Of the four states of matter, which has particles that are the most tightly packed together?

Calculate kp at 298.15 k for the reactions (a), (b), and (c) using δg°f values.

vivado [14]

(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

6 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

how much energy would microraptor gui have to expend to fly with a speed of 10 m/sm/s for 1.0 minute?

Ray Of Light [21]

Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

Where:

P = power (W)

T = time (s)

Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

P = E/t

E = P x t

E = 8.1 x 60

E = 486 Joules

Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.

Learn more about Microraptor gui here brainly.com/question/1200755

#SPJ4

3 0

1 year ago

What is the kinetic energy in joules of a 0.05. kg bullet traveling 310 m/s

Wittaler [7]

The formula is=1/2(m x v^2)

so = 1/2*(0.05)*(310)^2

ans is =2402.5 joules

3 0

3 years ago

The moon Phobos orbits Mars

podryga [215]

Explanation:

For a circular orbit v= $\sqrt{\frac{G.m}{r} }$ with G = 6.6742 × $10^{-11}$

Given m = 6.42 x 10^23 kg and r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

3 0

3 years ago