✦ ✦ ✦ Beep Boop - Blu Bot! At Your Service! Scanning Question . . . Code :
Green! Letters and Variables Received! ✦ ✦ ✦
--------------------------------------------------------------------------------------------------------------
Answer:
The entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process . The standard molar entropy, So, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure. (Below are listed tips)
- At absolute zero (0 K), the entropy of a pure, perfect crystal is zero.
- The entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process.
- The standard molar entropy, So, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure.
--------------------------------------------------------------------------------------------------------------
Answer:
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is ![5.23\times 10^{-5} m](https://tex.z-dn.net/?f=5.23%5Ctimes%2010%5E%7B-5%7D%20m)
Explanation:
Given :
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:
![E=\frac{R_y}{n^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BR_y%7D%7Bn%5E2%7D)
= Rydberg energy
n = principal quantum number of the orbital
Energy of 11th orbit = ![E_{11}](https://tex.z-dn.net/?f=E_%7B11%7D)
![E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J](https://tex.z-dn.net/?f=E_%7B11%7D%3D%5Cfrac%7B2.18%5Ctimes%2010%5E%7B-18%7D%20J%7D%7B11%5E2%7D%3D1.80%5Ctimes%2010%5E%7B-20%7D%20J)
Energy of 10th orbit = ![E_{10}](https://tex.z-dn.net/?f=E_%7B10%7D)
![E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J](https://tex.z-dn.net/?f=E_%7B10%7D%3D%5Cfrac%7B2.18%5Ctimes%2010%5E%7B-18%7D%20J%7D%7B10%5E2%7D%3D2.18%5Ctimes%2010%5E%7B-20%7D%20J)
Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.
![E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J](https://tex.z-dn.net/?f=E%27%3DE_%7B10%7D-E_%7B11%7D%3D2.18%5Ctimes%2010%5E%7B-20%7D%20J-1.80%5Ctimes%2010%5E%7B-20%7D%20J)
![=E'=0.38\times 10^{-20} J](https://tex.z-dn.net/?f=%3DE%27%3D0.38%5Ctimes%2010%5E%7B-20%7D%20J)
(Planck's' equation)
![\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-34%7D%20Js%5Ctimes%203%5Ctimes%2010%5E8%20m%2Fs%7D%7B0.38%5Ctimes%2010%5E%7B-20%7D%20J%7D)
![\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m](https://tex.z-dn.net/?f=%5Clambda%20%3D%205.2310%5Ctimes%2010%5E%7B-5%7D%20m%5Capprox%205.23%5Ctimes%2010%5E%7B-5%7D%20m)
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is ![5.23\times 10^{-5} m](https://tex.z-dn.net/?f=5.23%5Ctimes%2010%5E%7B-5%7D%20m)
Answer:
D. It is extremely reactive.
Explanation:
Hello!
In this case, since potassium is an alkali metal, it is known those are extremely reactive because the energy required to ionize them is very low, it means they react so easily. For instance, even in the presence of water, potassium is able to react and form a purple flame as a product of the reaction:
![2K+2H_2O\rightarrow 2KOH+H_2](https://tex.z-dn.net/?f=2K%2B2H_2O%5Crightarrow%202KOH%2BH_2)
As well as potassium, the rest of the elements belonging to the alkali metals series are extremely reactive; therefore the answer is D. It is extremely reactive.
Best regards!
Answer:
= 1 X 10⁻⁻¹²M
Explanation:
At 25°C & 1atm [H⁺][OH⁻] = 1 x 10⁻¹⁴ => [H⁺] = 1 X 10⁻¹⁴/[OH⁻] = 1 X 10⁻¹⁴/1 X 10⁻²
= 1 X 10⁻⁻¹²M