Catalysts are substances that speed up the rate of a chemical
reaction. It is important to know that reacting species must first possess a
minimum amount of energy equal to the activation energy for the reaction
to proceed. The catalyst works in a way that lowers the activation energy
required for the reaction to proceed. In this way, the reaction occurs at a
faster rate than without a catalyst. Catalysts do not react with
the chemical species in the reaction, thus, they are not consumed. However, over
time, catalysts tend to degrade and their function is also reduced. When this
happens, new catalysts replace the poisoned ones. Catalysts are widely used
today since they bring an increase in the production rate of commercial chemicals.
Can you post the question
Answer:
on N overlaps with 1s on H.
Explanation:
Hybridization of central atom generally can be found by finding the hybridization number.Hybridization number is the sum of number of lone pairs and number of sigma bonds.
If hybridization number is 2 then the central atom is sp hybridized.
If it is 3 it is
hybridized.
If it is 4 it is
hybridized and so on.
Here in
molecule there are 3 sigma bonds and one lone pair(see the figure attached).
Hence hybridization number is 4 and hence it is s
hybridized.
And in case of hydrogen atom the only electron of it resides in the 1s orbital.
Therefore
on N overlaps with 1s on H.
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