Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
where,
n = number of moles
Now put all the given values in this expression, we get
Therefore, the enthalpy change for this reaction is, -277.4 kJ
ANSWER I think its B
Explanation:
It is based on the size of seimetic waves
Lewis structure for each of the following N₂O₃ with no N¬N bond is attached below.
Even though pi symmetry occupies the antibonding orbitals of NO, this is unimportant after the dimer forms. A sigma connection exists. The enthalpy of the newly formed sigma bond in the dimer is low because the loss of a particularly distinctive set of single-electron resonance forms that were available for no monomer offset the net gain in bond. When the whole free energy is taken into account, there is no gain because the entropic effects are on the order of 1030kJ/mol, and dimerization is entropically disfavored at G=17kJ/mol. Therefore, any little increase in enthalpy is cancelled out by the loss of entropy.
Learn more about dimer here-
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What am I suppose to do???
Answer:
Since Q > Ksp, a precipitate of AgCl will form.
Explanation:
Step 1: Data given
Molarity AgNO3 = 0.10 M
Molarity of NaCl = 0.075 M
Ksp AgCl = 1.77 * 10^-10
Step 2: The balanced equation
AgNO3 + NaCl → AgCl(s) + NaNO3(aq)
For 0.10 moles AgNO3 we have 0.10 moles Ag+ ( molarity = 0.10 M)
For 0.075 moles NaCl we have 0.075 moles Cl- (molarity = 0.075 M)
Step 3: Calculate Q
Q = [Ag+][Cl-]
Q = (0.10 M )(0.075 M ) = 0.0075
Ksp = 1.77 * 10^-10
Q >>> Ksp
Since Q > Ksp, a precipitate of AgCl will form.