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adell [148]
4 years ago
5

These models show the electron structures of two different nonmetal elements. Element 1 at left has a purple circle at center wi

th 2 concentric black lines around it. The first line has 2 small green balls on it. The second line has 8 small green balls on it. Element 2 at right has a purple center with 5 concentric circles around it, with the first circle innermost. The first circle has 2 small green balls on it, and the second circle has 8 small green balls on it. The third circle has 18 small green balls on it, and the fourth circle has 18 small green balls on it. The fifth circle has 6 small green balls on it. Which element is likely more reactive, and why?
Chemistry
2 answers:
Daniel [21]4 years ago
7 0

Answer:

Element 2

Explanation:

If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.

The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.

OlgaM077 [116]4 years ago
7 0

Answer: D

Element 2 is more reactive because it does not have a full valence shell, so it will attract electrons.

Explanation: Answer on Edgenuity

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4NH3+5O2-4NO+6H2O<br>how many moles of NH3 must react to produce 5.0 moles of NO?​
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Calculate the pressure of 3.2 moles of an ideal gas that occupies a volume of 87 m3 at a temperature of 312 K.
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Answer:

                  P  =  0.0009417 atm

Or,

                 P  =  9.417 × 10⁻⁴ atm

Or,

                  P  =  0.0954157 kPa

Or,

                  P  =  0.715677 mmHg (Torr)

Explanation:

Data Given:

                 Moles  =  n  =  3.2 mol

                 Temperature  =  T  =  312 K

                  Pressure  =  P  =  ?

                  Volume  =  V  =  87 m³ = 87000 L

Formula Used:

Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,

                  P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for P,

                  P  =  n R T / V

Putting Values,

                  P  =  (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L

                  P  =  0.0009417 atm

Or,

                 P  =  9.417 × 10⁻⁴ atm

Or,

                  P  =  0.0954157 kPa

Or,

                  P  =  0.715677 mmHg (Torr)

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3 years ago
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