Answer:
is a simple, safe, and easy-to-create electrode
Explanation:
the hydrogen electrode is based on the redox half-cell:
building:
1) Platinium electrode
2) hydrogen pumping
3) acid solution [H+] = 1M
4) siphon to prevent oxygen presence
5) galvanic cell connector
this electrode is used as the basis for standard potential tables
Answer:
[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;
pH = 11.4; pOH = 2.6
Explanation:
The chemical equation is
For simplicity, let's re-write this as
1. Calculate [OH]⁻
(a) Set up an ICE table.
B + H₂O ⇌ BH⁺ + OH⁻
0.310 0 0
-x +x +x
0.310-x x x
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=K_%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%7B0.100%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:
(b) Solve for [OH⁻]
![\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.310%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.310%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5Cx%5E%7B2%7D%20%3D%205.58%20%5Ctimes%2010%5E%7B-6%7D%5C%5Cx%20%3D%20%5Csqrt%7B5.58%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%7B2.4%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Calculate the pOH
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}](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%282.4%20%5Ctimes%2010%5E%7B-3%7D%29%20%3D%20%5Cmathbf%7B2.6%7D)
3. Calculate the pH
4 Calculate [H₃O⁺]
When acids react with water, H ions are released which then combine with water molecules to form H₃O⁺
The ideal gas equation is;
PV = nRT; therefore making P the subject we get;
P = nRT/V
The total number of moles is 0.125 + 0.125 = 0.250 moles
Temperature in kelvin = 273.15 + 18 = 291.15 K
PV = nRT
P = (0.250 × 0.0821 )× 291.15 K ÷ (7.50 L) = 0.796 atm
Thus, the pressure in the container will be 0.796 atm
Answer:
The answer to your question is 7160 cm
Explanation:
Data
diameter = 1 mm
length = ?
amount of gold = 1 mol
density = 17 g/cm³
Process
1.- Get the atomic mass of gold
Atomic mass = 197 g
then, 197g ------------ 1 mol
2.- Calculate the volume of this wire
density = mass/volume
volume = mass/density
volume = 197/17
volume = 5.7 cm³
3.- Calculate the length of the wire
Volume = πr²h
solve for h
h = volume /πr²
radius = 0.05 cm
substitution
h = 5.7/(3.14 x 0.05²)
h = 5.7 / 0.0025
h = 7159.2 cm ≈ 7160 cm
