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padilas [110]
3 years ago
13

Alkenes can be converted to alcohols by hydroboration‑oxidation. Draw the structure of the alcohol or alcohols formed in the rea

ction sequence. Clearly indicate stereochemistry by drawing a wedged bond, a dashed bond and two in‑plane bonds per each chiral carbon. Draw hydrogen atoms that are connected to wedge‑and‑dash bonds.

Chemistry
1 answer:
OLEGan [10]3 years ago
6 0

Hydroboration-oxidation is a reaction which follow anti markonikov orientation.

The stereochemistry of hydroboration-oxidation follows syn-adition being a stereospecific reaction. That means that boron and hydrogen are added from the same side.  Then, when the oxidation occurs, the boron is replaced by an hydroxyl having the same geometric position.

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At 850°C, CaCO3 undergoes substantial decomposition to yield CaO and CO2. Assuming that the ΔH o f values of the reactant and pr
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Answer:

The enthalpy if 68.10 grams of CO2 is produced is  -189.04 kJ

Explanation:

<u>Step 1:</u> Data given

temperature = 850 °C

Mass of 68.10 grams of CO2

ΔH°f (CaO) = -635.6 kJ/mol

ΔH°f (CO2) = -693.5 kJ/mol

ΔH°f (CaCO3) =-1206.9 kJ/mol

<u>Step 2: </u>The balanced equation

CaCO3(s) → CaO(s) + CO2(g)

<u>Step 3:  </u>Calculate ΔH°reaction

ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)

ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) -  ΔH°f (CaCO3)

ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol

ΔH°reaction = -122.2 kJ /mol

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 68.10 grams / 44.01 g/mol

Moles CO2 = 1.547 moles

<u>Step 5:</u> Calculate the enthalpy change for 68.10 grams of CO2

-122.2 kJ/mol * 1.547 moles = -189.04 kJ

The enthalpy if 68.10 grams of CO2 is produced is  -189.04 kJ

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