Answer:
Option-B (2-methylpropene)
Explanation:
The reaction scheme is attached below,
In first step the alkene acts a nucleophile and adds H⁺ across double bond yielding a stable tertiary carbocation.
In the second step the oxygen atom of methanol acts as nucleophile and attacks the positive charge carrying carbon atom resulting in the formation of t-butyl methyl ether.
Answer:
It would get <u>colder</u>
Explanation:
The lattice energy is the energy involved in the disruption of interactions between the ions of the salt. In this case, we have: ΔHlat = 350 kJ/mol > 0, so it is an endothermic process (the energy is absorbed).
The solvation energy is the energy involved in forming interactions between water molecules and the ions of the salt. In this case, we have: ΔHsolv = 320 kJ/mol > 0, so it is an endothermic process (the energy is absorbed).
The dissolution process involve both processes: the disruption of ion-ion interactions of the salt and the solvation process. Thus, the enthalphy change (ΔHsol) in the preparation of the solution is calculated as the addition of the lattice energy and solvation energy:
ΔHsol= ΔHlat + ΔHsolv = 350 kJ/mol + 320 kJ/mol = 370 kJ/mol
370 kJ/mol > 0 ⇒ endothermic process
Since the preparation of the solution is an <u>endothermic</u> process, it will absorb energy from the surroundings, so <u>the solution would get colder</u>.
Answer:
0.41kg/sec
Explanation:
PV= nRT
Given : V= 505 L
P=0.88 atm
R= 0.08206 Latm/K*mol
T= 172 .0C = 172+273 = 445 K
n = PV /RT = 0.88 * 505 / 0.08206 * 445 = 12.17 moles per sec of N2 are consumed
As per reaction : N2 + 3H2 ----> 2NH3
1 mole N2 is consumed to produce 2 moles NH3
moles of NH3 produced per sec :
(2 moles NH3/1mol N2) * 12.17 moles N2 = 24.34 moles NH3 per sec
grams of NH3 produced per sec =
24.34 moles NH3 per sec * molar mass NH3 = 24.34 moles NH3 per sec * 17.031 g/mol = 414.5 g NH3 per sec
rate in Kg/sec = 414.5 g NH3 per sec * (1kg /1000g) = 0.4145 Kg/sec
= 0.41kg/sec
Carbon source is the answer for ur question
