Answer:
Answers with detail are given below
Explanation:
1) Given data:
Mass of Rb₃Rn = 76.19 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 478.43 g/mol
Number of moles = 76.19 g/ 478.43 g/mol
Number of moles = 0.16 mol
2) Given data:
Mass of FrBi₂ = 120.02 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 640.96 g/mol
Number of moles = 120.02 g/640.96 g/mol
Number of moles = 0.19 mol
3) Given data:
Mass of Zn₂F₃ = 88.24 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 187.73 g/mol
Number of moles = 88.24 g/ 187.73 g/mol
Number of moles = 0.47 mol
4) Given data:
Number of moles of Sb₄Cl = 1.20 mol
Mass of Sb₄Cl = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 522.49 g/mol
Mass = Number of moles × molar mass
Mass = 1.20 mol × 522.49 g/mol
Mass = 626.99 g
Answer:
7.03 g
Explanation:
Step 1: Write the balanced synthesis reaction
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
Step 2: Calculate the moles corresponding to 32.5 g of N₂
The molar mass of N₂ is 28.01 g/mol.
32.5 g × 1 mol/28.01 g = 1.16 mol
Step 3: Calculate the number of moles of H₂ needed to react with 1.16 moles of N₂
The molar ratio of N₂ to H₂ is 1:3. The moles of H₂ needed are 3/1 × 1.16 mol = 3.48 mol.
Step 4: Calculate the mass corresponding to 3.48 moles of H₂
The molar mass of H₂ is 2.02 g/mol.
3.48 mol × 2.02 g/mol = 7.03 g
Answer:
ΔH° of the reaction is -747.54kJ
Explanation:
Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.
Using the reactions:
<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ
<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ
<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ
<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ
<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ
The sum of 4×(4) + (5) gives:
4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = -74.81 kJ
×4 - 890.3 kJ = -1189.54kJ
Now, this reaction - 4×(1) gives:
4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)
ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>
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Thus <em>ΔH° of the reaction is -747.54kJ</em>
Answer:
1.225g
Explanation:
From the question given,
Volume of diamond = 0.35cm^3
Density of diamond = 3.5g/cm^3
Mass of diamond =?
Density = Mass /volume
Mass = Density x volume
Mass = 3.5 x 0.35
Mass of diamond = 1.225g