-130KJ is the standard heat of formation of CuO.
Explanation:
The standard heat of formation or enthalpy change can be calculated by using the formula:
standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation
Data given:
Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ
2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ
CuO + Cu ⇒ Cu2O (-11.3 KJ) ( Formation of Cu2O)
When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.
Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ
So standard heat of formation of formation of Cu0 as:
Cu + 1/2 02 ⇒ CuO
putting the values in the equation
ΔHf = ΔH1 + ΔH2 (ΔH1 + ΔH2 enthalapy of reactants)
heat of formation = -11.3 + (-119.35)
= - 130.65kJ
-130.65 KJ is the heat of formation of CuO in the given reaction.
Answer:
1384 kJ/mol
Explanation:
The heat absorbed by the calorimeter is equal to the heat released due to the combustion of the organic compound. C is the total heat capacity of the calorimeter and Δt is the change in temperature from intial to final:
Q = CΔt = (3576 J°C⁻¹)(30.589°C - 25.000°C) = 19986.264 J
Extra significant figures are kept to avoid round-off errors.
We then calculate the moles of the organic compound:
(0.6654 g)(mol/46.07) = 0.0144432 mol
We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)
(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol
Answer:The following statements are correct: 1,2 and 6
Explanation:
1.The cyclohexane ring adopts a chair conformation in order to minimize its torsional strain. In chair conformation 4 carbon atoms are in one plane 1 carbon atom is above that plane and the other 1 carbon atom is below that plane .This leads to chair conformation in which the bond angles are very close to the ideal tetrahedral angle of 109.5 degrees. The C-C-C bond angle in chair conformation is 110 degrees which is almost equal to the ideal tetrahedral angle.
2. In cyclohexane molecule as the molecule adopts a chair conformation in order to eliminate the torsional strain which would occur if the cyclohexane ring were to be planar. Torsional strain is basically the inter electronic repulsion between the atoms that do not share a bond. So this strain happens on account of eclipsing atoms. In case of eclipse structure there would be a lot of torsional strain. In case of chair conformation all the C-H bonds happen to be completely staggered in nature to eliminate the torsional strain.
3. The ring strain in case of cycloalkanes are dependent upon the number of CH₂ groups present as that would determine the size of the ring and subsequently its structure ,whether the ring would be 5 , 6 or 7 membered .Cyclohexane is a 6 -membered as there are 6CH₂ groups in it and the existence of chair conformation is only for Cyclohexane or for molecules having 6-membered ring . Any change in number of CH₂groups would lead to a different conformational structure.
4.All the bond angles in cyclohexane ring is approximately 110 degrees which is almost equal to the ideal terahedral bond angle. So the bond angles in cyclohexane are optimal.
5.The C-H bonds in cyclohexane are always staggered and never eclipsed in order to reduce there torsional strain.
6.All the bonds in cyclohexane ring are staggered to eliminate the torsional strain. It is quite evident that the cyclohexane ring is completely stable free of the ring strain.So there are no eclipsing bonds present in cyclohexane.
So the statements which are correct 1,2 and 6
I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
