(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
Answer:
A.) r = 2t
B.) V = 33.5t^3
Explanation:
Given that a spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cm/s
A) Express the radius (r) of the balloon as a function of the time (t).
Since the rate = 2 cm/s that is,
Rate = radius/ time
Therefore,
2 = r/t
Make r the subject of formula
r = 2t
(B) If V is the volume of the balloon as a function of the radius, find V or and interpret it.
Let assume that the balloon is spherical. Volume of a sphere is;
V = 4/3πr^3
Substitute r = 2t into the formula
V = 4/3π(2t)^3
V = 4/3π × 8t^3
V = 32/3 × πt^3
V = 33.5t^3
Explanation:
I remember that notation! The expression
is the 1st law of thermodynamics and it refers to the heat supplied to the system dQ which is also a change in its internal energy dU. The first term is the <u>partial</u> derivative of the internal energy U with respect to temperature T while the volume V is kept constant, as denoted by the subscript V. The 2nd term is similar but this time, temperature is kept constant while its volume partial derivative is being taken.
Ah, memories!
I think the answer is D, but I’m not at all sure :l
Answer:
im pretty sure its 10 m/s but its kinda hard sorry
Explanation:
