It is not good conductors of electricity or heat!
meat-eating carnivores have teeth for tearing and skulls capable of biting with great force, while the plant-eating herbivores have teeth and skulls equipped to grind tough vegetation
Answer:
194.6 mL of SO₂
Explanation:
The reaction that takes place is:
P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)
<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:
- 23.8 °C → 23.8 + 273.15 = 296.95 K
- 747 torr → 747/760 = 0.983 atm
We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):
0.576 g P₄S₃ *
= 7.85 * 10⁻³ mol SO₂ = n
PV=nRT
0.983 atm * V = 7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K
V = 0.1946 L
- Finally we convert L into mL:
0.1946 * 1000 = 194.6 mL
Answer:
Explanation:
Given that:
The flow rate Q = 0.3 m³/s
Volume (V) = 200 m³
Initial concentration
= 2.00 ms/l
reaction rate K = 5.09 hr⁻¹
Recall that:







where;







Thus; the concentration of species in the reactant = 102.98 mg/l
b). If the plug flow reactor has the same efficiency as CSTR, Then:
![t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]](https://tex.z-dn.net/?f=t%20_%7BPFR%7D%20%3D%20%5Cdfrac%7B1%7D%7Bk%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7BC_o%7D%7BC_e%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7Bk%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7BC_o%7D%7BC_e%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B5.09%7D%20%5CBig%20%5B%20In%20%28%20%5Cdfrac%7B200%7D%7B102.96%7D%29%20%5CBig%20%5D)
![\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7BV_%7BPFR%7D%7D%7BQ_%7BPFR%7D%7D%20%3D0.196%20%5CBig%20%5B%20In%20%28%201.942%29%20%5CBig%20%5D)





The volume of the PFR is ≅ 140 m³
1:2
The ratio of carbon, hydrogen, and oxygen in most carbohydrates is 1:2:1. This means for every one carbon atom there are two hydrogen atoms and one...