Ammonia that is used in the home has a pH of 11.9. What type of solution is ammonia?
1 answer:
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2-bromo-1-chloro-4-nitrobenzene is being synthesized in following sequence:
Step 1: Chlorination of Benzene:
This is Halogenation reaction of benzene. In this step benzene is reacted with Chlorine gas in the presence of lewis acid (i.e. FeCl₃). This results in the formation of Chlorobenzene as shown in red step below.
Step 2: Nitration of Chlorobenzene:
The chlorine atom on benzene has a ortho para directing effect. Therefore, the nitration of chlorobenzene will yield para nitro chlorobenzene as shown in blue step below.
Step 3: Bromination of 1-chloro-4-nitrobenzene:
In this step bromination is done by reacting bromine in the presence of lewis acid. The chlorine being ortho para directing in nature and nitro group being meta directing in nature will direct the incoming Br⁺ (electrophile) to the desired location. Hence, 2-bromo-1-chloro-4-nitrobenzene is synthesized in good yield.
Answer:
Q(total) = 283Kj
Explanation:
5 Heat Transitions …
Specific Heats => c(s) = 0.50cal/g∙⁰C, c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C
Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g
Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.
Q = mcΔT for temperature change regions of the heating curve (single phase only)
Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)
Solid (ice) => Melting Pt => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal
Melting (s/l) => Liquid (water) => Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal
Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal
Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal
Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal
Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)
= 478cal +7240cal + 9050 cal + 48,870cal + 2036cal
= 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj
Answer :
<h2>Multicellular organisms</h2>Explanation :
Coyotes are multicellular organisms.
Trees are multicellular organisms. They are made up of eukaryotic cells, which are complex cells full of organelles.
