Answer:
V = 44.85 L
Explanation:
Given data:
Volume of H₂ = ?
Number of moles of H₂ = 2.0 mol
Given temperature = 273.15 K
Given pressure = 1 atm
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 2.0 mol × 0.0821 atm.L/ mol.K × 273.15 K
V = 44.85 atm.L / 1 atm
V = 44.85 L
The correct is this because they said
Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O
Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...
Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...
Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....
Do the calculation
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
