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Interpretation:
The statements about the general properties of solutions are true or false has to be identified.
Concept introduction:
Solution: It is the homogeneous mixture of two or more substances. In such a mixture, solute is a substance (of lower proportion) dissolved in another substance (of larger proportion), called solvent. In aqueous solution, the solvent is water.
Homogeneous mixture:
- It is a mixture in which the components are uniformly distributed throughout the mixture.
- There is only one phase of matter observed in a homogeneous matter.
- Example- Air, Sugar water , Rain water
![Check Mark](/static/check-mark.png)
Explanation of Solution
Reason for true statements:
Option a):
In a solution, more than one substance can dissolve in a given solvent.
Hence, the statement is true.
Option b):
A solution is formed by dissolving two or more substances in a solvent. In a solution, the components are distributed uniformly through-out and hence it is homogeneous.
Hence, it is true.
Option c):
In a solution, the components are distributed uniformly throughout and thus every part is exactly the same.
Hence, it is true.
Reason for wrong statements:
Option d):
A solution is a homogenous mixture and it does not separate into parts. Therefore, the solutes present in a solution will not settle out with time if kept undisturbed.
Hence, the statement is wrong.
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Chapter 8 Solutions
General, Organic, and Biological Chemistry Seventh Edition
- टे Predict the major products of this organic reaction. Be sure to use wedge and dash bonds when necessary, for example to distinguish between different major products. ☐ ☐ : ☐ + NaOH HO 2 Click and drag to start drawing a structure.arrow_forwardShown below are five NMR spectra for five different C6H10O2 compounds. For each spectrum, draw the structure of the compound, and assign the spectrum by labeling H's in your structure (or in a second drawing of the structure) with the chemical shifts of the corresponding signals (which can be estimated to nearest 0.1 ppm). IR information is also provided. As a reminder, a peak near 1700 cm-1 is consistent with the presence of a carbonyl (C=O), and a peak near 3300 cm-1 is consistent with the presence of an O–H. Extra information: For C6H10O2 , there must be either 2 double bonds, or 1 triple bond, or two rings to account for the unsaturation. There is no two rings for this problem. A strong band was observed in the IR at 1717 cm-1arrow_forwardPredict the major products of the organic reaction below. : ☐ + Х ك OH 1. NaH 2. CH₂Br Click and drag to start drawing a structure.arrow_forward
- NG NC 15Show all the steps you would use to synthesize the following products shown below using benzene and any organic reagent 4 carbons or less as your starting material in addition to any inorganic reagents that you have learned. NO 2 NC SO3H NO2 OHarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardShow work...don't give Ai generated solutionarrow_forward
- 1 Please provide an efficient synthesis of the product below from the starting material. Use the starting material as the ONLY source of carbon atoms. Show the synthesis of each compound that would be used in the overall synthesis of the product. [This synthesis uses alkyne and alcohol chemistry.]arrow_forward10- 4000 20 20 30- %Reflectance 60 50- 09 60- 40- Date: Thu Feb 06 17:30:02 2025 (GMT-05:0(UnknownP Scans: 8 Resolution: 2.000 70 70 88 80 3500 3000 2500 90 100 00 Wavenumbers (cm-1) 2000 1500 2983.10 2359.13 1602.52 1584.22 1451.19 1391.87 1367.07 1314.37 1174.34 1070.13 1027.33 1714.16 1269.47 1000 1106.08 1001.14 937.02 873.60 850.20 780.22 686.91 674.38 643.09 617.98 02/06/25 16:38:20arrow_forwardd. Draw arrow-pushing mechanism for an enzymatic retro-aldol reaction of the following hexose. Use B: and/or HA as needed. OH OH سية HO OH OHarrow_forward
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