The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A
Answer:
d 2,4,3
Explanation:
2 Al2O3 ----> 4 Al + 3 O2
Answer:
749 grams CO₂
Explanation:
To find the amount of carbon dioxide produced, you need to (1) convert grams C₃H₈ to moles C₃H₈ (via molar mass from periodic table), then (2) convert moles C₃H₈ to moles CO₂ (via mole-to-mole ratio via reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The desired unit should be in the numerator. The final answer should have 3 significant figures because the given value (250. grams) has 3 sig figs.
Molar Mass (C₃H₈): 3(12.01 g/mol) + 8(1.008 g/mol)
Molar Mass (C₃H₈): 44.094 g/mol
1 C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O
Molar Mass (CO₂): 12.01 g/mol + 2(16.00 g/mol)
Molar Mass (CO₂): 44.01 g/mol
250. g C₃H₈ 1 mole C₃H₈ 3 moles CO₂ 44.01 g
------------------ x ---------------------- x ---------------------- x -------------------- =
44.094 g 1 mole C₃H₈ 1 mole CO₂
= 749 grams CO₂
I think the best answer to this is when the object is in high temperature and intense pressure that could make the molecule move and active. I hope you are satisfied with my answer and feel free to ask for more if you have further question and clarification.
Answer: the answer is C oxidizing