Answer:
The cell reaction reaches equilibrium quickly and the cell emf becomes zero.
Explanation:
The purpose of a salt bridge is not to move electrons from the electrolyte, its main function is to maintain charge balance because the electrons are moving from one-half cell to the other.
A solution of a salt that dissociates easily is normally used. Water is ineffective at functioning as a salt bridge. Hence the effect stated in the answer.
First, write the equation:
Cu + O2 -> CuO
Now, balance:
2 Cu + O2 -> 2 CuO
Answer:
In the calorimeter, water is the <u>exothermic</u>. The salt LiCI, which will dissolve, is the <u>endothermic</u>. The final temperature of the water after the dissolution of LiCI was <u>lower</u> than the initial temperature, meaning the process is <u>exothermic</u>. In the microscopic view of the disspolution of LiCI, water molecules were seen to move <u>slowly</u> as they <u>gained </u>energy.
Explanation:
Exothermic is a process in which heat is released during the process. Endothermic reactions absorbs heat from surrounding during a chemical process. The dissolution of salt into water is an exothermic reaction. During this process heat is release and water molecules are broken down which are surrounded by salt ions.
Answer:
0.0660 moles
Explanation:
Change torr to atm and change celcius to kelvin.
(1.35L)(.987 atm)/(0.08206 L atm /mol k)(290k)
= 0.0660 mol
Answer:
Here's what I get.
Explanation:
1. Brønsted-Lowry theory
An acid is a substance that can donate a proton to another substance.
A base is a substance that can accept a proton from another substance.
2. pH of ammonia
The chemical equation is

For simplicity, let's re-write this as

(a) Set up an ICE table.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.335 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.335 + x x x
![\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=%5Crm%20K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BBH%7D%5E%7B%2B%7D%5D%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.335%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:

(b) Solve for [OH⁻]
![\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.335%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.335%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5Cx%5E%7B2%7D%20%3D%206.03%20%5Ctimes%2010%5E%7B-6%7D%5C%5Cx%20%3D%20%5Csqrt%7B6.03%20%5Ctimes%2010%5E%7B-6%7D%7D%5C%5Cx%20%3D%20%5Ctext%7B%5BOH%5D%7D%5E%7B-%7D%20%3D%20%5Cmathbf%7B2.46%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
(c) Calculate the pOH
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61](https://tex.z-dn.net/?f=%5Ctext%7BpOH%7D%20%3D%20-%5Clog%20%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20-%5Clog%282.46%20%5Ctimes%2010%5E%7B-3%7D%29%20%3D%202.61)
(d) Calculate the pH
pH = 14.00 - pOH = 14.00 - 2.61 = 11.39