True,I think.(so don't really count on my answer sorry :D)
Sodium phosphide is the inorganic compound with the formula Na3P.
Answer:
5.5 L
Explanation:
Step 1: Given data
- Initial volume (V₁): 6.5 L
- Initial pressure (P₁): 840 mmHg
- Initial temperature (T₁): 84 °C
- Final pressure (P₂): 760 mmHg (standard pressure)
- Final temperature (T₂): 273.15 K (standard temperature)
Step 2: Convert T₁ to Kelvin
We will use the following expression.
K = °C + 273.15
K = 84 °C + 273.15 = 357 K
Step 3: Calculate the final volume of the gas
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂ / T₁ × P₂
V₂ = 840 mmHg × 6.5 L × 273.15 K / 357 K × 760 mmHg = 5.5 L
Answer:
Cl₂ is the limiting reactant.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3Cl₂ —> 2NCl₃
From the balanced equation above,
1 L of N₂ reacted with 3 L of Cl₂.
Finally, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
1 L of N₂ reacted with 3 L of Cl₂.
Therefore, 134 L of N₂ will react with = 134 × 3 = 402 L of Cl₂.
From the calculation made above, we can see that a higher volume (i.e 402 L) of Cl₂ than what was given (i.e 99 L) is needed to react completely with 134 L of N₂.
Therefore, Cl₂ is the limiting reactant and N₂ is the excess reactant.
The awnser to your question would be B
