1) A 50.0 gram sample of water is heated from 21.5 o C to 26.21 o C. How many Jo

Question

1) A 50.0 gram sample of water is heated from 21.5 oC to 26.21 oC. How many Joules of heat were added to this solution?

2) A 15.25 gram sample of copper initially at 92 degrees is added to a container containing water. The final temperature of the metal is 26.21 oC. What is the total amount of energy in Joules added to the water? What was the energy lost by the metal?

3) Mixing 25.0 mL of 1.2 M HClO4 and 25.0 mL of 1.1 M NaOH were mixed. The temperature of the initial solution was 22.4 oC. Assuming a Heat of Neutralization of 55.8 KJ/mol, what would the final temperature be if the specific heat for this solution is 4.032 J/g o?

Explanation / Answer

2 Molecular wt gas = .3167 grams times .082 times 300 degrees K over .,987 atm times .170 Liters volume

Molecular wt. gas = 46.43

Next assume you have l00 grams of this compound to work with, therefore you will have

52.13 grams Carbon, l3.15 grams H and 34.72 grams Oxygen

Then divide each of these element masses by their respective mass per mole to convert them into moles.

52.13 grams C over l2 grams/mole = 4.344 moles C

l3.15 grams H over l.008 grams/mole = 13.05 moles H

34.72 grams O over l6 grams/mole = 2.17 moles O

Next try to convert these mole values into whole number ratios by dividing all three values by the smallest value

4.34 moles C over 2.17 = 2 moles C

13.05 moles H over 2.17 = 6.01 moles H

2.17 moles O over 2.17 = one mole O so the empirical (lowest ratio between atoms is C2H6O

and the total mass of all the atoms in this empirical or simplest formula = 46

We calculated the actual molecular wt. of this substance above to be 46, so the empirical or simplest formula must also be the actual molecular formula.

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