Answer:
(a) ![V_B=11.68L](https://tex.z-dn.net/?f=V_B%3D11.68L)
(b) ![x_{He}=0.533](https://tex.z-dn.net/?f=x_%7BHe%7D%3D0.533)
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:
![n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol](https://tex.z-dn.net/?f=n_A%3D%5Cfrac%7B0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A298K%7D%7B1.974%20atm%2A6.00L%7D%3D2.063mol)
Thus, since the final pressure is 3.60 bar, we can write:
![P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar](https://tex.z-dn.net/?f=P%3Dx_%7BAr%7DP_A%2Bx_%7BHe%7DP_B%5C%5C%5C%5CP%3D%5Cfrac%7Bn_%7BAr%7D%7D%7Bn_%7BAr%7D%2Bn_%7BHe%7D%7D%20P_A%2B%5Cfrac%7Bn_%7BHe%7D%7D%7Bn_%7BAr%7D%2Bn_%7BHe%7D%7D%20P_B%5C%5C%5C%5C3.60bar%3D%5Cfrac%7B2.063mol%7D%7B2.063mol%2Bn_%7BHe%7D%7D%20%2A2.00bar%2B%5Cfrac%7Bn_%7BHe%7D%7D%7B2.063mol%2Bn_%7BHe%7D%7D%20%2A5.00bar)
The moles of helium could be computed via solver as:
![n_{He}=2.358mol](https://tex.z-dn.net/?f=n_%7BHe%7D%3D2.358mol)
Or algebraically:
![3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol](https://tex.z-dn.net/?f=3.60bar%3D%5Cfrac%7B1%7D%7B2.063mol%2Bn_%7BHe%7D%7D%20%2A%284.0126%2B5.00%2An_%7BHe%7D%29%5C%5C%5C%5C7.314%2B3.60n_%7BHe%7D%3D4.013%2B5.00%2An_%7BHe%7D%5C%5C%5C%5C7.314-4.013%3D5.00%2An_%7BHe%7D-3.60n_%7BHe%7D%5C%5C%5C%5Cn_%7BHe%7D%3D%5Cfrac%7B3.3%7D%7B1.4%7D%3D2.358mol)
In such a way, the volume of the compartment B is:
![V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L](https://tex.z-dn.net/?f=V_B%3D%5Cfrac%7Bn_%7BHe%7DRT%7D%7BP_B%7D%3D%5Cfrac%7B2.358mol%2A0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A298.15K%7D%7B4.935atm%7D%5C%5C%20%20%5C%5CV_B%3D11.68L)
Finally, he mole fraction of He is:
![x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533](https://tex.z-dn.net/?f=x_%7BHe%7D%3D%5Cfrac%7B2.358%7D%7B2.358%2B2.063%7D%5C%5C%20%5C%5Cx_%7BHe%7D%3D0.533)
Regards.
Half-life is the time required for the amount of something to fall to half its initial value. This term is usually used in nuclear physics to describe how quickly unstable atoms undergo decay, or how long stable atoms survive, radioactive decay, and it is also used more generally of any type of exponential or non-exponential decay.
In simpler terms: this is when an isotopes radioactivity is cut in half
Answer:
Explanation:
When an electron jumps from one energy level to a lower energy level some energy is released in the form of a photon.
The difference in energy between the two levels is the energy of the photon and that energy is related to the frequency of the photon by the Einstein - Planck equation:
Where,
- E = energy of the photon,
- h = 6.626×10⁻³⁴ J.s, Planck constant, and
- ν = frequency of the photon.
So, to find the frequency you must first find the energy.
The transition energy can be calculated using the formula:
Where E₀ = 13.6 eV ( 1 eV = 1.602×10⁻¹⁹ Joules) and n = 1,2,3,...
So, the transition energy between n = 4 and n = 3 will be:
- ΔE = - E₀ [ 1/4² - 1/3²] = - 13.6 eV [1/16 - 1/9] = 0.6611. . .eV
- ΔE = 1.602×10⁻¹⁹ Joules/eV × 0.6611... eV = 1.0591 ×10⁻¹⁹ Joules
Now you can use the Einstein - Planck equation:
- ν = 1.0591 ×10⁻¹⁹ J / 6.626×10⁻³⁴ J.s = 1.60×10¹⁴ s⁻¹ (rounded to 3 significant figures).
Answer:
noble gases from what i learnt
Answer: because ch4 is not considered a acid they said it is too weak
Explanation: