The group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from <u>Alkenes</u>
<h3>What are organic compounds?</h3>
Organic compounds are compounds which contains carbon and hydrogen
Some few classes or organic compounds or hydrocarbons are as follows:
- Alkanes
- Alkenes
- Alkynes
- Alkanols
- Alkanoic acid
- Ketones
- Esters
So therefore, the group of unsaturated hydrocarbons which 2 carbons are double bonded together, with H bonded to the left, and C H 2 bonded below left, above right, and below right is derived from <u>Alkenes</u>
Learn more about organic compounds:
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Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=
Density of octane= d = 
Mass of octane ,m= 
Moles of octane =
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%
Total moles = 11.212 mol
Moles of hexane :
Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D = 
Volume of hexane = V
(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.
Answer:
32.3 dm³
Explanation:
Data given:
no. of molecules of Cl₂ = 8.7 x 10²³
Volume of chlorine gas (Cl₂) = ?
Solution:
First we have to find number of moles
For this formula used
no. of moles = no. of molecules / Avogadros number
no. of moles = 8.7 x 10²³ / 6.022 x 10²³
no. of moles = 1.44 moles
Now we have to find volume of the gas
for this formula used
no. of moles = volume of gas / molar volume
molar volume = 22.4 dm³/mol
Put values in above equation
1.44 moles = volume of Cl₂ gas / 22.4 dm³/mol
rearrange the above equation
volume of Cl₂ gas = 1.44 moles x 22.4 dm³/mol
volume of Cl₂ gas = 32.3 dm³
Umm. . . Could you list some options? :)
Answer:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm
Total pressure is 0.966 atm
Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol
moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol
moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol
Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K
From ideal gas law equation, P = nRT/V
partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm
partial pressure of oxygen = (0.224 x 0.0821 x 332)/23.1 = 0.264 atm
partial pressure of carbon dioxide = (0.336 x 0.0821 x 332)/23.1 = 0.396 atm
Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm
