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2 years ago

Draw the Lewis structure for CH3CH2O2H?

Chemistry

1 answer:

kotegsom [21]2 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.

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Suppose a layer of oil is on the top of a beaker of water. Water in many oils is slightly soluble, so its concentration is so lo

rodikova [14]

Explanation:

It is given that energy to transfer one water molecule is $2.208 \times 10^{-20}$ J/molecule

As it is known that in 1 mole there are $6.022 \times 10^{23}$ atoms.

So, energy in 1 mole = $2.208 \times 10^{-20} \times 6.022 \times 10^{23}$ J/mol

= 13.3 kJ/mol

As, $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}$

Putting the given values in the above formula as follows.

$log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}$

$log \frac{k_{2}}{k_{1}} = \frac{13.3 kJ/mol}{2.303 \times 8.314 atm L/mol K} \times \frac{293 K - 283 K}{283K \times 293 K}$

$log \frac{k_{2}}{k_{1}}$ = 0.08377

$\frac{k_{2}}{k_{1}}$ = 1.213 = $\frac{Concentration_{2}}{Concentration_{1}}$

$Concentration_{2}$ = $1.213 \times Concentration_{1}$

= $1.213 \times 9 \times 10^{-4}$

= $10.915 \times 10^{-4}$ water molecules per oil molecule

Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is $10.915 \times 10^{-4}$ .

4 0

2 years ago

A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t

leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = $P_1=1.0 atm$

Initial temperature of the oxygen gas = $T_1=20^oC =293.15 K$

Final pressure of the oxygen gas = $P_2=?$

Final temperature of the oxygen gas = $T_2=100^oC =373.15 K$

Using Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm$

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

$P_1V_1=nRT_1$ (ideal gas equation)

$n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol$

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

6 0

2 years ago

You are asked to go into the lab and prepare an acetic acid - sodium acetate buffer solution with a ph of 4.00  0.02. what mola

Oxana [17]

Hello!

To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:

$pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )$

$\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75$

So, the mole ratio of CH₃COOH to CH₃COONa is 5,75

Have a nice day!

5 0

2 years ago

Convert 5.0x10^24 molecules to liters.

makvit [3.9K]

Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!

5 0

2 years ago

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

=670*4.1*40.3

=110704.1

8 0

2 years ago