Answer:
1.346 v
Explanation:
1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:
(oxidation) 
→
E°=0.337 v
(reduction) 
→
E°=1.679 v
(overall) 
+8H^{+}_{(aq)}→
E°=1.342 v
2) Nernst Equation
Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:
![E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}](https://tex.z-dn.net/?f=E%3DE%5E%7B0%7D%20-%5Cfrac%7BRT%7D%7BnF%7DLn%5Cfrac%7B%5Bred%5D%7D%7B%5Box%5D%7D)
Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.
The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give
![E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346](https://tex.z-dn.net/?f=E%3D1.342%20-%5Cfrac%7B298.15%2A8.314%7D%7B6%2A96500%7DLn%5Cfrac%7B%5B.66%5D%7D%7B%5B1.69%5D%7D%3D1.346)
E=1.346
Here you go:
Continent Polar - Cold and Dry
Maritime Polar - Cold and Humid
Arctic - Extremely Cold and Dry
Maritime Tropical- Warm and Humid
Hope this helps:)
Answer:
NaHCO₃
Explanation:
Sodium bicarbonate (baking soda) is a chemical compound with the formula NaHCO₃.
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
