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Andru [333]
3 years ago
12

What substance takes part in an enzymatic reaction, but is unchanged by the reaction?

Chemistry
1 answer:
Ierofanga [76]3 years ago
5 0
The answer to this question would be: enzyme 

<span>Enzyme is a kind of catalyst that </span>will take part in the chemical reaction but it won't be reacted. E<span>nzyme </span>only helps by reducing the energy needed to start the reaction.
Theoretically, the enzyme is not mandatory needed but if you have an enzyme the reaction will be easier, thus increasing reaction speed. But in most case, the reaction might be too slow without enzyme, making enzyme vital in living organism.
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What is the density of an object with 525 grams and a volume of 15cm3?
Hunter-Best [27]

Answer: 35 g/cm

Explanation:

Density equals mass over volume. 525 divided by 15 is 35

7 0
3 years ago
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How much copper is in 4.64 grams of copper
Romashka [77]

Answer:

4.64 grams.

Explanation:

without stating a desired unit, stating the answer in any unit is acceptable. So you can use grams and the problem is done for you

6 0
3 years ago
Predict the precipitate produced by mixing an Al(NO3)3 solution with a NaOH solution. Write the net ionic equation for the react
weqwewe [10]
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)

The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
4 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
A) The equilibrium shifts to the left, producing more H3O+.
Lunna [17]

Answer: these are just options, what is the main question, without it, we cannot determine which option is correct, so please repost the question.

5 0
3 years ago
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