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Zolol [24]
3 years ago
6

Which element in Figure 2 is a transition metal? Which is a noble gas?

Physics
1 answer:
quester [9]3 years ago
5 0
<span>V is the transition metal. He is the noble gas. The periodic chart is arranged with metals on the left and non-metals on the right and the noble (non-reactive) gases form the rightmost column. The transition metals are the group of elements from columns 3 through 12. Although a strict definition of transition metals is those elements that have an incomplete d sub-shell, in which case column 12 is excluded since zinc, cadmium, and mercury do have complete d sub-shells. With that in mind, since the element V is the only element listed in the range of a transition metal, that's the answer. And since He is the only element listed in the noble gas column, that too is the answer.</span>
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Explanation: opposite attracts, same repulse

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Sketch a reaction progress curve for a reaction that has an activation energy of 22 kj and the total energy change is -103kj.
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Answer:

Do find the answer in the attachment herein.

Explanation:

From the attached diagram:

I. Activation energy = Activated complex - ∆H(reactants)

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∆H(reaction) = 37 - 140 = -103Kj.

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3 years ago
A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
Evgen [1.6K]

Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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<span>action is the one car hitting the other, reaction is the other car being pushed away</span>
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