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3 years ago

Los isomeros ópticos

Chemistry

1 answer:

krek1111 [17]3 years ago

7 0

Answer:

Los isómeros ópticos son dos compuestos que contienen el mismo número y tipo de átomos, y enlaces (es decir, la conectividad entre los átomos es la misma), y diferentes disposiciones espaciales de los átomos, pero que tienen imágenes especulares no superponibles. Cada estructura de imagen especular no superponible se denomina enantiómero.

Explanation

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What determines the order of the elements in the modern periodic table ?

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3 years ago

The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol. How much heat would be absorbed or released upon freezing 25.6 grams

Verdich [7]

Answer:

There is 2.52 kJ of energy released (option 4)

Explanation:

Step 1: Data given

The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol

Mass of methanol = 25.6 grams

Molar mass of methanol = 32.04 g/mol

Step 2: Calculate moles of methanol

Moles methanol = mass methanol / molar mass methanol

Moles methanol = 25.6 grams / 32.04 g/mol

Moles methanol = 0.799 moles

Step 3: Calculate energy transfer

Energy transfer = moles * enthalpy of fusion

Energy = 0.799 moles * 3.16 kJ/mol

Energy = 2.52 kJ released

There is 2.52 kJ of energy released

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3 years ago

If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce

Natali5045456 [20]

2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
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3 years ago

Read 2 more answers

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

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3 years ago