Im not sure but i think its hbr. i hope im right and i hope this helped,
<span>There are factors that are important to be determined for the composition of ocean water. First factor is the temperatue, by this you will be able to know the rate of evaporation of the sea. Next is the salinity, through this you will be able to know salty the sea is which will help you identify the last factor- density. The density is the most common because when there is more salt in the sea, it is less dense.</span>
Answer: Neutron has no charge, electron has a charge and mass. Neutron occurs inside the nucleus where electron is seen outside the nucleus.
Explanation:
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:
5.2 x 10⁻⁴ M.
Explanation:
- The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:
<em>P = kC</em>
where P is the partial pressure of the gaseous solute above the solution.
k is a constant (Henry’s constant).
C is the concentration of the dissolved gas.
- At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:
<em>P₁C₂ = P₂C₁,</em>
P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.
P₂ = 0.76 atm, C₂ = ??? mol/L.
<em>∴ C₂ = (P₂C₁)/P₁ =</em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.</em>
