Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
Answer:
V₂ = 106.5 mL
Explanation:
Given data:
Initial volume =200 mL
Initial pressure = 2 atm
Initial temperature = 35 °C (35 +273 = 308 K)
Final temperature = 55°C (55+273 = 328 K)
Final volume = ?
Final pressure = 4 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2 atm ×200 mL × 328 K / 308 K ×4 atm
V₂ = 131200 atm .mL. K / 1232 K.atm
V₂ = 106.5 mL
Answer:
[Ar] 3d10 4s2 4p4
Explanation:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
Explanation:
It is given that,
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,
The amount of energy change during the transition is given by :
![\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5CDelta%20E%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
And
![\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5Cdfrac%7Bhc%7D%7B%5Clambda%7D%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
Plugging all the values we get :
![\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5](https://tex.z-dn.net/?f=%5Cdfrac%7B6.63%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7B3745%5Ctimes%2010%5E%7B-9%7D%7D%3D2.179%5Ctimes%2010%5E%7B-18%7D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C%5Cdfrac%7B5.31%5Ctimes%2010%5E%7B-20%7D%7D%7B2.179%5Ctimes%2010%5E%7B-18%7D%7D%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C0.0243%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B64%7D%5D%5C%5C%5C%5C0.0243%2B%5Cdfrac%7B1%7D%7B64%7D%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5C0.039925%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5Cn_f%5E2%3D25%5C%5C%5C%5Cn_f%3D5)
So, the final level of the electron is 5.
Answer:
V₂ = 530.5 mL
Explanation:
Given data:
Initial temperature = 20.0°C
Final temperature = 40.0 °C
Final volume = 585 mL
Initial volume = ?
Solution:
Initial temperature = 20.0°C (20+273 = 293 K)
Final temperature = 40.0 °C (40+273 = 323 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₂ = 585 mL × 293 K / 323 K
V₂ = 171405 mL.K / 323 K
V₂ = 530.5 mL
