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Anna71 [15]
3 years ago
5

In a system when energy is transformed from one form to another

Chemistry
1 answer:
maw [93]3 years ago
4 0
An energy transformation<span> is the change of </span>energy<span> from </span>one form to another<span>. </span>Energy<span> transformations occur everywhere every second of the day. There are many different </span>forms<span> of </span>energy<span> such as electrical, thermal, nuclear, mechanical, electromagnetic, sound, and chemical.</span>
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Which of the following will affect the rate of a chemical reaction?
dimulka [17.4K]

Answer:

Solute concentration will afect the rate of a chemical reaction, because you must work with molarity

Explanation:

I think that solute mass may be it can affect the rate of reaction, if you have more mass in a solute, you will also have more moles.

If you want to know more, you have to consider temperature in the reaction and the  presence of catalysts. They all, affect reactions.

7 0
3 years ago
Read 2 more answers
Write one common thing between condensation and hydrolysis ?​
serious [3.7K]

Answer:

The common thing is the compound water

Explanation:

in condensation h2O is expelled while in hydrolysis water is used or added

8 0
2 years ago
F(x) = 2|x+2| - 8 I need help ​
Korolek [52]

Answer: |x+2|=xF2+4

Explanation: not sure if this is right im just guessing

8 0
3 years ago
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
What is the percent by mass of oxygen in carbon dioxide (CO2)?
krok68 [10]

Answer:

I think that it will be

Explanation:

72.71%

8 0
3 years ago
Read 2 more answers
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