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Nostrana [21]
3 years ago
10

A section of the periodic table is shown below: A portion of two rows of the periodic table is shown. The first row reads two he

lium 4.003. The second row reads seven nitrogen 14.007, eight oxygen 15.999, nine flourine 18.998, and 10 neon 20.180. Helium and neon form the last column of the periodic table. Which of the following statements is true? Neon is more reactive than fluorine because neon needs only one electron to fill its outermost shell. Neon is more reactive than oxygen because neon has to lose only one electron to fill its outermost shell. Fluorine is more reactive than nitrogen because fluorine needs only one electron to fill its outermost shell. Fluorine is more reactive than neon because fluorine has to lose only one electron to fill its outermost shell.
Chemistry
1 answer:
cluponka [151]3 years ago
7 0

 The statement  which is true  is

Fluorine  is more reactive  than nitrogen because fluorine  needs  only  one electron to fill  its  outermost  shell.


     <u><em>Explanation</em></u>

Fluorine  has  electron configuration  of 1S²2S²2P⁵  while nitrogen  has 1S²2S²2P³  electron configuration.

The 2P sub shell  for nitrogen   is  half filled  therefore it is sable  than fluorine.

since  p orbital  can hold a maximum  of  6  electrons ,Fluorine  requires  1 electron  to  completely fill  it's 2P  sub shell  which make it    more  reactive than nitrogen.



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Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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