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2 years ago

Why was 1,2 dichlorobenzene used as the solvent for the diels alder reaction we performed in the lab?

Chemistry

1 answer:

pav-90 [236]2 years ago

8 0

Answer:

1,2 dichlorobenzene was used as the solvent for the diels alder reaction: because the co elimination part of the reaction needs high temp and a high boiling point solvent such as 1,2 dichlorobenzene

Explanation:

Diels-Alder Reaction is a useful synthetic tool to prepare cyclohexane rings. It is a process, which occurs in a single step that consists of a cyclic redistribution of its electrons. The two reagents are bond together through a cyclic transition state in which the two new C-C bonds are formed at the same time. For this to occur, most of the time, it is necessary a high temperature and high-pressure conditions. Since 1,2 dichlorobenzene has a boiling point of 180ºC is a good solvent for this type of reactions.

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Fill in the following: Sulfur--32 isotope notation: #p+ : #nº : #e- :

liq [111]

Answer:

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Explanation:

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2 years ago

Phosphine (PH3) can be prepared by the reaction of calcium phosphide , Ca3P2: based on this equation : Ca3P2 + 6H2O —-> 3 Ca(

MrMuchimi

Taking into account the reaction stoichiometry, you can observe that:

one mole of Ca₃P₂ produces 2 mol of PH₃.
the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Ca₃P₂ + 6 H₂O → 3 Ca(OH)₂ + 2 PH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Ca₃P₂:1 mole
H₂O: 6 moles
Ca(OH)₂: 3 moles
PH₃: 2 moles

The molar mass of the compounds is:

Ca₃P₂: 182 g/mole
H₂O: 18 g/mole
Ca(OH)₂: 74 g/mole
PH₃: 34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Ca₃P₂: 1 mole ×182 g/mole= 182 grams
H₂O: 6 moles× 18 g/mole= 108 grams
Ca(OH)₂: 3 moles ×74 g/mole= 222 grams
PH₃: 2 moles ×34 g/mole= 68 grams

<h3>Correct statements</h3>

Then, by reaction stoichiometry, you can observe that:

one mole of Ca₃P₂ produces 2 mol of PH₃.
the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.

Learn more about the reaction stoichiometry:

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2 years ago

What are the major difference between potential and kinetic energy?

Sholpan [36]

Answer:

The energy of a body or a system with respect to the motion of the body or of the particles in the system. Potential energy is the stored energy in an object or system because of its position or configuration. Kinetic energy of an object is relative to other moving and stationary objects in its immediate environment.

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3 years ago

What is true of saturated fatty acids?

Natali5045456 [20]

Answer:

The second answer, because when something saturated, it has the maximum possible number of hydrogen atoms.

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3 years ago

Read 2 more answers

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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3 years ago