Answer:
addition polymerization
Explanation:
In addition polymerization, the monomers are simply joined to each other to form a polymer having the same empirical formula as the monomer but of higher relative molecular mass. The monomers in addition polymerization are usually simple unsaturated molecules such as alkenes.
We can deduce the reaction to be an addition polymerization because of the the attachment of n to both the unsaturated monomer and the saturated polymer without the loss of any small molecule. If it was a condensation polymerization, there would have been an accompanying loss of a small molecule such as water.
Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Explanation:
mass of nitrogen = 37.8 g
mass of oxygen = (100-37.8) g = 62.2 g
Using the equation given by Raoult's law, we get:

= partial pressure of
= ?

= total pressure of mixture = 525 mmHg


Total moles = 1.94 + 1.35 = 3.29 moles


Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Answer:
heat flows from the object that has more thermal more energy to the object with less thermal energy
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
0.2g
Explanation:
All radiodecay follows the 1st order decay equation
A = A₀e^-kt
A => Activity at time (t)
A₀ => Initial Activity at time = 0
k => decay constant for isotope
T => time in units that match the decay constant
Half-Life Equation => kt(½) = 0.693 => k = 0.693/34 min = 0.0204min¹
A = A₀e^-kt = (26g)e^-(0.0204/min)(238min) = (26g)(0.0078) = 0.203g ~ 0.2g (1 sig fig).