The heat of formation of nitrogen (II) oxide : +91.3 kJ/mol
<h3>Further explanation</h3>
Given
Reaction
N2(g) + O2(g) + 182.6 kJ → 2 NO(g)
Required
The heat of formation
Solution
In the above reaction, the heat of the reaction is located on the reactant side which indicates that the formation of nitric oxide requires heat (endothermic reaction).
In the above reaction the heat required to form 2 moles of NO, so the heat required for each mole is:
+182.6 kJ : 2 = =+91.3 kJ/mol
Answer:
D-when the aim is to show electron distributions in shells
Explanation:
<u>Given:</u>
Volume of Na2CO3 = 250 ml = 0.250 L
Molarity of Na2CO3 = 6.0 M
Volume of CaF2 = 750 ml = 0.750 L
Molarity of CaF2 = 1.0 M
<u>To determine:</u>
The mass of CaCO3 produced
<u>Explanation:</u>
Na2CO3 + CaF2 → CaCO3 + 2NaF
Based on the reaction stoichiometry:
1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3
Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles
Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles
CaF2 is the limiting reagent
Thus, # moles of CaCO3 produced = 0.750 moles
Molar mass of CaCO3 = 100 g/mol
Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g
Ans: Mass of CaCO3 produced = 75 g
An apple should be cut into 4 equal pieces, then put each slice in a separate container and label accordingly with letters A, B, C, and Control. Put water, ginger ale, and lemon juice into containers A, B, and C respectively but leave the Control untouched. Observe which of the slices in containers A, B, C will stay the same color after the one in control turns brown, if the slice maintains its color then the liquid added prevents an apple slice from browning. The variables are the liquids added and the control is the slice that did not have anything added to it.
The heat of reaction : 50.6 kJ
<h3>Further explanation</h3>
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
N₂(g) + 2H₂(g) ⇒N₂H₄(l)
thermochemical data:
1. N₂H₄(l)+O₂(g)⇒N₂(g)+2H₂O(l) ΔH=-622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ
We arrange the position of the elements / compounds so that they correspond to the main reaction, and the enthalpy sign will also change
1. N₂(g)+H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ x 2 ⇒
2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
Add reaction 1 and reaction 2, and remove the same compound from different sides
1. N₂(g)+2H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2.2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
-------------------------------------------------------------------- +
N₂(g) + 2H₂(g) ⇒N₂H₄(l) ΔH=50.6 kJ