I believe that the choices for this question are:
C2H4O2, C4H8O4 CH2O, C6H12O6 C3H6O3, C6H12O6 C2H4O2, C6H12O6
The answer to this based on the molar masses given is:
C2H4O2, C6H12O6
To prove calculate the molar mass:
C2H4O2 = 2*12 + 4*1 + 2*16 = 60
C6H12O6 = 6*12 + 12*1 + 6*16 = 180
The IUPAC name for the organic compound CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol
This compound above belong to class of organic compound called the alkanols
<h3>What are organic compounds?</h3>
Organic compounds are compounds which contains carbon and hydrogen only.
Some few classes of organic compounds includes the following:
- Alkanes
- Alkenes
- Alkynes
- Alkanols
- Alkanals
- Alkanones
- Esters
- Ketones and other
So therefore, the IUPAC name for the organic compound is CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol
Learn more about organic compounds:
brainly.com/question/704297
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Answer:
C. the use of hydrogen gas as an electron donor.
Explanation:
Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.
I don't know if it is organized by reactivity
it is organized by families
it is organized by atomic number
it is NOT organized by atomic mass
it is organized by periods and groups
Pressure is 5.7 atm
<u>Explanation:</u>
P1 = Standard pressure = 1 atm
P2 = ?
V1 = Volume = 10L
V2= 2.4L
T1 = 0°C + 273 K = 273 K
T2 = 100°C + 273 K = 373 K
We have to find the pressure of the gas, by using the gas formula as,

P2 can be found by rewriting the above expression as,

Plugin the above values as,
