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Mashutka [201]
2 years ago
6

Atorvastatin is sold under the trade name Lipitor and is used for lowering cholesterol. Annual global sales of this compound exc

eed $13 billion. Assign a configuration to each chiral center in atorvastatin: In the boxes below, first input the numbers corresponding to the chiral centers in numerical order as a comma separated list (e.g., "1,3,4"). Then input the configurations of the chiral centers in the same order and format (e.g., "R,S,S"). chiral center(s): Configuration(s):

Chemistry
1 answer:
kotegsom [21]2 years ago
4 0

Answer:

Atorvastatin has two chiral centers. The question doesn't include the box where have to answer but I can show you in an image where are located and their configuration.

Explanation:

The first image shows the chemical structure of atorvastatin and their chiral centers identified as 1 and 2 respectively.

The second image shows the Fischer projections corresponding to every chiral carbon 1 and 2. I wrote R so suggest that there are more carbon atoms forward but not only corresponds to carbon atoms.

You can see that the chiral carbon 1 has R configuration due to the direction from the main substituent to the second follow the clockwise.

The chiral carbon 2 has S configuration due to the direction from the main substituent to the second one go anticlockwise.

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Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces
Shalnov [3]

Answer:  The concentration of excess [OH^-] in solution is 0.017 M.

Explanation:

1. Molarity=\frac{moles}{\text {Volume in L}}

moles of HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles

1 mole of HCl give = 1 mole of H^+

Thus 0.019 moles of HCl give = 0.019 mole of H^+

2. moles of Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles

According to stoichiometry:

1 mole of Ba(OH)_2 gives = 2 moles of OH^-

Thus 0.012 moles of Ba(OH)_2 give = 2 \times 0.012=0.024 moles of OH^-

H^++OH^-\rightarrow H_2O

As 1 mole of H^+ neutralize 1 mole of OH^-

0.019 mole of H^+ will neutralize 0.019 mole of OH^-

Thus (0.024-0.019)= 0.005 moles of OH^- will be left.

[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M

Thus molarity of [OH^-] in solution is 0.017 M.

4 0
2 years ago
Read 2 more answers
Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp
weeeeeb [17]

Answer : The grams of N_2 consumed is, 89.6 grams.

Solution : Given,

Mass of CaCN_2 = 265 g

Molar mass of CaCN_2 = 80 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate the moles of CaCN_2.

\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles

The given balanced reaction is,

CaC_2+N_2\rightarrow CaCN_2+C

from the reaction, we conclude that

As, 1 mole of CaCN_2 produces from 1 mole of N_2

So, 3.2 moles of CaCN_2 produces from 3.2 moles of N_2

Now we have to calculate the mass of N_2

\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2

\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g

Therefore, the grams of N_2 consumed is, 89.6 grams.

5 0
2 years ago
Read 2 more answers
Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi
ivann1987 [24]

Answer:The formulas of ionic compounds are:

a)CuBr

b)Mn_2O_3

c)Hg_2I_2

d)Mg_3(PO_4)_2

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains Cu^+ ion.

Cu^++Br^-\rightarrow CuBr

(b) Manganese oxide : Given that it contains Mn^{3+} ion.

Mn^{3+}+O^{2-}\rightarrow Mn_2O_3

(c)Mercury iodide :Given that it contains Hg_2^{2+}

Hg_2^{2+}+I^-\rightarrow Hg_2I_2

(d) Magnesium phosphate :Given that it contains PO_4^{3-}

Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2

4 0
2 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
2 years ago
At 10.°C, 20.g of oxygen gas exerts a pressure of 2.1atm in a rigid, 7.0L cylinder. Assuming ideal behavior, if the temperature
Alja [10]

Answer:

Final pressure = 2.3225 atm

Amontons’s law states that

At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature

Explanation:

Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure

Amontons’s law: P/T = Constant at constant V and n

That is P1/T1 = P2/T2

Where temperature is given in Kelvin

Hence T1 of 10°C = 273.15 + 10 = 283.15K

Also temperature T2 of 40°C = 313.15 K

Hence

P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm

3 0
2 years ago
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