<h3>

</h3><h3>

</h3><h3>

</h3><h3>

</h3>
<em>Hence</em><em>,</em><em> </em><em>84</em><em>°</em><em>F</em><em> </em><em>is</em><em> </em><em>30</em><em>°</em><em>C</em><em> </em><em>in</em><em> </em><em>Celsius</em><em>.</em><em>.</em>
Answer:
C) 712 KJ/mol
Explanation:
- ΔH°r = Σ Eb broken - Σ Eb formed
- 1/2Br2(g) + 3/2F2(g) → BrF3(g)
∴ ΔH°r = - 384 KJ/mol
∴ Br2 Eb = 193 KJ/mol
∴ F2 Eb = 154 KJ/mol
⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)
⇒ Σ Eb broken = (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol
∴ Eb formed: Br-F
⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r
⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )
⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol
It does not consist of a single element
Hope I helped! ( Smiles )
Answer:
powdered sugar
Explanation:
The higher is the exposed area of sugar, the faster is the dissolution process. Thus, to choose between the different types of sugar, we have to look at the volume occupied by the sugar.
In sugar cubes, the particles of sugar as compacted in a cube, so the particles inside the cube are not exposed to the solvent (water). So, sugar cubes have the slowest dissolution process. Then, in granulated sugar, the particles have more area exposed, so this type of sugar will dissolve faster than sugar cubes. Finally, powdered sugar is composed of tiny particles with more are exposed, so powdered sugar has the fastest dissolution process.
Therefore, powdered sugar will dissolve the fastest.
Answer:
Kindly check the explanation section.
Explanation:
From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.
In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;
∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).
That is the heat of formation of products minus the heat of formation of the reaction g specie(s).
Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.
−382.185 = x - 472.435kj/mol = 90.25 kJ/mol