Answer:
CH3CH3CH2CH3
Explanation:
Octane is a non-polar compound. It is a hydrocarbon with 8-carbon length along its chain.
It belongs to a special group of hydrocarbons called alkanes.
What makes a substance soluble in another?
It is a common phrase that "like dissolves like". This is applicable to solubility of substances in another.
- A polar solvent will freely and easily dissolve a polar solute. For example, water and salt.
- A non-polar solvent will also dissolve a non-polar solute. This case, hydrocarbons will dissolve themselves.
- The first option is a butane, a 4-carbon length hydrocarbon which will be dissolved in octane.
- Both compounds are non-polar.
Answer:
Ethanamine (also known as ethylamine)
Explanation:
The compound that is requested by the question is ethanamine. Its trivial name is ethylamine.
It is a compound that contained the ethyl moiety (CH3CH2-) as well as the amine moiety (-NH2).
Ethanamine has a structure that can easily be determined by the statements in the question.
The structure of ethanamine is shown in the image attached.
Answer:
c. It generates changes in the magnetic field of Earth.
Explanation:
please mark this answer as brainlest
Answer:Attemted Failed
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Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
