Enthalpy Question Urgent?

Question

U(s) + 3F2 (g) --> UF6 Delta H=-2113KJ

How much heat is released when 1.00 Kg of uranium reacts with flourine?

N2H4 (g) + O2 (g) --> N2 (g) + 2H20 (I) Delta H=-628 KJ

How many grams of Hydrazine (N2H4) would be required to produce 963 KJ of heat?

If the reaction was at standard temperature and pressure, how many liters of N2 gas would evolve if the amount of N2H4 found in part (a) was burned?

Only answer if you actually know what your doing and you actually have an answer

Also, please elaborate

Answer 1

By the first equation, we know that 1 mole of uranium reacting with fluorine yields -2113 kJ. To find out how much heat is produced when reacting 1.00 kg of uranium, you first must find how many moles of uranium are in 1.00 kg.

Looking at a periodic table, I see that uranium has an atomic mass of 238.029 grams per mole.

1.00 kilograms * (1000 grams / 1 kilogram) * (1 mole / 238.029 grams)
= 4.201 moles uranium

4.201 moles * (-2113 kJ / 1 mole)
= 8877 kJ

The second question is solved in a similar manner, except you work the steps backwords. You start off knowing how much heat is produced, and by equating heat produced with number of moles, then equating number of moles with mass, then equating mass with volume, you find out the volume (liters) of gas are needed.

Answer 2

I assume deltaH is expressed as KJ per mole. The negative sign indicates the reaction releases energy (Exothermic)


Divide 1000 g of U by its molecular mass, then multiply it by the deltaH, the result will be negative indicating a release of heat by the reaction.


Now, divide 963 KJ by 628 KJ/mole. You will get the moles of hydrazine needed. Multiply by hydrazine molecular weight and you'll get the mass of hydrazine needed.

Finally, use the ideal gas law to estimate the volume of N2 produced from the fact that moles of N2 = moles of Hydrazine.

There will be a reduction in the volume because there will be half of moles of gas than originally, but this is a different story.

Answer 3

a mol of ice is 18g and so 6000J/mol==333J/g; 330 ml of beverage have a mass of 330g warmth loss from beverage=330x25x4.18=warmth absorbed by ice=333xX X=103.6g in the different answer there is an blunders of ten on the warmth calculated for between the stairs

Answer 4

I honestly wish I was smart enough to answer this question,LOL.