<span>pv=nrt; Pressure and moles are constant.
p=nr(150k)/.5 L; Pressure initially
After temp change
pv=nrt; What is volume?
v=nr(350k)/p; p is constant so we can substitute from above
v=nr(350k)/(nr(150k)/.5 L))
v=350/150/.5 L
v=4.66 liters</span>
Answer:
the velocity is 25 m/s
Explanation:
The computation of the velocity is shown below:
As we know that
Magnitude of Momentum = (mass) × (speed)
75 kg. m/s = 3 kg × speed
So, the speed is
= 75 ÷ 3
= 25 m/s
hence, the velocity is 25 m/s
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
Answer:
Explanation:
This is a direct application of the equation for ideal gases.
Where:
- P = pressure = 1.25 atm
- V = volume = 25.2 liter
- R = Universal constant of gases = 0.08206 atm-liter/K-mol
- T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
- n = number of moles
Solving for n:
Substituting:
