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daser333 [38]
4 years ago
8

What are the electrons capable of Sharing accepting or donating called?

Chemistry
1 answer:
Oliga [24]4 years ago
8 0
Substances have different tendencies to donate or accept electrons. When a really good donor meets a great acceptor, the chemical reaction releases a lot of energy. Oxygen (O2) is the best electron acceptor and is used in many aerobic reactions (reactions with oxygen). Hydrogen gas (H2) is a good electron donor.

When O2 and H2 are combined, along with a catalyst, water (H2O) is formed. This example of a redox reaction can be written like this:




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The ability to put in a light bulb with greater ease is because the bottom half is what type of simple machine?
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A screw simple machine

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A sample of gas occupies a volume of 50.0 milliliters in a cylinder with a movable piston. The pressure of the sample is 0.90 at
alina1380 [7]
P₁ = 0.90 atm

V₁ = 50.0 mL

T₁ = 298 K

P₂ = 1 atm

T₂ = 273 K

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Answer (1)

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7 0
3 years ago
Which statement is a valid conclusion about the activation energy of the reverse reaction if the forward reaction is exothermic?
babunello [35]

Answer:

e) The activation energy of the reverse reaction is greater than that of the forward reaction.

Explanation:

  • Activation energy is the minimum amount of energy that is required by the reactants to start a reaction.
  • An exothermic reaction is a reaction that releases heat energy to the surrounding while an endothermic reactions is a reaction that absorbs heat from the surrounding.
  • <em><u>In reversible reactions, when the forward reaction is exothermic it means the reverse reaction will be endothermic, therefore the reverse reaction will have a higher activation energy than the forward reaction.</u></em> The activation energy of the reverse reaction will be the sum of the enthalpy and the activation energy of the forward reaction.
4 0
4 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
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