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SVETLANKA909090 [29]
3 years ago
7

7. List things that have LOTS of kinetic energy, and tell why.

Chemistry
1 answer:
vlabodo [156]3 years ago
7 0

Answers: -

For high kinetic energy, the object must have high speed of movement.

1) An airplane has a lot of kinetic energy. Airplanes move at high speed and thus posses a lot of kinetic energy.

2) A bullet from a gun has a lot of kinetic energy due to the high speed of bullet.

3) A formula one car moving at high speeds have a lot of kinetic energy.

4) A train moving at high speed has lots of kinetic energy.

5) An asteroid has a lot of kinetic energy due to it's high speed.

6) A roller coaster moving at high speeds have a lot of kinetic energy.

7) A missile fired from a fighter plane has lots of kinetic energy.

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Explanation:

an increase in concentration increases the rate of the reaction. This is because there are more reactant particles available which allows for more effective collisions between reactant particles in a given period of time. More effective collisions bring about a faster rate of reaction.

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What is the electron configuration of the element with 27 protons?
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A tank of gas has partial pressures of nitrogen and oxygen equal to 1.61 × 104 kPa and 4.34 × 103 kPa , respectively.What is t
Alchen [17]

Answer:

20.44\times 10^3 kPa is the total pressure of the tank.

Explanation:

Partial pressures of nitrogen = p_{N_2}=1.61\times 10^4 kPa

Partial pressure of oxygen =  p_{O_2}=4.34\times 10^3 kPa

Total pressure of gases in the tank = P

Applying Dalton's law of partial pressures :

P=p_{N_2}+p_{O_2}=1.61\times 10^4 kPa+4.34\times 10^3 kPa

P=20.44\times 10^3 kPa

20.44\times 10^3 kPa is the total pressure of the tank.

7 0
2 years ago
The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the
IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

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6547.9 feet to meters<br> Using dimensional analysis
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92.964

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