Mole of electron required by 
mole is 
- Faraday law expressed how the change that is been being produced by a current at an electrode-electrolyte interface is related and proportional to the quantity of electricity that is been used.
- There is one mole of electron required for 1 Faraday of electricity.
- Avogadro constant is
- Mole of electron can be calculated by dividing the number of electron by avogadro's constant.
=
= 
Therefore, it requires 
Faraday of electricity for the 
Learn more at: brainly.com/question/1640558?referrer=searchResults
Answer: False
Explanation:
Since the given equation is not balanced properly.
Since oxygen and hydrogen atoms are not balanced.
There should be 6 H2O (g) molecules and 14 mol H2 (g)
Answer:False
Explanation:co2 or carbon dioxide is a compound not an element, you can even check this on the periodic table
Answer:
CH3CH2CH2CH2CH2OH.
Explanation:
Hello.
In this case, since the vapor pressure is known to be the pressure exerted by the gaseous molecules in equilibrium with a liquid, we can infer that the higher the molecule, the lower the vapor pressure because the molecules tend to be help together more strongly and more energy is required to separate them and take them from liquid to gas.
In such a way, since CH3CH2CH2CH2CH2OH is the longest molecule (five carbon atoms) it would be more stable at liquid phase which means that it has less molecules moving to gaseous phase, which is also related with the lowest vapor pressure. Conversely, CH3CH2OH has the highest vapor pressure.
Best regards.
Answer:
1. 43.44g of HCl
2. 26.67 L of HCl
Explanation:
1) Molarity of a solution = number of moles (n) ÷ Volume (V)
According to the provided information in this question,
V = 350 mL = 350/1000 = 0.350L
Molarity = 3.4 M
Using Molarity = n/V
3.4 = n/0.350
n = 3.4 × 0.350
n = 1.19mol
Using the formula below to calculate the mass of HCl;
mole = mass/molar mass
Molar mass of HCl = 1 + 35.5 = 36.5g/mol
mole = mass/MM
mass = 1.19 mol × 36.5g/mol
mass = 43.44g of HCl
2) At STP, HCl has a pressure of 1atm, a temperature of 273K
V = ?
n = 1.19 mol
R = 0.0821 Latm/molK
Using PV = nRT
V = nRT/P
V = 1.19 × 0.0821 × 273/1
Volume = 26.67L
