Answer:
They break down materials into smaller pieces
Explanation:
Although using different methods
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.
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Answer:
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 25.0 L
V₂ = ?
P₁ = 2575 mm Hg
Also, P (atm) = P (mm Hg) / 760
P₁ = 2575 / 760 atm = 3.39 atm
P₂ = 1.35 atm
T₁ = 353 K
T₂ = 253 K
Using above equation as:

Solving for V₂ , we get:
<u>V₂ = 45.0 L</u>
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Equilibrium equation is
<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>
<span>From the reaction equation above, the formula for Ksp: </span>
<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>
Answer:
Percent yield = 82.5
Explanation:
Given data:
Actual yield of N₂ = 275 L
Theoretical yield = 333 L
Percent yield = ?
Solution:
Chemical equation:
4HCl + 6NO → 5N₂ + 6H₂O
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = (275 L/ 333 L ) × 100
Percent yield = 0.825× 100
Percent yield = 82.5%