<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.
Answer:
The Answer would be B (Materials can be seen in picometers which are trillionths of a meter or the size atoms.
Explanation:
A pure crystalline substance is a substance with an almost perfect regular and periodic pattern in a solid state. This makes this type of substance a hard one compared to an amorphous substance which is soft because of the irregular pattern within.
<h3>The enthalpy of the reaction : C.226.73 kJ</h3><h3>Further explanation</h3>
Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
Delta H reaction (ΔH) is the difference between product enthalpy and reactant enthalpy (Hp-Hr)
Reaction :
2C + H2 -> C2H2
(ΔH) reaction = (ΔH) product - (ΔH) reactants
(ΔH) reaction = (ΔH) C2H2 - ((ΔH) C + (ΔH) H2)
for free element (ΔH) it has zero value, so (ΔH) C and (ΔH) H2 have zero value
From the available data: Hf (kJ / mol) C2 H2 (g)= 226.73
then
(ΔH) reaction = ΔH) C2H2 - ((ΔH) C + (ΔH) H2)
(ΔH) reaction = 226.73 - (0 + 0)
(ΔH) reaction = 226.73 (answer C)
<h3>Learn more</h3>
(ΔH) reaction
brainly.com/question/1889660
Keywords : (ΔH) reaction, C2H2
Answer:
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<em>nakalimutan</em><em> </em><em>ko</em><em> </em><em>po</em><em> </em><em>sorry</em><em> </em><em>talaga</em>
Explanation:
