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Hannah J.
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SAT II Chemistry
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Question:

Molecular Geometry From the following molecules matched with molecular geometries, which is correct? A) $$BrF_3$$ - trigonal planar B) $$NF_4^+$$ - tetrahedral C) $$OCl_2$$ - linear D) $$ClF_5$$ - trigonal bipyramidal

Hannah J.

B) $$NF_4^+$$ - tetrahedral Draw out the Lewis dot structures for each molecule shown to solve (Google them here). A) $$BrF_3$$: $$Br$$ is the central atom bonded to three $$F$$ atoms. As $$Br$$ has 7 valence electrons and is bonded to three atoms, it will have four electrons left over, organized in two pairs. Three bonds with two electron pairs is T-shape. B) $$NF_4^+$$ - $$N$$ is the central atom bonded to four $$F$$ atoms. As $$N$$ has five valence electrons and is bonded to four atoms, it will have a single electron left over. The molecule has a $$+1$$ charge, meaning that it is less one electron than the neutral molecule. The electron lost will come from the central atom. Four bonds with no electron pairs is tetrahedral. C) $$OCl_2$$ - $$O$$ is the central atom bonded to two $$Cl$$ atoms. As $$O$$ has six valence electrons and is bonded to two atoms, it will have four electrons left over, organized in two pairs. Two bonds with two electron pairs is bent. D) $$ClF_5$$ - $$Cl$$ is the central atom bonded to five $$F$$ atoms. As $$Cl$$ has seven valence electrons and is bonded to five atoms, it will have an expanded octet and two electrons left over, organized in one pair. Five bonds with one electron pair is square pyramid.

Chemistry
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Question:

Thermochemistry and Work $$10.0gCu$$ react with excess $$HNO_{3}$$ at $$32^{\circ}C$$ and $$0.751atm$$. How much work is done by the system in $$L*atm$$, when $$R = 0.0821\frac{L*atm}{mol*K}$$? $$2 Cu + 6 HNO_{3} \rightarrow 3H_2 + 2Cu(NO_3)_3$$

Hannah J.

$$3.93 L*atm$$ 1. Start by evaluating the pieces you are given: a mass, a temperature, a pressure, and an $$R$$ variable. The presence of an $$R$$ variable in the question, in addition to the other pieces together, should make you think of the ideal gas law: $$PV=nRT$$. 2a. Now that you know your basic equation, you can start filling in the blanks. $$P$$ = pressure = $$0.751atm$$, this has the correct unit $$V$$ = volume = ??? $$n$$ = moles = $$10.0gCu$$, this must be converted into moles $$R$$ = ideal gas constant = $$0.0821\frac{L*atm}{mol*K}$$, this unit will tell you the units of the other variables $$T$$ = temperature = $$32^{\circ}C$$, this must be converted in Kelvins, $$K$$ You are missing only one variable, $$V$$ = volume, so volume is what you are solving for. 2b. Find your $$n$$ variable. Convert your $$10.0gCu$$ into moles of $$Cu(NO_3)_3$$ using the molar mass of $$Cu$$ on the periodic table and the mole ratio of $$Cu$$ to $$Cu(NO_3)_3$$ from the given reaction. $$10.0gCu * \frac{1 molCu}{63.55 g Cu} * \frac{2molCu(NO_3)_3}{2molCu} = 0.157molCu(NO_3)_3 = n$$ 2c. Find your $$T$$ variable. $$32^{\circ}C + 273 = 305K = T$$ 2d. Plug 'n chug into $$PV=nRT$$! $$(0.751atm)*(V)=(0.157mol)*(0.0821\frac{L*atm}{mol*K})*(305K)$$ $$V=5.23L$$ 3a. Remember that you are solving for work. The work equation is: $$w=-P \Delta V$$. 3b. You know $$P = 0.751atm$$, and $$V_f = 5.23L$$. Because you didn't have any $$Cu(NO_3)_3$$ on the reactants side of the reaction, $$V_o=0$$ and $$\Delta V = V_f - V_o = 5.23L-0L = 5.23L = V_f$$. 3c. Plug 'n chug into $$w=-P \Delta V$$! $$w = -(0.751atm)*(5.23L) = 3.93 L*atm$$ Note that even if you didn't remember the equation for work, you could still finish half the problem just by looking at the information given and piecing it together. If you're confused, try something you know, because you might remember what you're doing along the way!

Basic Chemistry
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Question:

Naming Compounds Write the formula name for tetraphosphorus triselenide.

Hannah J.

$$P_{4}Se_{3}$$ 1. The use of numerical prefixes in the given chemical name should immediately lead you to the conclusion that the compound must be molecular compound, even if you don't recall that is a compound of two or more nonmetals. 2. You must know your numerical prefixes: mono = one, di = two, tri = three, tetra = four, and etc. You are given TETRAphosphorus and TRIselenide, so you have 4 phosphorus and 3 selenides in your compound. 3. Identify the correct elemental symbol for the elements phosphorus and selenide from the periodic table. You should come up with P and Se (note these are both nonmetals, as they should be). 4. Put your pieces together to get the answer: 4 P and 3 Se = $$P_{4}Se_{3}$$

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