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Triss [41]
3 years ago
12

rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first

step? CO32- + H2O → __ + __ CO32- + H2O →
Chemistry
1 answer:
adoni [48]3 years ago
3 0

<u>Answer:</u> The products of the given chemical equation are HCO_3^-\text{ and }OH^-

<u>Explanation:</u>

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are HCO_3^-\text{ and }OH^-

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A chemist lectures about the following definition for acids and bases. The acid-base behavior is analyzed in terms of how electr
Y_Kistochka [10]

Answer:

the Lewis concept of acids and bases

Explanation:

8 0
3 years ago
Read 2 more answers
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
What method would you use to find the volume of an irregularly shaped object?
dimulka [17.4K]
Displacement, usually the method that is used to measure the volume of an irregularly shaped object.
4 0
3 years ago
Element X consist of two isotopes with masses of 62.9 and 64.9. The relative atomic mass of the element is 63.6. What is th
koban [17]

Answer:

it is 50%

Explanation:

62.9x50=3145

64.9x50=3245

3245+3145=6390

6390/100=63.9

4 0
2 years ago
The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b
____ [38]

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

\Delta S_{vap}=\frac{\Delta H_{vap}}{T}

We can solve for the temperature as follows:

T=\frac{\Delta H_{vap}}{\Delta S_{vap}}

Thus, with the proper units, we obtain:

T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C

Hence, answer is approximately 100 °C.

Best regards.

6 0
3 years ago
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