Answer:
1.99V
Explanation:
Balanced redox reaction equation:
3CU2+(aq) + 2Al(s) ------> 3Cu(s) + 2Al3+(aq)
E°cell= E°cathode- E°anode
E°cell= 0.34-(-1.66)
E°cell= 2.0V
From Nernst equation:
E= E°cell - 0.0592/n logQ
E= 2.0 - 0.0592/6 log [3.43]/[1.63]
E= 2.0- 0.0032
E= 1.99V
Explanation:
When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.
As sulfur (S) is a group 16 element and chlorine (Cl) is a group 17 element. Hence, sulfur (S) is more metallic in nature than chlorine.
This means that chlorine (S) is less metallic than chlorine (Cl).
Both indium (I) and aluminium (Al) are group 13 elements. And, when we move down a group then there occur an increase in non-metallic character of the elements. As indium belongs to group 13 and period 5 whereas aluminium belongs to group 13 and period 3.
Therefore, aluminium (Al) is more metallic than indium (In).
Arsenic (Ar) is a group 15 element and bromine (Br) is a group 17 element. Therefore, arsenic is more metallic than bromine.
The above question is incomplete, here is the complete question:
Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:
Answer:
The standard molar enthalpy of formation of NO is 90.25 kJ/mol.
Explanation:
To calculate the standard molar enthalpy of formation
...[3]
Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
[1] - [2] = [3]
According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :
=
1.49 moles of Al will be produced. Hope this helps you.
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