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Leto [7]
3 years ago
15

HELP PLEASE I BEG OF YOU

Chemistry
1 answer:
Vitek1552 [10]3 years ago
7 0

Answer:

37.7%

Explanation:

58% of 65 is 37.7%

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Part C<br> Number of molecules in 8.437x10-2 mol C6H6
N76 [4]

Answer:

There are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

Explanation:

In this problem, we need to find the number of molecules in 8.437\times 10^{-2} mol of C_6H_6.

The molar mass of C_6H_6 is 6\times 12+1\times 6=78\ g/mol

No of moles = mass/molar mass

We can find mass from above formula.

m=n\times M\\\\m=8.437\times 10^{-2}\ mol\times 78\ g/mol\\\\m=6.58\ g

Also,

No of moles = no of molecules/Avogadro number

N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}

Hence, there are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

5 0
3 years ago
What element is used to cross-link the hydrocarbon chains when natural rubber undergoes vulcanization?
makvit [3.9K]
Sulfur is the element used to cross-link the hydrocarbon chains when natural rubber undergoes vulcanization.
The invention relates to cross-linking hydrocarbon polymers with aromatic polyazides in the presence of sulfur. It has been discovered that hydrocarbon polymers can be cross-linked when heated with aromatic polyazides.
6 0
4 years ago
If you stretch back a rubber band and release it, it shoots across the room. What type of energy conversion has occurred?
MAVERICK [17]
<span>Elastic to mechanical is the energy conversion that occurs. Elastic, the pulling back and stretching of the rubber band, mechanical is the release!</span>
5 0
3 years ago
Read 2 more answers
How many grams of MgO are in 3.500 moles?
borishaifa [10]
(3.500 mol MgO)(40.3044g MgO/1mol MgO)
=141.0654
7 0
3 years ago
The decomposition of ozone in the upper atmosphere to dioxygen occurs by a two-step mechanism.The first step is a fast reversibl
alex41 [277]

<u>Answer:</u> Step 2 in the given mechanism is the rate determining step

<u>Explanation:</u>

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  O_3(g)\rightleftharpoons O_2(g)+O(g);\text{ (fast)}

<u>Step 2:</u>  O_3(g)+O(g)\rightarrow 2O_2(g);\text{(slow)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[O_3][O]

Hence, step 2 in the given mechanism is the rate determining step

8 0
4 years ago
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