Tutor profile: Pamela S.
Subject: Organic Chemistry
Which of the following compounds are water soluble: Benzene, Diethyl Ether, pentane, cyclohexanol, toluene, phenol
It is important to remember the "like-dissolves-like" rule here. Water is a very polar compound, and thus polar compounds will be water soluble. Any groups that donate a Hydrogen bond to water, such as alcohols or amines, contribute greatly to water solubility. Also, any group that accepts a Hydrogen bond from water increases the solubility, such as ketones, aldehydes, and ethers. Benzene is a non polar compound, and thus is not water soluble. Diethyl Ether is partially soluble in water. This is because while it is an ether, it is also a non-polar compound because of its symmetry. Pentane is not soluble in water because it is a non-polar compound, and has no functional groups. Cyclohexanol would be partially soluble in water. This is because while it is an alcohol, the compound is relatively non polar. Toluene is not soluble in water, as it is not polar and has no functional groups. Phenol is partially soluble in water because it is not polar, but has an -OH functional group.
Balance the following equation: __C2H6O + __O2 → __CO2 + __H2O
The most important thing to remember is the individual charge for each atom. Hydrogen is always +1. Oxygen is always +2. Also, note that if you are unsure of an atoms charge, you can figure it out by looking at the full, neutral molecule. For example, in C2H6O, we know that between the Hydrogens and the Oxygen there is a net charge of 4 (+6 - 2). With 2 Carbons, we can come to the conclusion that each one has a charge of +2 (+4 / 2). To balance the equation, we need 6 Hydrogens on both sides. The H2O on the right side can be made into 3H2O, so that there are 6 hydrogens on the right side. Now, we have 6 H's, 1 C, and 3 O's on the right side. In order to balance, we need 2 Carbons. Adding a 2 in front of the CO2 on the right side, we now have 2 carbons, and a total of *7* Oxygens on the right side. We have now exceeded the number of O's on the left side. This can be simply fixed by adding a 3 in front of the O2 to yield 7 total O's on the left side. We are left with C2H6O + 3O2 --> 2CO2 + 3H2O Counting all the atoms, we have 2C, 6H, and 7O on both the left and right. This reaction is now balanced.
Solve the equation: 3(-7x + 4) - (x - 5) = -3(2x + 7) + 5
It is important to follow PEMDAS when solving these equations. The first step here would be to multiply the numbers outside of the parentheses by what is inside the parentheses. 3(-7x + 4) would come out to -21x + 12. Remember to carry the negative sign that is on the 7. -3(2x + 7) would come out to -6x - 21. Remember when doing this, that each term inside the parentheses must be multiplied by the number directly outside. Note that the negative sign on the 3 follows to both terms in the parentheses. -(x - 5) would come to -x + 5. We are now left with: -21x + 12 - x + 5 = -6x - 21 + 5 At this point, like terms must be added together. The more steps this is done in, the less chance there is for error. Simplifying the terms on the left and the right alone we have: -22x + 17 = -6x - 16 The next simplification can be: -16x = -33 Dividing the term on the x, we have x = 33/16 or 2.0625 as an answer
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