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Mekhanik [1.2K]
3 years ago
6

What are the products of the following reaction C6H6+O2

Chemistry
1 answer:
Stels [109]3 years ago
5 0

Answer: The answer is C

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I need help solving this chemistry question ​
Arturiano [62]

Answer:

I think the answer is 22.2

Explanation: What i DID was adding 12.0 + 10.0 and than that gave me 22 the I had added the to 0.200 and that how i got 22.2. Sorry if i got is wrong. :(

3 0
3 years ago
If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of
irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of C_{18}H_{21}NO_3 = 46.85 g

Molar mass of C_{18}H_{21}NO_3 = 299.4 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{46.85\ g}{299.4\ g/mol}

Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol

Given, Mass of C_{2}H_{5}OH = 125.5 g

Molar mass of C_{2}H_{5}OH = 46.07 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{125.5\ g}{46.07\ g/mol}

Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}

Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22

4 0
3 years ago
According to solubility rules, which of the following compounds is soluble in water?
Nata [24]
The answer is A hope this help
6 0
3 years ago
Duncan takes a break from studying and goes to the gym to swim laps If swimming burns amount of time ? 85 * 10 ^ 5 cal per hour
Volgvan

When 6.85×10⁵ cal is converted to kilojoules, the result obtained is 2866.04 KJ

<h3>Data obtained from the question </h3>
  • Energy (cal) = 6.85×10⁵ cal
  • Energy (KJ) =?

<h3>Conversion scale </h3>

1 cal = 0.004184 KJ

<h3>How to convert 6.85×10⁵ cal to kilojoules</h3>

1 cal = 0.004184 KJ

Therefore,

6.85×10⁵ cal = 6.85×10⁵ × 0.004184

6.85×10⁵ cal = 2866.04 KJ

Thus, 6.85×10⁵ cal is equivalent to 2866.04 KJ

Learn more about conversion:

brainly.com/question/2139943

6 0
2 years ago
Describe how you could use two LB/agar plates, some E. coli, and some ampicillin to determine how E. coli cells are affected by
KiRa [710]

Answer:

For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin.  Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.

Explanation:

If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.

8 0
3 years ago
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