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Gennadij [26K]
3 years ago
12

What can be determined about the electron of an atom

Chemistry
1 answer:
swat323 years ago
6 0
By the electrons in the atoms you can know how many protons are in the atom and you can see what elements to mix with which without the risk of a fatal accident
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Which option is an example of physical change?
Murljashka [212]

Answer:

They are all physical exept c

Explanation:

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3 years ago
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Which metal will gain heat faster. Brass or Gold?
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here is your answer >>>

In order to find out which metal gains heat faster, we will have to find out their specific heat!. Specific heat is the heat required to raise the temperature of the unit substance, The specific heat of gold and brass are 0.126 for gold and 0.380 for brass!. The specific heat of gold is less than brass!. So, gold will gain heat faster!.

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3 years ago
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Soil that are light in color have a great amount of humus in<br> them. True or False
GaryK [48]

Answer:

please the answer Is false

6 0
2 years ago
Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.
N76 [4]

Answer:

1 . Al_2O_3

2. TiO_2

Explanation:

The more stable the ionic compound, the more is it lattice energy.

  • The more the charge on the cation and the anion, the greater is the lattice energy.
  • The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide (Sc_2O_3) is an oxide in which Sc^{3+} behaves as cation and O^{2-} behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . Al_2O_3

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation Al^{3+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

2. TiO_2

This is because the charge on the cation Ti^{4+} is greater than that of Sc^{3+} and also the size of the cation Ti^{4+} is smaller than Sc^{3+}. Thus, this corresponds to higher lattice energy.

3 0
3 years ago
10. Isolation of a pure sample of the third product, which has been determined to be an isomer of the major and minor products,
Ymorist [56]

Answer:

Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.

Explanation:

Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.

7 0
3 years ago
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