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Papessa [141]
3 years ago
13

Ruben's Series

Chemistry
1 answer:
andrew11 [14]3 years ago
7 0

Explanation:

Four elements, hydrogen, carbon, oxygen and nitrogen, are the major components of most organic compounds. Consequently, our understanding of organic chemistry must have, as a foundation, an appreciation of the electronic structure and properties of these elements. The truncated periodic table shown above provides the orbital electronic structure for the first eighteen elements (hydrogen through argon). According to the Aufbau principle, the electrons of an atom occupy quantum levels or orbitals starting from the lowest energy level, and proceeding to the highest, with each orbital holding a maximum of two paired electrons (opposite spins).



Electron shell #1 has the lowest energy and its s-orbital is the first to be filled. Shell #2 has four higher energy orbitals, the 2s-orbital being lower in energy than the three 2p-orbitals. (x, y & z). As we progress from lithium (atomic number=3) to neon (atomic number=10) across the second row or period of the table, all these atoms start with a filled 1s-orbital, and the 2s-orbital is occupied with an electron pair before the 2p-orbitals are filled. In the third period of the table, the atoms all have a neon-like core of 10 electrons, and shell #3 is occupied progressively with eight electrons, starting with the 3s-orbital. The highest occupied electron shell is called the valence shell, and the electrons occupying this shell are called valence electrons.

The chemical properties of the elements reflect their electron configurations. For example, helium, neon and argon are exceptionally stable and unreactive monoatomic gases. Helium is unique since its valence shell consists of a single s-orbital. The other members of group 8 have a characteristic valence shell electron octet (ns2 + npx2 + npy2 + npz2). This group of inert (or noble) gases also includes krypton (Kr: 4s2, 4p6), xenon (Xe: 5s2, 5p6) and radon (Rn: 6s2, 6p6). In the periodic table above these elements are colored beige.

The halogens (F, Cl, Br etc.) are one electron short of a valence shell octet, and are among the most reactive of the elements (they are colored red in this periodic table). In their chemical reactions halogen atoms achieve a valence shell octet by capturing or borrowing the eighth electron from another atom or molecule. The alkali metals Li, Na, K etc. (colored violet above) are also exceptionally reactive, but for the opposite reason. These atoms have only one electron in the valence shell, and on losing this electron arrive at the lower shell valence octet. As a consequence of this electron loss, these elements are commonly encountered as cations (positively charged atoms).

The elements in groups 2 through 7 all exhibit characteristic reactivities and bonding patterns that can in large part be rationalized by their electron configurations. It should be noted that hydrogen is unique. Its location in the periodic table should not suggest a kinship to the chemistry of the alkali metals, and its role in the structure and properties of organic compounds is unlike that of any other element.

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The formula of nitrogen oxide is NO, of nitrogen dioxide is NO_2. Write a balanced equation for the reaction of nitrogen oxide w
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a) 0,5 mol O₂; 1 mol NO₂

b)

Liters of NO = 22,4 L

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d) 7,41g HCl

e) 107,7 g/mol

f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL

ii. 51%

g) 32,6L

Explanation:

a) For the reaction:

2 NO + O₂ → 2 NO₂

For 1 mole of NO there are consumed:

1 mol NO ×\frac{1molO_{2}}{2molNO} = <em>0,5 mol of O₂</em>

And produced:

1 mol NO ×\frac{2molNO_{2}}{2molNO} = <em>1 mol of NO₂</em>

b) By ideal gas law:

V = nRT/P

Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) The mass of each compound are:

1 mol NO×\frac{30 g}{1molNO} = <em>30g</em>

0,5 mol O₂×\frac{32 g}{1molO_{2}} = <em>16g</em>

1 mol NO₂×\frac{46 g}{1molNO_{2}} = <em>46g</em>

d) Using:

n = PV / RT

Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:

0,203 moles of HCl. In grams:

0,203 mol HCl×\frac{36,46 g}{1molHCl} = <em>7,41 g of HCl</em>

e) Using:

δRT/P = MW

Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)

And MW is molecular mass: <em>107,7 g/mol</em>

f) For the reaction:

2 KClO₃ → 2 KCl + 3 O₂

2,550 g of KClO₃ are:

2,550 g of KClO₃×\frac{1mol}{122,55 gKClO_{3}} = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃×\frac{3 mol O_{2}}{2 molKClO_{3}} = <em>0,0312 moles of O₂</em>

In grams:

0,0312 moles of O₂ ×\frac{32g}{1 molO_{2}} = <em>0,999g of O₂</em>

V = nRT/P

At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; <em>V = 0,699L ≡ 699mL</em>

At 29 °C (302,15K) and 732 torr (0,963 atm)

<em>V = 0,803L ≡ 803mL</em>

ii. 182 mL ≡ 0,182L of O₂ are:

n = PV/RT

moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:

7,07x10⁻³ moles of O₂×\frac{2 mol KClO_{3}}{3 molO_{2}} = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃×\frac{122,55 g}{1molKClO_{3}} = 1,300 g of KClO₃.

Thus, percent by mass of KClO₃ in the mixture is:

1,300g/2,550g ×100 = <em>51%</em>

g. Combined gas law says that:

\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}

Where:

P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) <em>V₂ = 32,6 L</em>

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