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Anit [1.1K]
3 years ago
5

Write the autoionization reaction for methanol, ch3oh.

Chemistry
1 answer:
aliina [53]3 years ago
4 0
Autoionization Reactions are those reactions in which ions or molecules ionizes spontaneously without adding any external reagent.

For Example,
                    Autoionization of water.

                               H₂O  +  H₂O   ⇆   H₃O⁺  +  OH⁻

Autoionization reaction of Methanol is shown below,

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A student has a large number of coins of different diameters, all made by the same metal. She wishes to find the density of the
Elenna [48]

Answer:

A. In a graduated cylinder, put some quantity of water and measure the initial volume. Then put a coin and measure the volume. To find the volume of the coin, simply subtract the initial volume (water only) from the ending volume (water + coin). To measure the mass, take a dry coin and place it on an electronic scale. Density = mass / volume, so divide the mass by the volume to calculate the density of the coin.

B. When measuring the volume, make sure to look at the graduated cylinder at eye level and read from the bottom of the meniscus.

6 0
2 years ago
A chemical bond results from the mutual attrac- tion of the nuclei for
aleksandrvk [35]

for what????? complete the question

8 0
3 years ago
Read 2 more answers
For each set of values, calculate the missing variable using the ideal gas law.
allsm [11]

Answer:

1. n = 0.174mol

2. T= 26.8K

3. P = 1.02atm

4. V = 126.88L

Explanation:

1. P= 2.61atm

V = 1.69L

T = 36.1 °C = 36.1 + 273= 309.1K

R = 0.082atm.L/mol /K

n =?

n = PV / RT = (2.61x1.69)/(0.082x309.1)

n = 0.174mol

2. P = 302 kPa = 302000Pa

101325Pa = 1atm

302000Pa = 302000/101325 = 2.98atm

V = 2382 mL = 2.382L

T =?

n = 3.23 mol

R = 0.082atm.L/mol /K

T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K

3. P =?

V = 0.0250 m³ = 25L

T = 288K

n = 1.08mol

R = 0.082atm.L/mol /K

P = nRT/V = (1.08x0.082x288)/25 = 1.02atm

4. P = 782 torr

760Torr = 1 atm

782 torr = 782/760 = 1.03atm

V =?

T = 303K

n = 5.26 mol

R = 0.082atm.L/mol /K

V = nRT/P

V = (5.26x0.082x303)/1.03 = 126.88L

8 0
3 years ago
Part A
velikii [3]

Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.

Explanation:

Plato

5 0
3 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

7 0
2 years ago
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