Answer:
Explanation:
We must use the Nernst equation
1. Write the equation for the cell reaction
If you want the reduction potential, the pH 5.65 solution is the cathode, and the cell reaction is
<u> E°/V</u>
Anode: H₂(1 bar) ⇌ 2H⁺(1 mol·L⁻¹) + 2e⁻; 0
Cathode: <u>2H⁺(pH 5.65) + 2e⁻ ⇌ H₂(1 bar)</u>; <u> 0 </u>
Overall: 2H⁺ (pH 5.65) ⇌ 2H⁺(1 mol·L⁻¹); 0
Step 2. Calculate E°
(a) Data
E° = 0
R = 8.314 J·K⁻¹mol⁻¹
T = 25 °C
n = 2
F = 96 485 C/mol
pH = 5.65
Calculations:
T = 25 + 273.15 = 298.15 K
![\text{H}^{+} = 10^{\text{-pH}} = 2.24 \times 10^{-5}\text{ mol/\L}\\\\Q = \dfrac{\text{[H}^{+}]_{\text{prod}}^{2}}{\text{[H}^{+}]_{\text{react}}^{2}} = \dfrac{(1.00)^{2}}{(2.24 \times 10^{-5})^{2}} =2.00 \times 10^{9}\\\\\\E = 0 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln{2.00 \times 10^{9}}\\\\= -0.01285 \times 21.41 = \textbf{-0.275 V}\\\text{The cell potential for the cell as written is }\boxed{\textbf{-0.275 V}}](https://tex.z-dn.net/?f=%5Ctext%7BH%7D%5E%7B%2B%7D%20%3D%2010%5E%7B%5Ctext%7B-pH%7D%7D%20%3D%202.24%20%5Ctimes%2010%5E%7B-5%7D%5Ctext%7B%20mol%2F%5CL%7D%5C%5C%5C%5CQ%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BH%7D%5E%7B%2B%7D%5D_%7B%5Ctext%7Bprod%7D%7D%5E%7B2%7D%7D%7B%5Ctext%7B%5BH%7D%5E%7B%2B%7D%5D_%7B%5Ctext%7Breact%7D%7D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B%281.00%29%5E%7B2%7D%7D%7B%282.24%20%5Ctimes%2010%5E%7B-5%7D%29%5E%7B2%7D%7D%20%3D2.00%20%5Ctimes%2010%5E%7B9%7D%5C%5C%5C%5C%5C%5CE%20%3D%200%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298.15%20%7D%7B2%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%7B2.00%20%5Ctimes%2010%5E%7B9%7D%7D%5C%5C%5C%5C%3D%20-0.01285%20%5Ctimes%2021.41%20%3D%20%5Ctextbf%7B-0.275%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20for%20the%20cell%20as%20written%20is%20%7D%5Cboxed%7B%5Ctextbf%7B-0.275%20V%7D%7D)
Answer: 234.4K
Explanation:
Given that,
Original volume of gas (V1) = 5.00 L
Original temperature of gas (T1) = 20.0°C
[Convert 20.0°C to Kelvin by adding 273
20.0°C + 273 = 293K]
New volume of gas (V2) = 4.0L
New temperature of gas (T2) = ?
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
5.00L/293K = 4.0L/T2
To get the value of T2, cross multiply
5.00L x T2 = 293K x 4.0L
5.00L•T2 = 1172L•K
Divide both sides by 5.00L
5.00L•T2/5.00L = 1172L•K/5.00L
T2 = 234.4K
Thus, the new temperature of the gas is 234.4 Kelvin
Thus we can balance the oxygen atoms by putting a prefix of 25/2 on the left side. To obtain a equation containing whole numbers, we multiply the entire equation by 2. This gives the final equation. 2 C8H18 + 25 O2 ---> 16 CO2 +18 H2O.
Carbon has the highest ionization energy as its energy 1086KJ\Mol and the rest are between 500 and 800.
Answer:
334.2× 10²³ molecules
Explanation:
Given data:
Mass of water = 1 Kg ( 1000 g )
Number of molecules = ?
Solution:
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 1000 g/ 18 g/mol
Number of moles = 55.5 mol
1 mole contain 6.022× 10²³ molecules
55.5 mol×6.022× 10²³ molecules
334.2× 10²³ molecules
