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zheka24 [161]
3 years ago
8

Give the structure corresponding to the name (s)−3−iodo−2−methylnonane.

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
7 0
The (s) in the chemical name of (s)-3-iodo-2-methylnonane indicates an S-configuration using the Cahn-Ingold-Prelog system of stereochemical nomenclature. The S-configuration means that an "imaginary" rotation from the highest priority substituent group to the lowest priority substituent group of the chiral center moves counterclockwise (to the left), provided that the lowest priority group is oriented "towards the back" (symbolized by dashed lines). 

The highest priority group (iodine in this case) is the one with the highest atomic number and the lowest priority (hydrogen in this case) is one with the lowest atomic number.

If the atoms directly beside the chiral center have the same atomic number (Carbon-2 and Carbon-4 in this case), the atoms next to them will be evaluated until a point of difference is found. Carbon-2 is connected to 2 other carbon atoms and 1 hydrogen atom, while Carbon-4 is connected to only 1 carbon atom and 2 hydrogen atoms. Thus, Carbon-2 has a higher priority, with the point of difference being the carbon atom of the methyl group attached to Carbon-2. Both Carbon-2 and Carbon-4 are connected to one carbon atom from the main nonane chain, but the other atoms connected to Carbon-4 are hydrogen atoms only. Carbon-2 has an extra carbon connected to it and carbon has a higher atomic number than hydrogen.

If there is no point of difference, the central atom is not chiral and cannot be named using the Cahn-Ingold-Prelog system. 

Thus, the structure of (s)-3-iodo-2-methylnonane is


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Answer:

9.6 mol AgCl2

Explanation:

You have to use Avogadro's number: 6.023 x 10^23

5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2

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3 years ago
If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

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8 0
3 years ago
A mixture that has two or more substances that are spread out evenly is called _____
svlad2 [7]

Answer:

D

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heterogeneous mixture you can see the different substances that go into it.

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If 0.600mol of chloride gas reacted with 0.500mol of aluminium metal to produce aluminium chloride,which reactant is in excess?h
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Answer:

The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3

Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3

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Required: Excess reactant___; Number of moles of AlCl3 produced__

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Based on the given:

0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)

0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)

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4 0
3 years ago
A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch
kozerog [31]

Answer:

\large \boxed{\text{4.63 atm}}

Explanation:

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\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ =    1 °C

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Calculations:

(a) Convert the temperatures to kelvins

T₁ = (   1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The  new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}

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2 years ago
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