Answer and Explanation:
<em>A funnel is in the top of the buret and a beaker is positioned underneath the buret:</em> This is correct and is necessary to fill the buret, but the funnel and the beaker has to be removed before the titration starts. The calculation for moles of analyte does not affect.
<em>A solution is being poured from a bottle into the buret via the funnel:</em> Using a funnel helps to fill the burette but it must be removed to filling the buret at 0.0 mL. In this case, the calculation for moles of analyte do not affect.
<em>Adding titrant past the color change of the analyte solution</em>: In this case, an excess of titrant is added, thus the calculation for moles of anality will be higher than it should be.
<em>Recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M</em>: In this case, the titrant is considered more concentrated than it is hence, the calculation for moles of anality will be higher than it should be.
<em>Spilling some analyte out of the flask during the titration</em>: The excess of titrant spilled out of the flask higher up the volume of titrant measured. Therefore, the calculation for moles of anality will be higher than it should be.
<em>Starting the titration with air bubbles in the buret</em>: The air inside the burette occupies measured volume, thus the volume of titrant measured will be higher than the real volume spilled in the flask. Hence the calculation for moles of anality will be higher than it should be.
<em>Filling the buret above the 0.0 mL volume mark</em>: Some volume of titrant will be spilled inside the flask but will no be measured since the buret measures the titrant below the 0.0mL mark, thus the calculation for moles of anality will be lower than it should be.
The volume of the 0.130 M sulfuric acid, H₂SO₄ required to react completely with 65.9 g sodium hydroxide, NaOH is 6.34 L
- We'll begin by calculating the number of mole of in 65.9 g sodium hydroxide, NaOH. This can be obtained as follow:
Mass of NaOH = 65.9 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
<h3>Mole of NaOH =?</h3>
Mole = mass / molar mass
Mole of NaOH = 65.9 / 40
<h3>Mole of NaOH = 1.6475 mole</h3>
- Next, we shall determine the number of mole of H₂SO₄ needed to react with 1.6475 mole of NaOH. This can be obtained as follow:
2NaOH + H₂SO₄ —> Na₂SO₄ + 2H₂O
From the balanced equation above,
2 moles of NaOH reacted with 1 mole of H₂SO₄.
Therefore,
1.6475 mole of NaOH will react with = 
= 0.82375 mole of H₂SO₄.
- Finally, we shall determine the volume of 0.130 M sulfuric acid, H₂SO₄ required for the reaction.
Molarity of H₂SO₄ = 0.130 M
Mole of H₂SO₄ = 0.82375 mole
<h3>Volume of H₂SO₄ =? </h3>
Volume = mole / Molarity
Volume of H₂SO₄ = 0.82375 / 0.130
<h3>Volume of H₂SO₄ = 6.34 L </h3>
Therefore, the volume of the 0.130 M sulfuric acid, H₂SO₄ required for the reaction is 6.34 L
Learn more: brainly.com/question/7882345
It is part of oceans, air, rocks, soil and all living things. Carbon doesn't stay in one place. It is always on the move! In the atmosphere, carbon is attached to oxygen in a gas called carbon dioxide.
Hope this helps.
Examining the given reaction:
Li2O + H2O ........> 2LiOH
we find that, 1 mole of Li2O is required to react with one mole of H2O in order to produce two moles of LiOH.
Therefore, the ration between the required Li2O and H2O is 1:1
Based on this, 2.2 moles of H2O (water) are required to react with 2.2 moles of Li2O
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