We are tasked to solve for the volume of the gas that occupies when pressure and temperature changes to 400 Torr and 200 Kelvin from Torr and 400 Kelvin. We can use ideal gas law assuming constant gas composition and close system. The solution is shown below:
P1V1 / T1 = P2V2 / T2
V2 = P1V1T2 / T1P2
V2 = 800*72*200 / 400*400
V2 = 72 ml
The answer for the volume is 72 ml.
0.6137 g of KHP contains 1.086 × 10^21 acidic protons.
Number of moles of KHP = mass of KHP/molar mass of KHP
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.6137 g
Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP
Now, 1 each molecule of KHP contains 1 acidic proton.
For 0.003 moles of KHP there are; 0.003 × 1 × NA
Where NA is Avogadro's number.
So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23
= 1.086 × 10^21 acidic protons.
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Answer:
The mass of Mg consumed is 21.42g
Explanation:
The reaction is

As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.
as given that mass of nitrogen reacted = 8.33g
So moles of nitrogen reacted = 
moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol
Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g
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
The answer would A. Sand grain I got 100 on the sedimentary assignment.