Answer: The energy of a Br–F bond is 110 kJ/mol
Explanation:
The balanced chemical reaction is,
![Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)](https://tex.z-dn.net/?f=Br_2%28g%29%2B3F_2%28g%29%5Crightarrow%202BrF_3%28g%29)
The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E%28reactant%29%5D-%5Csum%20%5Bn%5Ctimes%20B.E%28product%29%5D)
![\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BBr_2%7D%5Ctimes%20B.E_%7BBr_2%7D%29%2B%28n_%7BF_2%7D%5Ctimes%20B.E_%7BF_2%7D%29%20%5D-%5B%28n_%7BBrF_3%7D%5Ctimes%20B.E_%7BBrF_3%7D%29%5D)
![\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BBr_2%7D%5Ctimes%20B.E_%7BBr-Br%7D%29%2B%28n_%7BF_2%7D%5Ctimes%20B.E_%7BF_F%7D%29%20%5D-%5B%28n_%7BBrF_3%7D%5Ctimes%203%5Ctimes%20B.E_%7BBr-F%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20193%29%2B%283%5Ctimes%20155%29%5D-%5B%282%5Ctimes%203%5Ctimes%20B.E_%7BBr-F%7D%29%5D)
![B.E_{Br-F}=110kJ/mol](https://tex.z-dn.net/?f=B.E_%7BBr-F%7D%3D110kJ%2Fmol)
Thus the energy, in kJ/mol, of a Br–F bond is 110
Answer:
troposphere
he troposphere begins at the Earth's surface and extends up to the stratosphere. It contains roughly 80 percent of the mass of the entire atmosphere and is where most familiar weather patterns form.
Answer:
I. FALSE
II. FALSE
III. TRUE
IV. TRUE
Explanation:
<em>I. A set of sp² orbitals can be thought of as one s orbital one-third of the time and two p orbitals two-thirds of the time.</em> FALSE. 1 s orbital hybridize with 2 p orbitals to form 3 sp² permanent orbitals.
<em>II. A set of sp orbitals can accommodate a maximum of six electrons. </em>FALSE. sp orbitals, as any orbital, can accommodate a maximum of 2 electrons.
<em>III. The orbitals resulting from sp³d² hybridization are directed toward the corners of an octahedron.</em> TRUE. 1 s orbital, 3 p orbitals and 2 d orbitals hybridize to form 6 sp³d² orbitals, which are directed toward the corners of an octahedron.
<em>IV. A set of sp³ orbitals results from the mixing of one s orbital and three p orbitals.</em> TRUE. 1 s orbital hybridize with 3 p orbitals to form 4 sp³ orbitals.
Q must be supplied = 119523.3~J
<h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
So from the question :
Q to the system(Q supplied) = Q water + Q steel vessel(Q gained)
![\tt Q~supplied=m.c\Delta T(water)+m.c\Delta T(vessel)\\\\Q~supplied=375\times 4.18\times (100-35)+532\times 0.51(100-35)\\\\Q~supplied=101887.5+17635.8=119523.3~J](https://tex.z-dn.net/?f=%5Ctt%20Q~supplied%3Dm.c%5CDelta%20T%28water%29%2Bm.c%5CDelta%20T%28vessel%29%5C%5C%5C%5CQ~supplied%3D375%5Ctimes%204.18%5Ctimes%20%28100-35%29%2B532%5Ctimes%200.51%28100-35%29%5C%5C%5C%5CQ~supplied%3D101887.5%2B17635.8%3D119523.3~J)
Answer:
½N 2(g) + O 2(g) + 8.1 kcal → NO 2(g)
½N 2(g) + ½O 2(g) → NO(g), ΔH = +21.6 kcal/mole
NH 3(g) → ½N 2(g) + 3/2H 2(g), ΔH = +11.0 kcal/mole
General Formulas and Concepts:
<u>Thermochemistry</u>
- Endothermic vs Exothermic
- Enthalpy
Explanation:
An endothermic reaction would have heat or energy gained in the final stage of the reaction.
This means that our enthalpy would be positive and heat/energy would be added as a reactant.
Out of all the answer choices, we see that only these have heat/energy added in the reactant side or the enthalpy being positive:
½N 2(g) + O 2(g) + 8.1 kcal → NO 2(g)
½N 2(g) + ½O 2(g) → NO(g), ΔH = +21.6 kcal/mole
NH 3(g) → ½N 2(g) + 3/2H 2(g), ΔH = +11.0 kcal/mole
Topic: AP Chemistry
Unit: Thermochemistry