In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
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
Answer:
a. 53.5 g/mol
b. 80.06 g/mol
c. 133.33 g/mol
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Molar Mass - 1 mol per <em>x</em> grams substance
Explanation:
<u>Step 1: Define</u>
a. NH₄Cl
b. NH₄NO₃
c. AlCl₃
<u>Step 2: Find masses</u>
Molar Mass of N - 14.01 g/mol
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of Cl - 35.45 g/mol
<u>Step 3: Calculate compound masses</u>
Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol
Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol
Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol
