equal to 1
Explanation:
All conversion factors used in any calculation is usually a ratio that is equal to 1. This is because when 1 is used to multiply or divide any number, the number stays the same.
- Conversion factors are useful in making the simplification of an expression very simple. They are usually ratios of 1.
- It follows that a conversion factor is a number that when multiplied or divided by itself will stay the same.
Learn more:
conversion factor: brainly.com/question/555814
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Answer:
[CaCl₂·2H₂O] = 1.43 m
Explanation:
Molality is mol of solute / kg of solvent.
Mass of solvent = 40 g
Let's convert g to kg → 40 g / 1000 = 0.04 kg
Let's determine the moles of solute (mass / molar mass)
8.43 g / 146.98 g/mol = 0.057 mol
Molality = 0.057 mol / 0.04 kg → 1.43
<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.
