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OlgaM077 [116]
2 years ago
6

Mention the type of Reaction common to organic compound-

Chemistry
1 answer:
Anna11 [10]2 years ago
6 0

Answer:

addition reactions, elimination reactions, substitution reactions, pericyclic reactions, rearrangement reactions, photochemical reactions and redox reactions

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The percent composition of carbon in C6H12O6 is:
nata0808 [166]
To calculate percent composition, you first need to find the molar mass of C (carbon), H (hydrogen) and O (oxygen).
C is 12.01
H is 1.00
O is 16

Then multiply each by the number of atoms of each element in the formula (the number that comes after each element in the equation for example C6 means 6 carbon atoms.

C: 12.01 x 6= 72.06
H: 1x12= 12
O: 16x6= 96

Then add them up.
72.06+ 12+ 96= 180.06

Now find the percent composition of carbon.

72.06/ 180.06 x 100= 40.01%

So the answer is C 40%.
8 0
3 years ago
Which statement about the atomic nucleus is correct
Sladkaya [172]
The nucleus is made of protons and neutrons. It has a positive charge.
8 0
3 years ago
Which molecule plays the greatest role in the thermal regulation of the troposphere?
makkiz [27]

Answer: The moon was 20 farms away from earth so If I ate a couch then went to my treadmile outside of my cow, I could add the letter Grammar to my checklist.

Explanation: You see, If there was a man who had no arms or legs he would be walking with his feet right? SO If I added chickens to my head and ate the eifel tower am I 50% a choclate bar. Hope this helps

5 0
3 years ago
An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated
pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we<u> calculate the moles of the gas</u>, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

  • 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
  • n = 0.0256 mol

Now we<u> calculate the molar mass of the compound</u>:

  • 2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:

  • C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
  • H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
  • N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

6 0
3 years ago
Calculate the pH of a 3.58x10^-9 M Nitric acid (a strong acid) solution
lubasha [3.4K]

Answer:

pH= 8.45

Explanation:

when working with strong accids pH = -log(Concentration)

so -log(3.58e-9) = 8.446

7 0
3 years ago
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