ionic bond:
Bond formed when an atom donates its electron and other atom receives those electrons.
polar covalent:
Bond formed by equal sharing of electrons between both the atoms and there is an electronegativity difference between the two atoms.
Nonpolar covalent:
Bond formed by equal sharing of electrons between both the atoms and there is no electronegativity difference between the two atoms.
Metallic:
Formed between two metals.
So the bond between :
Phosphorus and chlorine-polar covalent bond as it is formed by equal sharing of electrons between both the atoms and there is an electronegativity difference between the two atoms.
Potassium and oxygen -ionic bond as here potassium donates its electron and oxygen receives those electrons
Fluorine and fluorine -Non polar covalent bond as formed by equal sharing of electrons between both the atoms and there is no electronegativity difference between the two atoms.
Copper and aluminum-metallic bond as Formed between two metals.
Carbon and fluorine -polar covalent bond as it is formed by equal sharing of electrons between both the atoms and there is an electronegativity difference between the two atoms.
Carbon and hydrogen --Non polar covalent bond as formed by equal sharing of electrons between both the atoms and there is no electronegativity difference between the two atoms.
Aluminum and oxygen--ionic bond as here aluminum donates its electron and oxygen receives those electrons
Silver and copper --metallic bond as Formed between two metals.
Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:


Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:

Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.
Answer:
ΔS=0.148 KJ/K
Explanation:
Given that
Q = 100 KJ
T₁=200°C
T₁=200+273 = 437 K
T₂=5°C
T₂=5 + 273 = 278 K
Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.
So the total change in entropy given as
ΔS= - Q/T₁ + Q/T₂
ΔS= - 100/473 + 100/278 KJ/K
ΔS=0.148 KJ/K
Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.
Explanation : Given,
=21 torr/min
The balanced chemical reaction is,

The rate of disappearance of
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of
= ![-\frac{d[Cl_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D)
The rate of formation of
= ![\frac{1}{2}\frac{d[NOCl]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D)
As we know that,
=21 torr/min
So,
![-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BCl_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2021torr%2Fmin%3D10.5torr%2Fmin)
And,
![\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D)
![\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNOCl%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%3D21torr%2Fmin)
Now we have to calculate the rate change.
Rate change = Reactant rate - Product rate
Rate change = (21 + 10.5) - 21 = 10.5 torr/min
Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.
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