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Alona [7]
3 years ago
6

In which model are atoms are imagined as tiny balls?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
5 0
Dalton's Billiard Ball Model
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Increasing technology addiction among the youth write a letter to editor
ruslelena [56]

Answer:

these two would help you

Explanation:

have a good day ᕕ( ᐛ )ᕗ

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2 years ago
Use the drop-down menus to complete the statements
Kipish [7]

Answer:

First one: group

Second one: period

Third one: number of valence electrons

Last one: increases

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3 years ago
Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting mate
Troyanec [42]

Answer:

Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)  

Explanation:

Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.

In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.

8 0
3 years ago
Read 2 more answers
What can you say about the strength of the intermolecular forces in neon and argon based on the critical points of Ne and Ar? (c
algol13

Explanation:

Since, it is given that critical temperature of Argon is 150.9 K and critical pressure of Argon is 48.0 atm.

It is known that gas phase of neon occurs at 50 K. As the boiling point of Ar is more than the boiling point of neon which means that there is strong intermolecular force of attraction between argon molecules as compared to neon molecules.

This is also because argon is larger in size. As a result, induced dipole-induced dipole forces leads to more strength in Ar as compared to Ne.

8 0
3 years ago
The solubility of NaCH3CO2 in water is ~1.23 g/mL. What would be the best method for preparing a supersaturated NaCH3CO2 solutio
Len [333]

Answer:

b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.

Explanation:

The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).

<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>

<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.

<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.

<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.

5 0
3 years ago
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