(a) 2NO(g) + O₂(g) ⇄2NO₂(g)kp
(b) 2N₂O(g)⇄2NO(g) + N₂(g) kp
(c) N₂(g) + O₂(g)⇄ 2NO(g) kp
Now A is
2NO +O₂⇄2NO₂
ΔG° =ΔG° products - ΔG reactants
=2× 51.3-(256.6)
-70.6kJ/mol.
ΔG° = -RT Inkp
-70.6 = -8.314 ×10⁻³ ˣ 298.15 ˣInkJ
InkJ = 28.48
kp=2.34 ˣ 10¹²
B is
ΔG° = 2× 86.6 - 2 × 104.2 = -35.2
-35.2 = 8.314 × 10⁻³ ˣ 298.15 ˣInkJ
InkJ = 14.2
kp = 1.47ˣ 10⁶
C is
It is also similar
kp = 4.62 ˣ 10⁻³I
Explanation:
The table is level and there are no other forces on the book, so the normal force is equal to the weight.
N = mg
N = (2.3 kg) (9.8 m/s²)
N = 22.5 N
Answer:
e. 400 Hz
Explanation:
In closed organ pipe, only odd harmonics of fundamental note is possible .
The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .
200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc
600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .
Frequency not possible is 400 Hz .
Answer:
T= 5.18N
Explanation:
u = mass of chord / length of chord
u = 0.49/ 7.3
u = 0.067 kg/m
Velocity of sound waves (v) =length of chord / time taken for wave to travel
v = 7.3 / 0.83 = 8.795m/s
Tension is calculated below using the formula
T = v² * u
T = (8.795)² x 0.067
T= 5.18N
