Answer:
697 g
Explanation:
Ethanol (C₂H₅OH) and butanoic acid (C₃H₇COOH) react to form ethyl butanoate (C₃H₇COOC₂H₅) and water (H₂O).
C₂H₅OH + C₃H₇COOH → C₃H₇COOC₂H₅ + H₂O
The molar ratio of C₂H₅OH to C₃H₇COOC₂H₅ is 1:1. The moles of C₃H₇COOC₂H₅ produced from 6.00 moles of C₂H₅OH are:
6.00 mol C₂H₅OH × (1 mol C₃H₇COOC₂H₅/1 mol C₂H₅OH) = 6.00 mol C₃H₇COOC₂H₅
The molar mass of C₃H₇COOC₂H₅ is 116.16 g/mol. The mass corresponding to 6.00 mol is:
6.00 mol × (116.16 g/mol) = 697 g
Chlorine (Cl) forms a salt when it is combined with a metal. This element belongs in <span>halogens.
</span>
Elements in the same group have D. Same number of valence electrons.
Answer:
C) H2S
Explanation:
In chemistry, the dissolution of one substance in another is dependent on the magnitude of intermolecular interaction between the two substances. Hence, if two substances do not interact in one way or the other, then one can not dissolve the other.
Let us consider the fact that NH3 is a polar molecule and it is a general principle that like dissolves like. Hence, only H2S which is also a polar molecule can effectively interact with NH3 due to dipole-dipole interaction between the two molecules.
Also, ammonia reacts with hydrogen sulphide as follows;
2NH3 + H2S → (NH4)2S
Hence H2S is more likely to dissolve in NH3.
<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.
