Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
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Explanation:
SADMEP
-2(bx-5) = 16 distribute
-2bx +10 = 16 subtracte
-10 -10
-2bx = 6
divide by -2x (on both sides)
b = -3x
Answer:
785
Explanation:
Molecular. X. W
Weight
8000-16000 0.05 0.03
16000-24000. 0.017. 0.08
24000-32000. 0.22. 0.18
32000-40000. 0.25. 0.35
40000-48000. 0.22. 0.27
48000-56000. 0.09. 0.09
Mean weight X*M. W*M
12000. 600. 240
20000. 3200. 2000
28000. 6720. 5600
36000. 10080. 10800
44000. 8800. 11880
52000. 3640 3640
Total=33040g\mol 36240
Note before repeat molecular weight m= 3*12.01+6*1.008=
42.08g/mol
Degree of polymerization= total W*M/w=33040/42.08 =785
