Concept explainers
(a)
Interpretation:
The molar mass of the compound is to be calculated.
Concept introduction:
The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.
The formula to calculate the change in freezing point is as follows:
Here,
(b)
Interpretation:
The empirical and the molecular formula of the compound are to be determined.
Concept introduction:
An empirical formula gives the simplest whole number ratio of atoms of each element present in a molecule. The molecular formula tells the exact number of atoms of each element present in a molecule.
(c)
Interpretation:
The Lewis structures for the compound that forms hydrogen bonds and one that does not form hydrogen bonds are to be drawn.
Concept introduction:
Lewis structure is basically a simplified representation of the structure of any molecule or atom. Lewis structure shows the bonding between the atoms and the lone pairs of electrons as dot.
The steps to draw the Lewis structure of any molecule are as follows:
1. Write the letter
2. Count the total number of valence electrons in the molecule. In case of charged molecules subtract the positive charge from the total number of valence electrons and add the negative charge to the total number of valence electrons.
3. Assign two electrons between two atoms and join them via a single bond. Place the remaining valence electrons as lone pairs such that octet of each element is achieved. Use multiple bonds to complete the octet.
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Chapter 13 Solutions
Chemistry The Molecular Nature Of Matter And Change 9th
- Provide steps and explanation please.arrow_forwardDraw a structural formula for the major product of the acid-base reaction shown. H 0 N + HCI (1 mole) CH3 N' (1 mole) CH3 You do not have to consider stereochemistry. ● • Do not include counter-ions, e.g., Na+, I, in your answer. . In those cases in which there are two reactants, draw only the product from 989 CH3 344 ? [Farrow_forwardQuestion 15 What is the major neutral organic product for the following sequence? 1. POCI₂ pyridine ? 2. OsO4 OH 3. NaHSO Major Organic Product ✓ OH OH 'OH OH 'OH 'CIarrow_forward
- Could you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but (color-coded) and step by step so I can understand it better? Thank you! I want to see what they are doingarrow_forward
- Can you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.arrow_forwardPart 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forwardPart 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forward
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