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
The reactants in the neutralization reaction are an acid and a base while the products are a salt and water.
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
Learn more about limiting reagent here ;
brainly.com/question/11848702
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Molecules are cooler so they bocme more dense.. move very slow, have less kinetic energy and are not in random motion.
