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Nina [5.8K]
3 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]3 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]3 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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Explanation:

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The atomic number of an element gives the number of __ and __in the atom while the mass number gives the total number of__and__
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6 0
2 years ago
Read 2 more answers
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
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