Answer:
So you need to look for the average speed of each car, which would allow you to then enter the numbers.
Explanation:
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The enthalpy change in a reaction is given by-
ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants
This can be expressed in terms of bond energy as-
ΔH°rxn = BEreactants - BEproducts
Therefore, the calculated bond energy according to the above equation will be-
ΔH°rxn = [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) ] - [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) = 0 kJ/mol
<h3>What is enthalpy change?</h3>
Enthalpy change is a measure of the energy emitted or consumed in a reaction. This can be determined using the following equation which involves standard enthalpy of reactant and product formation:
ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants
<h3>What is bond energy?</h3>
Bond energy is defined as the amount of energy needed to dissociate a mole of molecules into their individual atoms.
Learn more about the Enthalpy Change here:
brainly.com/question/14047927
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Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
