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musickatia [10]
4 years ago
13

5. What is the major difference between the two-dimensional and three- dimensional models of ethane and methane? When might you

choose to use a two-dimensional model rather than a three-dimensional model when representing one of the molecules?
Chemistry
2 answers:
maksim [4K]4 years ago
8 0
<span>The major difference is that the two-dimensional model doesn't show the C-H angles. For example, three-dimensional models are needed to show stereo-isomerism</span>
nordsb [41]4 years ago
3 0

Two-dimensional structures of methane show that it is a carbon atom bonded to four (4) hydrogen atoms while ethane is two (2) carbon bound to six (6) hydrogens. The three-dimensional structures show that methane is tetrahedral is shape while ethane is trigonal pyramidal. The tw0-dimensional structure does not show bond angles and the position in space of the atoms within the molecule.

Two dimension structure can be used to illustrate a chemical reaction and the exchange of atoms by donors and acceptors in the reaction. Three-dimensional structures are significant in illustrating the crystal lattice of molecules since their 3-D structures determine how the molecules interact with each other.


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A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant Kf = 5.32 °C.
Citrus2011 [14]

Answer : The freezing point of a solution is -15.4^oC

Explanation : Given,

Molal-freezing-point-depression constant (K_f) = 5.32^oC/m

Mass of urea (solute) = 29.82 g

Mass of solvent = 500 g  = 0.500 kg

Molar mass of urea = 60.06 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of solvent in Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = ?

\Delta T^o = freezing point of solvent = -10.1^oC

i = Van't Hoff factor = 1 (for urea non-electrolyte)

K_f = freezing point constant = 5.32^oC/m

m = molality

Now put all the given values in this formula, we get

-10.1^oC-T_s=1\times (5.32^oC/m)\times \frac{29.82g}{60.06g/mol\times 0.500kg}

T_s=-15.4^oC

Therefore, the freezing point of a solution is -15.4^oC

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f piece of copper has a density of 13.29 g/cm³ and a volume of 25 cm, what is the mass of the liquid?
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13.29 multipled by 25
the answer is 332.25 grams
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Fill in the blank: <br> A correctly written hypothesis uses an ____ ______ statement
myrzilka [38]

Answer:

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Explanation:

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MatroZZZ [7]

Answer:

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Explanation:

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