Answer:
The gas was N₂
Explanation:
V = 3.6L
P = 2.0 atm
T = 24.0°C = 297K
R = 0.0821 L.atm/K.mol
m = 8.3g
M = molar mass = ?
Using ideal gas equation;
PV = nRT
n = no. Of moles = mass / molar mass
n = m/M
PV = m/M * RT
M = mRT / PV
M = (8.3*0.0821*297) / (2.0*3.6)
M = 28.10
Since X is a diatomic molecule
M = 28.10 / 2 = 14.05 g/mol
M = Nitrogen
X = N₂
Answer:
Bueno,
Explanation:
Tres pesnamientos negativos sobre el C0Vid-19 son,
- Encerrados en la casa
- Tener que ponerse mascara
- Todos los lugares divertidos cerrados
Convertidos en positivos:
- Pasamos mas tiempo de calidad en familia.
- Nos protejemos unos a otros.
- Descansan un poco todas las personas que se han pasado toda la vida trabajando.
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Answer: The amount of energy lost by the metal is equal to the amount of energy gained by the water
Explanation:
In nuclear physics and nuclear chemistry, nuclear fission is either a nuclear reaction or a radioactive decay process in which the nucleus of an atom splits<span> into smaller parts (lighter </span>nuclei<span>). Hope this helps</span>
