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Maurinko [17]
3 years ago
6

In nature, rings with ____ carbon atoms are the most abundant

Chemistry
2 answers:
iogann1982 [59]3 years ago
8 0

Answer:  Five and Six

Explanation:   In nature, usually we can observe the rings with 5 or 6 carbon atoms are most abundant.

First of all, the rings with 5 or 6 carbon atoms are the most stable rings which are not affected by any free energy factor or enthalpy factor or entropy factor.

Moreover 6 membered rings are found to be the most stable ones because of the extra stability provided by the term known as Aromaticity like Benzne is known to be most stable and most aromatic compound.

5 membered rings are found in our body as ribose sugar which makes the important nucleic acids which are RNA (ribose nucleic acids)and DNA(de oxyribose nucleic acids) .

notsponge [240]3 years ago
4 0
<span>In nature, rings with five to six carbon atoms are the most abundant</span>
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You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of
SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

Moles=1.60 \times {300.0\times 10^{-3}}\ moles

<u>Moles of potassium iodide = 0.48 moles </u>

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=1.20\times {285\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.342 moles </u>

According to the given reaction:

2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

<u>Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.</u>

Also, consumed lead(II) nitrate = 0.24 moles  (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.102/0.585 M = 1.174 M</u>

<u>Statement A is correct.</u>

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

<u>Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g </u>

<u>Statement B is correct.</u>

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.48/0.585 M = 0.821 M</u>

<u>Statement C is correct.</u>

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate  produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate  produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>

<u>Statement D is incorrect.</u>

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3 years ago
What atoms in water are attracted to Chlorine?
aniked [119]

Answer: hydrogen and oxygen

Explanation:

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Convert 150 g/L to the unit g/mL.
inn [45]
The correct answer is C 0.15 g/mL.
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Use reactivity series to predict the products of the following reactions. complete and balance the equation
Gre4nikov [31]

Answer:

products and reaction

Explanation:

Products

1. FeSO4 and Cu

2. ZnSO4 and Fe

3. CaCl2 and H2

Reactions

1.    Fe + CuSO4 → FeSO4 + Cu

2.   FeSO4 + Zn → ZnSO4 + Fe

3.   2HCl + Ca → CaCl2 + H2

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3 years ago
What branch of science is, matter, energy, motion, forces, time
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These are terms that we use in physics
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