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Mama L [17]
2 years ago
11

Why would an atom emit particles or waves from its nucleus?

Chemistry
1 answer:
Neko [114]2 years ago
5 0

Answer:

It’s A non of these

Explanation:

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Calculate the vapor pressure of water at T=90°C
LenKa [72]

The equilibrium vapour pressure is typically the pressure exerted by a liquid .... it is A FUNCTION of temperature...

Explanation:

By way of example, chemists and physicists habitually use

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saturated vapour pressure

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100

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,

P

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Well, because this is the normal boiling point of water: i.e. the conditions of pressure (i.e. here

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100

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, at which the VAPOUR PRESSURE of the liquid is ONE ATMOSPHERE...and bubbles of vapour form directly in the liquid. As an undergraduate you should commit this definition, or your text definition, to memory...

At lower temperatures, water exerts a much lower vapour pressure...but these should often be used in calculations...especially when a gas is collected by water displacement. Tables of

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4 0
3 years ago
Mr. Smith wanted to see if the color of light shined on a plant had an effect on the number of leaves it had. He gathered a 2
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3 0
2 years ago
HELP ASAP PLEASE<br> Use the word Precision in a sentence relating to chemistry!
Leno4ka [110]

Answer:

I needed to use good precision in my measurement for my chemistry lab

Explanation:

7 0
3 years ago
What's the compound name of Si+ C2H3O1/2?
pochemuha
What is the name of CuC2H3O2?
The answer is Copper (I) Acetate. (:
3 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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