All categories

3 years ago

Draw the Lewis structure for CH3CH2O2H?

Chemistry

1 answer:

kotegsom [21]3 years ago

6 0

Answer:

hydroperoxyethane

Explanation:

tomsFor the Lewis structure we have to remember that all the atoms must have 8 electrons (except for hydrogen). In this structure, we have three types of atoms, Carbon, Hydrogen and Oxygen. So, we have to remember the valence electrons for each atom:

-) Carbon : 4 electrons

-) Hydrogen: 1 electron

-) Oxygen: 6 electrons

We can start with the " $CH_3$ " part. We can put 3 hydrogen bond arroun the carbon. We can use this same logic with " $CH_2$ ". Finally for oxygens, we can put it one bond with $CH_2$ and a bond between oxygens with a final bond with hydrogen to obtain hydroperoxyethane.

See figure 1 for further explanations.

You might be interested in

5. How can you tell the difference between CuS and Cu2S

puteri [66]

Answer:

Density, melting point. and magnetic properties

Explanation:

I can think of three ways.

1. Density

The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.

It should be possible to distinguish these even with high school equipment.

2. Melting point

Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).

A Bunsen burner can easily reach these temperatures.

3. Magnetic properties

You can use a Gouy balance to measure the magnetic susceptibilities.

In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.

In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.

A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.

A sample of CuS will be attracted by the magnetic field and show an increase in weight.

In the picture below, you can see the sample partially suspended between the poles of an electromagnet.

6 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$