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3 years ago

10

Notice that there are actually 3 chiral centers in the (1S)-[endo]-(-)borneol substrate. C1 and C2 are obvious. What is the othe

r center? Is the absolute configuration at this center R or S?

1 answer:

lesantik [10]3 years ago

5 0

Answer:

C5 and its absolute configuration is R

Explanation:

<u>A chiral center is a carbon atom (C) which is bonded to four different groups, generating optical activity in the molecule. </u>

The first two chiral centers are:

C1: bonded to OH, H, the carbon chain (seen from left to right) and the carbon chain (seen fron rigth to left).
C2: bonded to CH3, the above C and its CH3, the carbon chain (seen from left to right) and the carbon chain (seen fron rigth to left).

The thrid chiral center is C5, which is bonded to H, the above C and its CH3, the carbon chain (seen from left to right) and the carbon chain (seen fron rigth to left).

The absolute configuration of it is R (clockwise)

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At 50 degrees Celsius and standard pressure inter-molecular forces of attraction are strongest in a sample of ethanoic acid.
Ethanoic acid has hydrogen atom bonded with a more electronegative atom; Oxygen. As a result, the molecule possesses strong intermolecular Hydrogen Bonds. Therefore; ethanoic acid, and all other carboxyllic acids have the tendency to form dimers.

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What was the 'elixir of life'

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A magical or medicinal potion/solution

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3 years ago

How many grams of Mg are needed to produce 224 g of MgO in the complete reaction of Mg

zmey [24]

Answer:

134.4 g of Mg

Explanation:

reaction:

2Mg + O2 ➡️ 2MgO

1) find the mol of MgO

mol = mass / molar mass

mass = 224 g

molar mass = 24+16 = 40

mol = 224 / 40

= 5.6 moles

2 mol = 5.6 moles

2) find the mass of Mg

mass = mol × molar mass

mol = 5.6

molar mass = 24

mass = 5.6 × 24

= 134.4 g

7 0

3 years ago

Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq

nekit [7.7K]

Answer:

1.0 L

Explanation:

Given that:-

Mass of $CaC_2$ = $2.54\ g$

Molar mass of $CaC_2$ = 64.099 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.54\ g}{64.099\ g/mol}$

$Moles_{CaC_2}= 0.0396\ mol$

According to the given reaction:-

$CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}$

1 mole of $CaC_2$ on reaction forms 1 mole of $C_2H_2$

0.0396 mole of $CaC_2$ on reaction forms 0.0396 mole of $C_2H_2$

Moles of $C_2H_2$ = 0.0396 moles

Considering ideal gas equation as:-

$PV=nRT$

where,

P = pressure of the gas = 742 mmHg

V = Volume of the gas = ?

T = Temperature of the gas = $26^oC=[26+273]K=299K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

$742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L$

<u>1.0 L of acetylene can be produced from 2.54 g $CaC_2$ .</u>

4 0

2 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago