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:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin
Answer:
100 °C, because it is the boiling point of water.
Explanation:
Answer:
Explanation:
Water vapor, carbon dioxide, methane, and other trace gases in Earth's atmosphere absorb the longer wavelengths of outgoing infrared radiation from Earth's surface. These gases then emit the infrared radiation in all directions, both outward toward space and downward toward Earth
The minimum mass of NaHCO3 that must be added to the spill to neutralize the acid is 27.216 grams
<h3>calculation </h3>
write the balanced chemical equation
2NaHCO3 +H2SO4 → Na2SO4 +2H2O +2CO2
find the moles of H2SO4 = molarity x volume in liters
volume in liters = 27/1000=0.027 l
moles is therefore= 0.027 x6=0.162 moles
by use of mole ratio of NaHCO3: H2SO4 which is 2:1 the moles of NaHCO3=0.162 x2=0.324 moles
mass of NaHCO3= moles of NaHCO3 x molar mass of NaHCO3(84g/mol)
= 84g/mol x 0.324=27.216 grams