Methyl methacrylate was polymerized at a mass concentration of 200 g/dm3 in toluene using azobisisobutyronitrile (AIBN) as initi
ator at a mass concentration of 1.64 x 10-2 g/dm3 and a reaction temperature of 60°C. Calculate the initial rate of polymerization and the number average molar mass of poly(methyl methacrylate) formed in the initial stages of the reaction given that the relevant rate coefficients at 60°C are:
Initiator dissociation, kd = 8.5 x 10-6 s-1
Propagation, kp = 833 dm3 mol-1 s-1
Termination, kt = 9.3 x 106 dm3 mol-1 s-1
Transfer to monomer, ktrM = 3.93 x 10-3 dm3 mol-1 s-1
Transfer to solvent, ktrS = 7.34 x 10-3 dm3 mol-1 s-1
Initiator efficiency, f = 0.7 Termination by combination is negligible Density of initial solution of methyl methacrylate in toluene is 860 g/dm3
Initiator dissociation, kd = 8.5 x 10-6 s-1
Propagation, kp = 833 dm3 mol-1 s-1
Termination, kt = 9.3 x 106 dm3 mol-1 s-1
Transfer to monomer, ktrM = 3.93 x 10-3 dm3 mol-1 s-1
Transfer to solvent, ktrS = 7.34 x 10-3 dm3 mol-1 s-1
Initiator efficiency, f = 0.7 Termination by combination is negligible Density of initial solution of methyl methacrylate in toluene is 860 g/dm3
1 answer:
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Answer:
ŷ = -1.618X + 215.636
191
Step-by-step explanation:
From the graph:
The data below could be obtained to produce the best fit line :
Week, x :
0
1
2
3
4
5
6
7
8
9
10
Weight :
216
214
213
210
210
207
206
202
203
202
200
Using a linear regression calculator. The best fit line obtained is :
ŷ = -1.618X + 215.636 ; which shows decrease in weight has the number of weeks grow.
Carters weight 15 Weeks into his diet :
X = 15
ŷ = -1.618(15) + 215.636
= 191.366
= 191 (nearest whole number)
Answer:
Addition property of equality
Step-by-step explanation:
The given expression is :
We need to find the first step to solve this expression.
For this, we add 3/4 to both sides of the equation i.e.
So, we have used the addition property of equality.