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dlinn [17]
3 years ago
6

Which term refers to the speed at which reactants are converted into products?

Chemistry
1 answer:
shutvik [7]3 years ago
6 0

Answer:

activation energy hope this is right

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What is the definition of relief in science terms
SOVA2 [1]
The apparent topography exhibited by minerals in thin section as a consequence of refractive index.
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4 years ago
HEYYY FAM IS THIS RIGHT
Otrada [13]

Answer: A reversible reaction is a reaction that takes place in back and front directions. If the reaction were to reach equilibrium, the rate of forward direction would be equal to that of the reverse reaction.

Explanation:

Reversible reactions :

These are the reaction in which reactants reacts to give product and products reacts to give reactants as a product in return.

        A+B\leftrightharpoons C+D

  • In above equation, 'A' and 'B' are reacting together to give 'C' and 'D', as products and vice-versa.
  • When the above reaction reaches equilibrium the rate of forward and backward reaction becomes equal.




3 0
3 years ago
Convert 31.9 g of N2 to moles
pashok25 [27]

Answer: 1.14mol

Nitrogen molar mass=14g/mol

N2 molar mass=14*2=28g/mol

31.9g(1/28)=1.14mol N2

5 0
4 years ago
POINTS!!!!
Setler79 [48]
C) 3 Moles
This is because of the molecular structure shown in the image

5 0
3 years ago
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer usi
insens350 [35]

Explanation:

Formula to calculate standard electrode potential is as follows.

          E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between E^{o}_{cell} and K is as follows.

            E^{o}_{cell} = \frac{RT}{nF} \times ln K

                 ln K = \frac{nFE^{0}_{cell}}{RT}      

Substituting the given values into the above formula as follows.

                 ln K = \frac{nFE^{0}_{cell}}{RT}    

                        =  \frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}  

                ln K = -41.28

                    K = e^{-41.28}    

                        = 1 \times 10^{-18}

Thus, we can conclude that the value of the equilibrium constant for the given reaction is 1 \times 10^{-18}.      

5 0
3 years ago
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