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Radda [10]
2 years ago
4

Which general trend is demonstrated by the Group 17 elements as they are considered in order from top to bottom on the Periodic

Table?
(1) a decrease in atomic radius
(2) a decrease in electronegativity
(3) an increase in first ionization energy
(4) an increase in nonmetallic behavior
Chemistry
2 answers:
lord [1]2 years ago
8 0
(2) a decrease in electronegativity can be noted while traveling down the periodic table, especially since Fluorine exists in Group 17 with the highest electronegativity of 4.0. The rest are all the opposite as you travel down the periodic table in general.
tatiyna2 years ago
8 0

Answer:

2. decrease in electronegativity

Explanation:

Reference Table S shows a decrease in electronegativity as the Group 17 elements are considered from top to bottom on the Periodic Table of the Elements.

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The total energy can be found by adding the different energies:
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8 0
2 years ago
A liquid has a volume of 34.6 ml and a mass of 46.0 g. what is the density of the liquid?
Minchanka [31]

The density of a substance can simply be calculated by dividing the mass by the volume:

density = mass / volume

 

Therefore calculating for the density since mass and volume are given:

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5 0
2 years ago
Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)
Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0
2 years ago
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