<span>global wind patterns, rotation of the earth, shape of ocean basins.</span>
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
<h3>
Answer:</h3>
78.75 K
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial pressure, P₁ = 500 torr
- Initial temperature,T₁ = 225 K
- Initial volume, V₁ = 3.3 L
- Final volume, V₂ = 2.75 L
- Final pressure, P₂ = 210 torr
We are required to calculate the new temperature, T₂
- To find the new temperature, T₂ we are going to use the combined gas law;
- According to the combined gas law;
P₁V₁/T₁ = P₂V₂/T₂
We can calculate the new temperature, T₂;
Rearranging the formula;
T₂ =(P₂V₂T₁) ÷ (P₁V₁)
= (210 torr × 2.75 L × 225 K) ÷ (500 torr × 3.3 L)
= 78.75 K
Therefore, the new volume of the sample is 78.75 K
Transform the g into kg by multiplying 0.74 by 10^-3. then divide 2.4 kg by the density
Organism<span>. ' s rate of mutation is directly proportional to its adaptability. .... D. The cell membrane contains </span>genetic<span> information of the cell. .... </span>What<span> cellular structure was the dialysis tubing most likely</span>representing<span> in this experiment? ...... C </span>Rr<span> and </span>rr<span> only ... A healthy </span>individual<span> is a carrier of a lethal allele but is unaffected by it.</span>
