A physical change doesn't change what the substance is but for chemical change a new substance is formed and energy is either given off or absorbed.
Answer:
The PH2 becomes attached to the alkyl halide with inversion of configuration
Explanation:
SN2 reaction involves the simultaneous attachment of the nucleophile as the leaving group departs. This leads to an inversion of configuration because the sterochemistry around the carbon bearing the leaving group is reversed. The reaction proceed via a crowded five coordinate transition state. Details are shown in the image attached.
Answer: The statement is not correct because the decrease in entropy of the hot solid CANNOT BE equal to the increase in entropy of the cold one
Explanation:
Let us start by stating the second law of thermodynamics and it the second law of thermodynamics states that there is an entity called entropy and entropy will always increase all the time. Also, the second law of thermodynamics states that the change in entropy can never be negative. The second law of thermodynamics can be said to be equal to Change in the transfer of heat, all divided by temperature.
So, the first law of thermodynamics deals with the conservation of energy. But there is nothing like conservation of entropy.
Therefore, the decrease in entropy of the hot solid CANNOT BE equal to the increase in entropy of the cold one because entropy is not a conserved property.
Answer:
<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>
Explanation:
If we know the ΔG° of a chemical reaction it is possible to calculate the equilibrium constant (k) of this procedure with the next equation:
ln Keq = -ΔG° / RT (1)
Where: Keq is equilibrium contant, ΔG° is standard state free energy change, R is gas constant and T is temperature.
Watching (1), it is possible to know that the large negative ΔG° the largest equilibrium constant. That is because R and T are always positive and to cancel the negative of equation it is necessary that ΔG° be negative.
Knowing this, is the oxidation of Hg the reaction that has the largely negative ΔG°. So, this reaction will have the largest equilibrium constant.
<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>
CaCO₃(s) → CaO(s) + CO₂(g) ΔG° =+131.1 kJ
3 O₂(g) → 2 O₃(g) ΔG° = +326 kJ
Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) ΔG° = -28.0 kJ
I hope it helps!
Answer:
V= 0.031L
Explanation:
P= 0.97atm, V= ?, n= 0.12/98 =0.00122mol, R= 0.082, T= 22.4+273= 295.4
Applying
PV=nRT
0.970×V = 0.00122×0.082×295.4
Simplify the above equation
V= 0.031L
