Explanation:
Part of life is the role we certainly play during our role in life.Like a quote of shakespeare,we have our roles in earth to perform or we are like a chess whom we don't want to waste a move.
Art of life signifies that our life is wonderful with mysteries and hypotenic beauty and wonders and with different moments in a journey.
<em>Keep</em><em> </em><em>smiling </em><em>and</em><em> </em><em>hope</em><em> </em><em>u</em><em> </em><em>r</em><em> </em><em>satisfied </em><em>with</em><em> </em><em>my</em><em> </em><em>answer</em><em>.</em><em>Have</em><em> </em><em>a</em><em> </em><em>great</em><em> </em><em>day</em><em> </em><em>:</em><em>)</em>
Don’t eat or drink in labs
Dress for the lab; don’t wear open toed shoes
Dispose of lab waste properly
No horse play
Don’t taste or sniff things in the lab
Tire your hair back
<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.
