Yo sup??
we can solve this problem by applying Newton's 2nd law
F*t=Δp
p=momentum
pi=mu=1500*30
pf=mv=m*0=0
Therefore
F*3=1500*30
F=15000 N
Hope this helps.
Answer:
Explanation:
The correct answer of this questions is 38°c
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:
last choice
Explanation:
Specific heat of a metal is j/kg-c
It says nothing about melting point or heat of fusion
A = twice B = 2 * j/kg-c B = j/kg-c
if j is constant and mass is the same we can write this more like
A = 2 /C and B = 1/C
or A ~ C/2 and B~ C
now you can see that the temp change in A will be 1/2 that of B
Answer:
Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments
Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments
Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)
Explanation:
Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL
Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL. All produced in three cut sites.
