The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
Answer:
a. 92.4%
Explanation:
Based on the reaction:
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)
To obtain the percent yield you need to obtain moles of trona and calculate thoeretical moles of Na₂CO₃, and the ratio of obtained moles / theoretical moles of Na₂CO₃ give percent yield, thus:
Moles of trona:
1.00 metric ton × (1x10³kg / 1 metric ton) × ( 1000moles /226.03 kg) = <em>4424 moles</em>
The theoretical moles of Na₂CO₃ that produce 4424 moles of trona are (Based on the reaction, 2 moles of trona produce 3 moles of Na₂CO₃):
4424 moles trona × (3 moles Na₂CO₃ / 2 moles trona) = <em>6636 moles of Na₂CO₃.</em>
The obtained moles of Na₂CO₃:
0.650 metric ton × (1x10³kg / 1 metric ton) × (1000 moles / 105.99kg) = <em>6133 moles</em>
The ratio of obtained moles / theoretical moles gives:
6133 moles / 6636 moles = 0.924 = <em>92.4%</em>
I hope it helps!
Answer:
Nitrogen gas (chemical symbol N) is generally inert, nonmetallic, colorless, odorless and tasteless. Its atomic number is 7, and it has an atomic weight of 14.0067. Nitrogen has a density of 1.251 grams/liter at 0 C and a specific gravity of 0.96737, making it slightly lighter than air.
Explanation:
They like cheese because it is easy to eat and it tases really good
