbed during this step can be calculated as: q3 = 5.00g * 4.18 J/g·°C * 25°C = 522.5 J.
To find the total energy change, we need to sum up the heat absorbed for each step:
Total energy change = q1 + q2 + q3
= 1670 J + 1670 J + 522.5 J
= 3862.5 J
Therefore, the total energy change when warming 5.00g of ice at -15 degrees Celsius to 25 degrees Celsius is 3862.5 J.
For the second problem, to calculate the thermal energy change as 175.0g of water drops from 75.00 degrees Celsius to 15.50 degrees Celsius, we can use the same formula:
q = m * c * ΔT
where q is the heat absorbed or released, m is the mass, c is the specific heat capacity of water (also 4.18 J/g·°C), and ΔT is the temperature change.
ΔT = 75.00°C - 15.50°C = 59.50°C
Now, substitute the values into the formula:
q = 175.0g * 4.18 J/g·°C * 59.50°C
Calculating this expression will give you the thermal energy change for this specific case.