Answer:
the dog barked is the answer
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Covalent compounds have a low boiling point. All the other answers apply to ionic compounds.
Answer:
Explanation:
<u>1. Molecular chemical equation:</u>
- 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
<u>2. Mole ratios:</u>
- 2 mol KClO₃ : 2 mol KCl : 3 mol O₂
<u>3. Number of moles of KClO₃</u>
- Number of moles = mass in grams / molar mass
- Molar mass of KClO₃ = 122.55 g/mol
- Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol
<u>3. Number of moles of O₂</u>
As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.
- 3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃
- x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂
Round to 3 significant figures: 0.665 mol of O₂ ← answer
Answer:
1360kJ are evolved
Explanation:
When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.
To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:
<em>Moles H2 -Molar mass: 2g/mol-</em>
90g H2 * (1mol / 2g) = 45 moles
<em>Moles O2 -Molar mass: 32g/mol-</em>
90g * (1mol / 32g) = 2.81moles
For a complete reaction of 2.81 moles of O2 are needed:
2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2
As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.
As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:
2.81 moles O2 * (241.8kJ / 1/2moles O2) =
<h3>1360kJ are evolved</h3>
