Answer:
The value is 
Explanation:
From the question we are told that
The equation is 
The temperature is ![T = 25^oC = 298 K [room \ temperature ]](https://tex.z-dn.net/?f=%20T%20%3D%2025%5EoC%20%3D%20%20298%20K%20%20%20%5Broom%20%20%5C%20temperature%20%5D)
The emf at standard condition is 
Generally at the cathode

At the anode

Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as

Here n is the no of electron with value n = 6
F is the Faraday's constant with value 96487 J/V
=>
=> 
This Gibbs free energy can also be represented mathematically as

Here R is the cell constant with value 8.314J/K
K is the equilibrium constant
From above
=> 
Generally antilog = 2.718
=>
=> 
Explanation:
Different atoms binds their outermost shell electrons with different amount of energy.
The amount of energy required to remove an electron from an atom is the ionization energy.
- Ionization energy measures the readiness of an atom to lose electrons.
- From the given problem, we can infer that in group O the ionization energy decreases down the group.
- Helium has the highest ionization energy.
- Down a group on the periodic table, ionization energy decrease because:
- atomic radii increases down the group.
- there is an increasing shielding/screening effect of inner shell electrons on the outermost shell electrons.
Learn more:
Ionization energy brainly.com/question/2153804
#learnwithBrainly
Answer:
-145.2kJ
Explanation:
Enthalpy is an extensive property as its value depends on the amount of substance present in the system.
If the enthalpy for one mole of methanol = -726 kJ/mol;
The Enthalpy for 0.2 mol is given as;
Enthalpy = 0.200 * 726
Enthalpy = -145.2kJ
It would take -145.2kJ for 0.200 mol of methanol to undego the combustion reaction.
Answer:
When a number is written in scientific notation, the exponent tells you if the term is a large or a small number. A positive exponent indicates a large number and a negative exponent indicates a small number that is between 0 and 1.