Answer: Naprox 350 is the anwser
Explanation: because to fourmals mixed makes naprox know as gr27
Ernest Rutherford was a famous British physicist who lived and worked in the 19th and the first half of the 20th century. He is now considered to be the father of nuclear physics, given that he was the one responsible for figuring out about the nuclear structure of an atom (proton, electron, neutron) and about the nuclear processes in general. The element rutherfordium (Rf) is named after this scientist. When it comes to his personal life, his parents are both British, however, Ernest was born in New Zealand because his mom and dad before he was born. He worked as a professor in several universities, and was even knighted in 1914. He is a Nobel prize winner, which he won for Chemistry in 1908.
Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red,cat - E°red,an
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
![E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V](https://tex.z-dn.net/?f=E%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20logQ%5C%5CE%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20log%5Cfrac%7B%5BSn%5E%7B%2B4%7D%5D.%5BFe%5E%7B2%2B%7D%5D%7D%7B%5BSn%5E%7B2%2B%7D%5D.%5BFe%5E%7B3%2B%7D%20%5D%7D%20%5C%5CE%3D0.617V-%5Cfrac%7B0.05916%7D%7B2%7D%20log%5Cfrac%7B%280.13%29.%280.0037%29%7D%7B%280.0023%29.%280.11%29%7D%20%5C%5CE%3D0.609V)
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Answer:
(a) Density of the air = 1.204 kg/m3
(b) Pressure = 93772 Pa or 0.703 mmHg
(c) Force needed to open the door = 15106 N
Explanation:
(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:

First, we change the units of P to Pa and T to deg K:

Then we have

(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:

(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.
