Answer:
154 g
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP
At STP, 1 mole of N₂ occupies 22.4 L.
79.5 L × 1 mol/22.4 L = 3.55 mol
Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂
The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.
Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃
The molar mass of NaN₃ is 65.01 g/mol.
2.37 mol × 65.01 g/mol = 154 g
Answer:
221 °C
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 4.1 L
Initial temperature (T₁) = 25 °C
= 25 °C + 273
= 298 K
Final volume (V₂) = 6.8 L
Final temperature (T₂) =?
The final temperature of the gas can be obtained as follow:
V₁ / T₁ = V₂ / T₂
4.1 / 298 = 6.8 / T₂
Cross multiply
4.1 × T₂ = 298 × 6.8
4.1 × T₂ = 2026.4
Divide both side by 4.1
T₂ = 2026.4 / 4.1
T₂ ≈ 494 K
Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:
°C = K – 273
K = 494
°C = 494 – 273
°C = 221 °C
Thus the final temperature of the gas is 221 °C
The mass of iron (III) oxide (Fe2O3) : 85.12 g
<h3>Further explanation</h3>
Given
3.20x10²³ formula units
Required
The mass
Solution
1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mol of Fe₂O₃ :
mass of Fe₂O₃ (MW=160 g/mol)
Answer:
Explanation:
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