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Peter V.
Graduate Student
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Chemistry
TutorMe
Question:

Why Does Ice Float?

Peter V.
Answer:

A substance floats if it is less dense, or has less mass per unit volume, than other components in a mixture. Water reaches its maximum density at 4°C (40°F). As it cools further and freezes into ice, it actually becomes less dense because of hydrogen bond. The hydrogen bonds adjust to hold the negatively charged oxygen atoms apart. This produces a crystal lattice, which is commonly known as 'ice'. Ice floats because it is about 9% less dense than liquid water.

Calculus
TutorMe
Question:

What is the slope of the line tangent to the graph of the function f(x)=ln(sin(x)+3) at the point where x=0 ?

Peter V.
Answer:

To find the slope of the tangent line of a curve, we first need to determine the derivative of function f(x): f'(x)= cos(x)/ (sin(x)+3). At the point where x=0, the slope of the tangent line is f'(0)= cos(0)/(sin(0)+3)= 1/(0+3) =1 /3. Thus, the final answer is 1/3.

Chemical Engineering
TutorMe
Question:

The rate constant for a second-order reaction at 300 K is 2.5 and the activation energy is 20,000 J/mole. a) What are the units of the rate constant? b) At what temperature would the rate be triple (of the rate at 300K) for this reaction?

Peter V.
Answer:

a) Reaction: A->P, Reaction Rate: d[A]/dt = -k*[A]^2. The unit of d[A]/dt is M/s (Molarity/second) while the unit of [A] is M (Molarity). For the above reaction rate, both left hand side and right hand side of this rate must have the same unit, which means [M/s]=[k]*[M]^2. Thus, the rate constant k must have the units: [k]= [M/s]/[M^2]= 1/[Ms] or [M.s]^-1. b) Applying the Arrhenius equation, we have: k(T)=k0*exp(-Ea/(R*T)), which Ea is the activation energy (20,000 J/mole), R is gas constant (8.314 J/mole-K) and T is temperature in Kelvin (K). At 300K, k(300)=k0*exp(-20,000/(8.314*300))=3.2928e-4*k0. At the temperature when the rate is triple of the rate at 300K, we have: k(T)=3* k(300K)= 9.8784e-4 *k0 (1) Also, from Arrhenius equation k(T)= k0*exp(-20,000/(8.314*T)) (2) Combine (1) &(2), we have: 9.8784e-4 *k0 = k0*exp(-20,000/(8.314*T)) or ln(9.8784e-4)= -2405.6/T or T= 347.63 K. Therefore, at T=347.63 K, the reaction rate is triple of the rate at 300K.

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