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Nina [5.8K]
2 years ago
13

For a given chemical reaction, the addition of a catalyst provides a different pathway that(1) decreases the reaction rate and h

as a higher activation energy
(2) decreases the reaction rate and has a lower activation energy
(3) increases the reaction rate and has a higher activation energy
(4) increases the reaction rate and has a lower activation energy
Chemistry
2 answers:
kykrilka [37]2 years ago
6 0
For a given chemical reaction, the addition of a catalyst provides a different pathway that "<span>(4) increases the reaction rate and has a lower activation energy"</span>
Komok [63]2 years ago
3 0

Answer: Option (4) is the correct answer.

Explanation:

Activation energy is the minimum amount of energy required by the reactants to participate in the chemical reaction.

Therefore, reactant molecules whose energy is less than the activation energy are not able to participate in the reaction. Hence, a catalyst lowers the activation energy so that molecules with less energy can also participate in the reaction.

Thus, a catalyst is a specie that increases the reaction rate and has a lower activation energy.

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Answer:

I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.

n = 5 4th excited state

n = 4 3rd excited state

n = 3 2nd excited state

n = 2 1st excited state

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n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.

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n = 3 to 2, 3 to 1 = 2 lines

n = 2 to 1 = 1 line. Add 'em up. I get 10.

b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.

c.The E for any level is -21.8E-19 Joules/n^2

To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.

So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.

Explanation:

8 0
2 years ago
Calculate the pH of 1.00 X 10^-6 M HCI *​
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7 0
3 years ago
2C2H6 + 7O2 ------&gt; 4CO2 + 6H2O
Romashka-Z-Leto [24]

We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.

Dividing the mass of each reactant by its molar mass:

(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6

(10 g O2)(31.999 g/mol) = 0.3125 mol O2.

Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.

Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).

So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.

Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.

8 0
2 years ago
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