Answer:
The hydrogen ion concentration in a solution, [H+], in mol L-1, can be calculated if the pH of the solution is known.
pH is defined as the negative logarithm (to base 10) of the hydrogen ion concentration in mol L-1 pH = -log10[H+] ...
[H+] in mol L-1 can be calculated using the equation (formula): [H+] = 10-pH
Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
As the given reaction is as follows.
(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.
Thus, formation of 
will increase.
- (b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.
Hence, formation of will decrease with decrease in volume.
- When we increase the mount of
then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.
Thus, we can conclude that formation of will increase then.
The steam releases 39.9 kJ when it condenses..
The steam condenses and transfers its energy to the skin.
<em>q = m</em>Δ<em>H</em>_cond = 17.7 g × (-2257 J/1 g) = -39 900 kJ = -39.9 kJ
The negative sign shows that the steam is releasing energy
<h3>
Answer:</h3>
2.809 L of H₂SO₄
<h3>
Explanation:</h3>
Concept tested: Moles and Molarity
In this case we are give;
Mass of solid sodium hydroxide as 13.20 g
Molarity of H₂SO₄ as 0.235 M
We are required to determine the volume of H₂SO₄ required
<h3>First: We need to write the balanced equation for the reaction.</h3>
- The reaction between NaOH and H₂SO₄ is a neutralization reaction.
- The balanced equation for the reaction is;
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
<h3>Second: We calculate the umber of moles of NaOH used </h3>
- Number of moles = Mass ÷ Molar mass
- Molar mass of NaOH is 40.0 g/mol
Moles of NaOH = 13.20 g ÷ 40.0 g/mol
= 0.33 moles
<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
- From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
- Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
- Thus, Moles of H₂SO₄ = moles of NaOH × 2
= 0.33 moles × 2
= 0.66 moles of H₂SO₄
<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
- When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
- That is; Volume = Number of moles ÷ Molarity
In this case;
Volume of the acid = 0.66 moles ÷ 0.235 M
= 2.809 L
Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.
