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3 years ago

5

The graph represents the reaction 3H2 + N2 2NH3 as it reaches equilibrium. Based on the graph, which two statements about this r

eaction are true?
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1 answer:

nata0808 [166]3 years ago

8 0

Answer:A and C

Explanation: I took the test lol

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A thermally insulated vessel containing a gas whose molar mass is equal to M and the ratio of specific heats cP /cV = γ moves wi

Vika [28.1K]

Answer:

∆T = Mv^2Y/2Cp

Explanation:

Formula for Kinetic energy of the vessel = 1/2mv^2

Increase in internal energy Δu = nCVΔT

where n is the number of moles of the gas in vessel.

When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas

We say

1/2mv^2 = ∆u

1/2mv^2 = nCv∆T

Since n = m/M

1/2mv^2 = mCv∆T/M

Making ∆T subject of the formula we have

∆T = Mv^2/2Cv

Multiple the RHS by Cp/Cp

∆T = Mv^2/2Cv *Cp/Cp

Since Y = Cp/CV

∆T = Mv^2Y/2Cp k

Since CV = R/Y - 1

We could also have

∆T = Mv^2(Y - 1)/2R k

6 0

4 years ago

How was Bohr's atomic model different from those of previous scientists?

exis [7]

Answer:Bohr placed the electrons in distinct energy levels. Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits.

Explanation: also rutherfords was just a hypothesis while Bhor took the time to make his an experiment

4 0

3 years ago

A block of mass 9.2kg rests on a slope an angle of 26.9∘ relative to the horizontal. What is the size of the contact force norma

Olegator [25]

Answer:

80.4 N

Explanation:

As the block is at rest on the slope, it means that all the forces acting on it are balanced.

We are only interested in the forces that act on the block along the direction perpendicular to the slope. Along this direction, we have two forces acting on the block:

- The normal reaction N (contact force), upward

- The component of the weight of the block, $mg cos \theta$ , downward, where m is the mass of the block, g is the gravitational acceleration and $\theta$ is the angle of the incline

Since the block is in equilibrium along this direction, the two forces must balance each other, so they must be equal in magnitude:

$N=mg cos \theta$

And by substituting the numbers into the equation, we find the size of the contact force normal to the slope:

$N=(9.2 kg)(9.8 m/s^2)(cos 26.9^{\circ})=80.4 N$

6 0

3 years ago

1. An object has a mass of 10g and a volume of 50mL. What is the density?

Alona [7]

1. 0.2 g/mL

The relationship between mass, density and volume of an object is

$d=\frac{m}{V}$

where

d is the density

m is the mass

V is the volume

For the object in this problem, we have

m = 10 g

V = 50 mL

Substituting into the equation,

$d=\frac{10}{50}=0.2 g/mL$

2. 10 mL

In this exercise we know:

- The density of the object: d = 2 g/mL

- The mass of the object: m = 20 g

Therefore, we can re-arrange the previous equation to find the volume:

$V=\frac{m}{d}$

And substituting values into the equation, we find

$V=\frac{20}{2}=10 mL$

7 0

3 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

4 years ago