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bazaltina [42]
3 years ago
7

The motion of objects can be quite complex. The motion of this frog is an example. Suppose that the frog moves from one lily pad

to the other in 1 second.
Describe the motion of the frog as it makes one round trip between the two lily pads. Use the terms speed, velocity, displacement, and distance in your answer.
Chemistry
1 answer:
olga2289 [7]3 years ago
8 0

Answer:

The speed of the frog is 1m per second the velocity is he can turn and jump in a different direction in 1 second the distance is 1m and there is no displacement. I'm pretty sure this is right

Explanation:

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The strongest intermolecular forces are in ion-ion bonds which happen when a metal bonds to another metal. 2. The next strongest forces are ion-dipole bonds which happen when metals bond to nonmetals. 3.

4 0
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If you know the number of protons, how do you find the number of electrons
iren [92.7K]

Answer:

to find the number of electrons you need the charge of the element.

4 0
3 years ago
What is the equilibrium expression for the reatcion bellow<br><br> 2SO3(g)&lt;=&gt;O2(g)+2SO2(g)
Luda [366]

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

8 0
3 years ago
What are the formulas of the acids: <br>AsO4<br>CIO4 ( l )<br>S ( ll )<br>F ( l )<br>PO4 (lll)​
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Mg3(AsO4)2
Ca(ClO4)2
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[F (I) not sure]
PO₄³

Sorry I don’t know all of them, good luck though! :)
4 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
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