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mihalych1998 [28]
2 years ago
7

Think Critically. Describe two ways you could influence the following equilibrium to produce more ethanal (CH3CH). Use Le Chatel

ier's principle to explain why each of your methods would produce the desired result.
Chemistry
1 answer:
ch4aika [34]2 years ago
8 0
Ethanal can be by oxidation of ethanol, as shown below:

CH3CH2OH    +       [O]            ↔          CH3CHO + H2

Above reaction is exothermic in nature, hence later amount of heat is released during the course of reaction.

Now, Le Chatelier's principle is stated as 'any changes in the temperature, volume, or concentration of a system will result in predictable and opposing changes in the system in order minimize this change and achieve a new equilibrium state.' 

Thus, the yield of product can be enhanced by following two ways:

1) Increasing the concentration of reactant i.e. CH3CH2OH. Due to this, equibrium will shift to right and hence, favor the formation of product.
2) Decreasing the temperature of system. Since, the reaction is exoothermic in nature, according to Le Chatelier's principle, lowering the temperature of system will shift the equibrium will shift to right and hence, favor the formation of product. 

Thus, concentration of reactant and temperature of system will <span>influence the equilibrium to produce more ethanal.</span>
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Find the molarity of 186.55 g of sugar (C12H22O11) in 250. mL of water.
Anna [14]

Answer:

The molarity of this sugar solution in water is 2.18 M

Explanation:

Step 1: Data given

Mass of sugar (C12H22O11) = 186.55 grams

Molar mass of C12H22O11 = 342.3 g/mol

Volume of water = 250.0 mL = 0.250 L

Step 2: Calculate moles sugar

Moles sugar = mass sugar / molar mass sugar

Moles sugar = 186.55 grams / 342.3 g/mol

Moles sugar = 0.545 moles

Step 3: Calculate molarity of the sugar solution

Molarity = moles sugar / volume of water

Molarity = 0.545 moles / 0.250 L

Molarity = 2.18 MThe molarity of this sugar solution in water is 2.18 M

6 0
2 years ago
his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round yo
Alinara [238K]

Answer:

\%\ Composition\ of\ iron=69.92\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is Cr(NO_3)_3

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100

\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100

\%\ Composition\ of\ iron=69.92\ \%

5 0
2 years ago
Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii
Lana71 [14]

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

5 0
2 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
2 years ago
State whether each of the following will be more soluble in water or hexane i. Butane ii. Ch3cooh iii. K2so4
ioda

Explanation:

Solubility is determined by the principle , "like dissolves like" .

i.e. , if a compound is polar then it will dissolve in a polar compound only , and

if a compound is non - polar then it will dissolve in a non - polar compound only .

Hence , from the question ,

Water is a polar molecule , and hence it will dissolve only the polar molecule , i.e. , from the given options the polar molecule is , iii. K₂SO₄

Hexane , is a non - polar molecules ,  hence it will dissolve only the non polar molecule , i.e. , from the given options the non polar molecule is i. Butane .

3 0
2 years ago
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