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2 years ago

Acetone major species present when dissolved in water

Chemistry

1 answer:

jek_recluse [69]2 years ago

3 0

Answer: acetone molecule ( CH₃-CO-CH₃)

Explanation:

1) Acetone is CH₃-CO-CH₃

2) That is a molecule (build up of covalent bonds).

3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.

This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.

That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).

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The first step in the Ostwald process for producing nitric acid is 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g). If the reaction of

wel

Answer:

a. 77%

Explanation:

To solve the exercise, you need to know the limiting reagent. The limiting reagent is one that is consumed first in the chemical reaction. The other reagent is known as an excess reagent, and it is the one left over in a chemical reaction.

To know if ammonia or oxygen is the limiting reagent, we should relate the weights of both compounds in the reaction. We know that 4NH3 (68 g) react with 5O2 (160 g), so we should find out how much NH3 is required to react with 150 g of O2:

160 g O2 _______ 68 g NH3

150 g O2 _______ x = 150 g * 68 g / 160 g = 63.75 g NH3

As we have 150 g of NH3 but only 63.75 g are required, we know that ammonia is the excess reagent, so oxygen is the reagent that limits the reaction and that we should use to calculate the percentage yield of the reaction.

In the reaction we observe that 5O2 (160 g) produces 4NO (120g), so:

160 g O2 ______ 120 g NO

150 g O2 ______ x = 150 g * 120 g / 160 g = 112.5 g NO

If the percent yield of the reaction were 100%, it would produce 112.5 g of NO, however only 87 g of NO were obtained, so we should find out what the percentage of reaction yield was:

112.5 g NO ______ 100%

87 g NO _______ x = 87g * 100% / 112.5 g = 77%

Thus we observe that the reaction had a yield of 77%.

7 0

2 years ago

Read 2 more answers

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon

Ulleksa [173]

Answer:

Theoretical yield = 31.8 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $Al_2O_3$

Mass of $Al_2O_3$ = 60.0 g

Molar mass of $Al_2O_3$ = 101.96128 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{60.0\ g}{101.96128\ g/mol}$

$Moles_{Al_2O_3}= 0.5885\ mol$

Given: For $C$

Given mass = 30.0 g

Molar mass of $C$ = 12.0107 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{30.0\ g}{12.0107\ g/mol}$

$Moles_{C}= 2.4978\ mol$

According to the given reaction:

$Al_2O_3+3C\rightarrow 2Al+3CO$

1 mole of aluminium oxide react with 3 moles of carbon

0.5885 mole of aluminium oxide react with $3\times 0.5885$ moles of carbon

Moles of carbon = 1.7655 moles

Available moles of carbon = 2.4978 moles

Limiting reagent is the one which is present in small amount. Thus, aluminium oxide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of aluminium oxide on reaction forms 2 moles of aluminium.

0.5885 mole of aluminium oxide on reaction forms $2\times 0.5885$ moles of aluminium.

Moles of aluminium = 1.177 moles

Molar mass of aluminium = 26.981539 g/mol

Mass of sodium sulfate = Moles × Molar mass = 1.177 × 26.981539 g = 31.8 g

<u> Theoretical yield = 31.8 g</u>

3 0

2 years ago

When elemental sodium is added to water, the sodium atoms ionize spontaneously. uncharged na becomes na+. this means that the na

Temka [501]

Ionized, become charged, become a cation

hopefully that helps

5 0

2 years ago

You are given three bottles labeled a, b, and

ElenaW [278]

Answer :Solid in bottle a is ionic, solid in bottle b is molecular and solid in bottle c is ionic.

Explanation :

Ionic compound is formed when a metal atom donates one or more electrons to a non metal. This results in the formation of a cation ( a positive ion) and an anion ( a negative ion). These ions are bonded to each other by electrostatic attraction.

The intermolecular forces in case of a an ionic compound are very strong.

The melting point of a substance depends on how strongly the molecules are attracted to each other. Stronger the forces, higher is the melting point.

Therefore ionic compounds always have very high melting points.

On the other hand, covalent compounds have weak intermolecular forces. Therefore they have low melting points.

Based on above discussion, we can classify the given compounds as follows.

a) Solid in bottle a is Ionic as it has high melting point.

b) Solid in bottle b is molecular as it has low melting point.

c) Solid in bottle c is Ionic as it has high melting point.

6 0

2 years ago

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0

2 years ago