A reaction known to release 2.00 kJ of heat takes place in a calorimeter contain
ID: 916548 • Letter: A
Question
A reaction known to release 2.00 kJ of heat takes place in a calorimeter containing 0.200 L of solution and the temperature rose by 4.46 degree C. When 100 mL of nitric acid and 100 mL of sodium hydroxide were mixed in the same calorimeter, the temperature rose 2.01 degree C. What is the heat output for the neutralization reaction? Assume the heat capacity of the solutions is the same as water. When the electron in a hydrogen atom falls from the n = 3 energy level to the ground state energy level, a photo is emitted. An electron having this wavelength would have a velocity of IF the average speed of a carbon dioxide molecule is 410 m s^-1 at 25 degree C, what is the average speed for a molecule of methane at the same temperature?Explanation / Answer
7.
A reaction known to release 2.00kJ of heat takes place in a calorimeter containing 0.2L of solution and
the temperature rose by 4.46C.
When 100 mL of nitric acid and 100 mL of NaOH were mixed in the same calorimeter, the temperature rose 2.01C.
What is the heat output for the neutralization reactions?
Assume the heat capacity of the solutions is the same as water
Answer: (E)
Heat released in the first neutralization process, q1 = (mw*cpw+cpcal)*dT1
where mw = mass of solution, cpw = specific heat of solution, cpcal = heat capacity of calorimeter, dT1 = change in temperature = 4.46C
q1 = 4.46*(mw*cpw+cpcal) = 2.0 kJ
(mw*cpw+cpcal) = 2/4.46 kJ/C
Heat released in the second neutralization reaction, q2:
mw, mass of solution remains same as volume is same 100ml+100ml = 200 ml = 0.2 q2 = m*cp*dT2
cpw and cpcal also remains same
only dT changes to dT2 = 2.01C
q1 = 2.01*(mw*cpw+cpcal)=2.01*2/4.46 = 0.901 kJ
Answer: (E)
8.
When the electron in a hydrogen atom falls from the n=3 energy level to the ground energy level, a photon is emitted.
An electron having this wavelength would have a velocity of
Answer:(A)
For finding wavelength we use Rydberg's constant, RH = 1.09E7 m-1
Wavelength, lambda due to transition from n = 3 to n = 1(ground level),
1/lambda = -RH*(1/3^2-1/1^2) = 8/9*RH = 8/9*1.09E7 = 0.97E7 m-1
lambda = 1/0.97E7 = 1.032E-7 m
The deBrogile wavelength of electron is related to velocity v by
v = h/(m*lambda)
mass of electron, m = 9.1E-31 Kg
Planck's constant, h = 6.626E-34 J·s
Velocity, v = h/(m*lambda) = 6.626E-34/(9.1E-31*1.032E-7) =7.1 m/s(A)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.