Answer:
Therefore, the partial pressure of oxygen at high altitudes is less than at sea level.
Explanation:
The deep ocean thus has higher oxygen because rates of oxygen consumption are low compared with the supply of cold, oxygen-rich deep waters from polar regions. In the surface layers, oxygen is supplied by exchange with the atmosphere.
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
Answer:
Kₐ = 4.06 × 10⁻⁷
Explanation:
Step 1. <em>Calculate [H₃O⁺]
</em>

pH = 3.94

[H₃O⁺] = 1.15 × 10⁻⁴ mol·L⁻¹
Calculate 
HF + H₂O ⇌ H₃O⁺ + F⁻
I/mol·L⁻¹: 0.0326 0 0
C/mol·L⁻¹: 0.0326-1.15 × 10⁻⁴ +1.15 × 10⁻⁴ +1.15 × 10⁻⁴
E/mol·L⁻¹: 0.0325 1.15 × 10⁻⁴ 1.15 × 10⁻⁴
So, at equilibrium,
[H₃O⁺] = [F⁻] = 1.15 × 10⁻⁴ mol·L⁻¹
[HF] = 0.326 – 1.15 × 10⁻⁴ mol·L⁻¹ = 0.0325 mol·L⁻¹
Kₐ = {[H₃O⁺][F⁻]}/[HF]
Kₐ = (1.15 × 10⁻⁴ × 1.15 × 10⁻⁴)/0.0325
Kₐ = 1.32 × 10⁻⁸/0.0325
Kₐ = 4.06 × 10⁻⁷
This is <em>NOT</em> a solution of HF (Kₐ = 7.2 × 10⁻⁴). It is more likely a solution of carbonic acid (H₂CO₃; Kₐ₁ = 4.27 × 10⁻⁷).
Solids have a solid, fixed shape, while liquids can change shape as they are put into a different container or cup. Gases also take the shape of the object they are in.
30% of 107.75 g is 32.325g so there are about 8.08125 mol of He which is
4.86 x 10^24 particles
70% of 107.75 g is 75.425g so there are about .9 mol of Kr which is
5.42 x 10^23 paticles
In the whole sample, there are about 5.41 x 10^24 particles