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Gnesinka [82]
3 years ago
5

Hydrophobic molecules (like oil) tend to self-associate in water rather than dissolve. What is the MAJOR energetic contribution

to this property.
Chemistry
2 answers:
Inga [223]3 years ago
8 0

Answer:

The self association maximizes the entropy of the water molecules

Explanation:

Nonpolar molecules, like Oil, do not dissolve easily in water. They are described as hydrophobic, or water fearing. When put into polar environments, such as water, nonpolar molecules stick together and form a tight membrane, preventing water from surrounding the molecule.

Moreso, the hydrophobic effect was found to be entropy-driven at room temperature because of the reduced mobility of water molecules in the solvation shell of the non-polar solute.

anyanavicka [17]3 years ago
4 0

Answer: The MAJOR energetic contribution to this property is the self association that maximizes the entropy of the water molecules.

Explanation:

Hydrophobic molecules (like oil) tend to self-associate in water rather than dissolve in it. The MAJOR energetic contribution to this property is the self association of this oil which increases the degree of disorderliness of the water molecules.

You might be interested in
4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.
ikadub [295]
<h3>Answer:</h3>

0.127 mol Au

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 25.0 g Au

[Solve] moles Au

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})
  2. [DA] Multiply/Divide [Cancel out units:                                                          \displaystyle 0.126923 \ mol \ Au

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.126923 mol Au ≈ 0.127 mol Au

3 0
3 years ago
Read 2 more answers
Your teacher asks you to follow the instructions in a lab to produce several known chemical reactions. During these procedures,
vova2212 [387]

Answer:

There are five evidences that tell whether a chemical change has occurred. These are change of color, change of odor, change in temperature or energy, formation of gas and formation of a precipitate.

Explanation:

Chemical Change- This is a type of chemical reaction which occurs when the properties of one or more atoms change and results into a<u> newly formed substance. </u>

Let's have a further discussion of the evidences.

1. Change of Color- Color change is caused by the combination of two or more substance with different molecular structures. A popular example of this is the Statue of the Liberty, which is made of copper plates. Due to the exposure of copper to elements like water, it changed color.

2. Change of Odor- This can be best presented with rotting food. During the rotting process, the food undergoes a chemical reaction. The result is a rotten smell.

3. Change in Temperature or Energy- An example of this is the burning of wood. Its change is considered non-reversible.

4. Formation of Gas- This can be best presented with the cake batter (the one being used to make cakes or pancakes). The batter rises which means it is forming gas. This is caused by the reaction of the baking soda and the acid.

5. Formation of a Precipitate- This occurs when two soluble salts combine and their outcome is an insoluble salt (this is the precipitate).

Take note that if any of these evidences occur, then there's definitely a chemical reaction.

3 0
3 years ago
I AM SO STUCK!
ankoles [38]
1kg is about 2.2 pounds. So 4kg = 8.8 pounds. If that represent 19% of the total, here is the answer.
0.19*x=8.8
x=8.8/0.19
x=46.31 pounds
3 0
3 years ago
Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su
Free_Kalibri [48]

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

Also,

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,,

Volume =\frac{Moles\ of\ solute}{Molarity}

Given, Molarity = 0.00231 M

So,

Volume =\frac{1.7234}{0.00231}\ L

<u>Volume = 746 L</u>

7 0
3 years ago
In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?
dedylja [7]

<u>Answer:</u> The correct answer is 2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

<u>Explanation:</u>

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

  • <u>Reaction 1:</u>  2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 2:</u>  CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 3:</u>  HCl(aq.)+NaOH(aq.)\rightarrow NaCl(aq.)+H_2O(l)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

  • <u>Reaction 4:</u>  2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: 0

Thus, the oxidation state of hydrogen is changing.

  • <u>Reaction 5:</u>  SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq.)

Oxidation state of hydrogen on reactant side: +1

Oxidation state of hydrogen on product side: +1

Thus, the oxidation state of hydrogen is not changing.

Hence, the correct answer is 2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)

6 0
3 years ago
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