1. a. FeS + 2HCl --> FeCl2 + H2S
b. This is a double- replacement reaction. (The elements "switch partners".)
2. a.2Na + F2 --> 2NaF
b. This is a composition/synthesis reaction. (The two reactants are combining or synthesizing to make one product.)
3. a. 2HgO --> 2Hg + O2
b. This is a decomposition reaction. (The single reactant is breaking down or "decomposing" into multiple reactants.)
4. Hydrogen gas reacts with oxygen gas to create water in a composition/synthesis reaction.
I hope these answers will help you! If you need any explanation, ask and hopefully I can get back to you. ;)
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Intermolecular forces are the forces of attraction or repulsion which act between neighboring particles (atoms, molecules, or ions ). These forces are weak compared to the intramolecular forces, such as the covalent or ionic bonds between atoms in a molecule.
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK
</span>
Now, energy required for melting of ICE = <span> 334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC = 4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span> 2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2) + (3)
= 1753.5 + 2195.18 + 11849.25
= </span><span>15797.93 J
</span><span> = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
