Answer:
A. 6atm
Explanation:
Using pressure law equation:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
T1 = initial temperature (K)
P2 = final pressure (atm)
T2 = final temperature (K)
According to this question;
P1 = 3 atm
P2 = ?
T1 = 120K
T2 = 240K
Using P1/T1 = P2/T2
3/120 = P2/240
Cross multiply
240 × 3 = P2 × 120
720 = 120P2
P2 = 720/120
P2 = 6atm
Answer:
![Q = 375\,cal](https://tex.z-dn.net/?f=Q%20%3D%20375%5C%2Ccal)
Explanation:
The quantity of heat transfered from the jellybean to the water is:
![Q = \rho\cdot V \cdot c\cdot \Delta T](https://tex.z-dn.net/?f=Q%20%3D%20%5Crho%5Ccdot%20V%20%5Ccdot%20c%5Ccdot%20%5CDelta%20T)
![Q = \left(1\,\frac{g}{cm^{3}}\right)\cdot (15\,cm^{3})\cdot \left(1\,\frac{cal}{g\cdot ^{\circ} C} \right)\cdot (25\,^{\circ}C )](https://tex.z-dn.net/?f=Q%20%3D%20%5Cleft%281%5C%2C%5Cfrac%7Bg%7D%7Bcm%5E%7B3%7D%7D%5Cright%29%5Ccdot%20%2815%5C%2Ccm%5E%7B3%7D%29%5Ccdot%20%5Cleft%281%5C%2C%5Cfrac%7Bcal%7D%7Bg%5Ccdot%20%5E%7B%5Ccirc%7D%20C%7D%20%5Cright%29%5Ccdot%20%2825%5C%2C%5E%7B%5Ccirc%7DC%20%29)
![Q = 375\,cal](https://tex.z-dn.net/?f=Q%20%3D%20375%5C%2Ccal)
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
Answer:
Your answer is :- C [OH-] = 10 x 10-7 mol dm-3
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