Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x <u> 1 mole CH4 </u> = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
We can use the formula P=IV to calculate the current, where “P” is power (measured in watts), “I” is current (measured in Amps), and “V” is voltage. Simply plug and solve:
P = IV
(3.5 Watts) = I(120 volts)
I = 0.0292 Amps
The current flowing through the bulb is approximately 0.0292 Amps.
Hope this helps!
Answer: 127.5ml
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
Thus 127.5 ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH
