Answer:
solid
Explanation:
they are all stuck together
40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.
<em>Step 1</em>. Calculate the pOH of the solution
pOH = 14.00 – pH = 14.00 -13 = 1
<em>Step 2</em>. Calculate the concentration of NaOH
[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH
<em>Step 4</em>. Calculate the mass of NaOH
Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH
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
Answer:
A water molecule can react with the carbonyl group of an aldehyde or a ketone to form a substance known as a carbonyl hydrate, as shown in the first reaction below. The carbonyl hydrates usually form a very small percentage of the molecules in a sample of a specific aldehyde or ketone.
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