Answer:
E = 1.33 MeV = 2.13 x 
J
v = wavelength = E / h = 2.13 x / 6.626 x 
= 3.2 x 
m
f = frequency = c / 3.2 x m = 3 x 
/ 3.2 x = 9.375 x Hz
Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>
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Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>
<em></em>
W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work
Answer: Density is defined as the ratio between mass and volume or mass per unit volume.
Explanation:
<span>Water soaks into it easily and during freeze-thaw cycles would be more likely to break apart because of the water freezing and thawing inside the rock. The water freezes and expands inside the rock, causing the surface to break off in flakes from the pressure of the expanding ice inside.</span>
Answer:
1) Endothermic.
2) 
3) 
Explanation:
Hello there!
1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.
2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:
3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:
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