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riadik2000 [5.3K]
2 years ago
3

Classify the molecules as hydrophilic, hydrophobic, or amphipathic (amphiphilic).

Chemistry
1 answer:
adelina 88 [10]2 years ago
8 0
Since there are no molecules to classify, the best I could do is to discuss as much what are the characteristics of molecules that are classified as hydrophilic, hydrophobic, or amphipathic (amphiphilic). Hydrophilic molecules are molecules which are "water-loving" which means they are dissolved in water. These molecules are polar molecules. Examples are salt, milk and soap. Hydrophobic molecules are molecules which are "water-fearing" which means they do not dissolve in water. Examples are oils, fatty acids and soap. As you can see from the examples, soap is written in both. This is because soap has both hydrophobic end and hydrophilic end which makes it an amphiphatic or amphiphilic molecule.
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A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
elena55 [62]

Answer:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Explanation:

Mass of an alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Mass of water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

m_{a} C_{a}  [T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This is the specific heat of the alloy.

4 0
2 years ago
You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
What is the benefit of having a limiting reagent when performing a lab experiment
Shalnov [3]

Answer : The role of limiting reagent or reactant is important in a chemical reaction because it can help the chemist to predict that complete amount of reactant is consumed, as it is limiting the reaction, only required moles of products can get formed instead of the theoretical yield where the perfect amount is used.


In short, Limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is found to be complete.

3 0
2 years ago
Read 2 more answers
Which of the following species contains manganese with the highest oxidation number?
ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0
2 years ago
If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration
Angelina_Jolie [31]
Diluted by a factor of two means that we double the volume of the solution by adding an equal volume of the water.
if we diluted it by a factor of one so the new concentration = 0.1/2=0.05 M and diluted by a factor of two so, the new concentration will be 0.05/2 = 0.025 M
7 0
2 years ago
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