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
Answer:
The reaction will be slower because the leaving group will be poorer.
Explanation:
One Important thing to put at the back of our mind is that weak bases are good leaving groups. Another thing to take note is that in the halogen series(which is our main subject in this question) as you go down the group the greater or the more heavy the halide is and the heavier halides are more stable that is Bromine will be more stable than chlorine.
Now, we are now told that the bromo halide is been replaced by the chloro halide which means that the rate of Reaction will surely decrease because the leaving group. The cl^- is a stronger base.
Answer:
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Explanation:
Step 1: Data given
tetraphosphorus decoxide = P4O10
Molar mass of P4O10 = 283.89 g/mol
Mass of P4O10 = 4.5 grams
Number of Avogadro = 6.022 * 10^23 / mol
Step 2: Calculate moles of P4O10
Moles P4O10 = mass P4O10 / molar mass P4O10
Moles P4O10 = 4.5 grams / 283.89 g/mol
Moles = 0.016 moles
Step 3: Calculate moles of P
For 1 mol P4O10 we have 4 moles of phosphorus
For 0.016 moles P4O10 we have 4*0.016 = 0.064 moles P
Step 4: Calculate number of P atoms
Number of P atoms = moles P * number of Avogadro
Number of P atoms = 0.064 moles * 6.022*10^23
Number of P atoms = 3.85 * 10^22 atoms
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Answer:
In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.
Explanation:
In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.
