A smaller gear driving a larger gear gives a blank advantage (called blank gears)

Construct a three-step synthesis of 1,2-epoxycyclopentane from cyclopentanol by dragging the appropriate formulas into the bins.

zubka84 [21]

Answer:

(1) Bromination, (2) E2 elimination and (3) epoxidation

Explanation:

In the first step, -OH group in cyclopentanol is replaced by more facile leaving group Br by treating cyclopentanol with $PBr_{3}$
In the second step, E2 elimination in presence of strong base e.g. NaOEt/EtOH produce cyclopentene
In the third step, treatment of cyclopentene with mCPBA produces 1,2-epoxycyclopentane
Full reaction scheme has been shown below

3 0

2 years ago

In the following reaction, what is the quantity of heat (in kJ) released when 5.87 moles of CH₄ are burned?

IRISSAK [1]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

The enthalpy of a chemical reaction as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

In this case, the balanced reaction is:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g)

and the enthalpy reaction ∆H° has a value of -802 kJ/mol.

This equation indicates that when 1 mole of CH₄ reacts with 2 moles of O2, 802 kJ of heat is released.

When 5.87 moles of CH₄ are burned, then you can apply the following rule of three: if 1 mole of CH₄ releases 802 kJ of heat, 5.87 moles of CH₄ releases how much heat?

$heat=\frac{5.87 molesof CH_{4}x802 kJ}{1 mol of CH_{4} }$

Finally, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

Learn more:

5 0

2 years ago

Do step 3 as outlined in the lab guide. Record your results in the appropriate blanks.

PolarNik [594]

Explanation:

Do the step 3 as outlined in the lab guide. record your results in the appropriate blank.

D

8 0

3 years ago

How many moles of electrons are transferred when 2.0 moles of aluminum metal react with excess copper(II) nitrate in aqueous sol

mojhsa [17]

Moles of electrons:

The moles of electrons that are transferred are 12F

A balanced equation:

2 moles of Aluminium metal react with excess copper(II) nitrate.

$2Al + 3Cu{(NO_3)}_2 \rightarrow 2Al{(NO_3)}_3 +3 Cu$

Given:

Moles of Aluminium = 2

As Aluminium goes from 0 to +3 oxidation state

$Al \rightarrow Al^{3+} + 3e^-$

And copper goes from +2 to 0

$Cu^{2+} + 2e^-\rightarrow Cu$

On balancing the number of electrons we get:

For 1 mole of Al $6e^-$ is required.

Therefore for 2 moles of Al,

Total $(2\times6)$ F mole of electrons

Where F= Faraday's constant= 96500 C

So, 12F moles of electrons are transferred.

Learn more about Faraday's Law here,

brainly.com/question/27985929

#SPJ4

4 0

1 year ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

2 years ago

A smaller gear driving a larger gear gives a blank advantage (called blank gears) ￼

A smaller gear driving a larger gear gives a blank advantage (called blank gears)