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3 years ago

Solubility (polar vs. Nonpolar)

Chemistry

2 answers:

OverLord2011 [107]3 years ago

8 0

#.3 Ethanol as a Solvent. Ethanol is a very polar molecule due to its hydroxyl (OH) group, with the high electronegativity of oxygen allowing hydrogen bonding to take place with other molecules. The ethyl (C2H5) group in ethanol is non-polar. Thus, ethanol can dissolve both polar and non-polar substances.
4.Benzene is non-polar Electric charge in the molecules of non-polar solvents is evenly distributed, therefore the molecules have low dielectric constant. Non-polar solvents are hydrophobic (immiscible with water). Non-polar solvents are liphophilic as they dissolve non-polar substances such as oils, fats, greases.

eduard3 years ago

4 0

Explanation:

It is given that polar solutes can be dissolved in polar solvents and non-polar solutes can be dissolved in non-polar solvent.

Alcohol being polar, does not dissolves ionic salt in it.

$CCl_4$ is a non-polar solvent.

From the given options:

1. NaCl: This is an ionic salt and hence, it will be soluble in water only.

2. $I_2$ : Iodine gas is a non-polar solute and hence, will be dissolved in non-polar solvent which is $CCl_4$

3. Ethanol: As, it is a polar molecule and is not an ionic salt, therefore it can be soluble in both water and alcohol.

4. Benzene: It is a non-polar molecule and hence, it will be dissolved in $CCl_4$

5. $Br_2$ : Bromine gas is a non-polar solute and hence, will be dissolved in $CCl_4$

6. $KNO_3$ : This is an ionic salt and hence, it will be soluble in water only.

7. Toluene: This is a non-polar solute and hence, will be dissolved in $CCl_4$

8. $Ca(OH)_2$ : This is an ionic salt and hence, it will be soluble in water only.

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How many atoms and elements are there in C2H5OH

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Total atoms is 9 ( 2 carbon atoms, 5 hydrogen atoms, 1 oxygen atom and 1 hydrogen atom = 9 atoms)

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3 years ago

Describe how you would set up an experiment to test the rela-tionship between completion of assigned homework and the fi-nal gra

ValentinkaMS [17]

To start this test, you need to identify the variables it presents. As you may already know, there are independent and dependent variables. Independent variables are those that act on a factor, influencing it to generate a result. In the case of this experiment, the independent variable is the completion of the homework. The dependent variable, in turn, is the factor that receives the influence of the independent variable, in this experiment this variable is the final grade you received in the course.

After that you must select a number of students, give them their homework and ask each student to complete a percentage of that amount. An example of this could be that you select 11 students and ask the first to complete 0% of the homework, the second student must complete 10%, the third 20% and so on, and the 11th student must complete 100% of the homework.

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3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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3 years ago

. One of the essential minerals in the human body is salt. How much salt (NaCl) is in the average adult human body?

son4ous [18]

Answer:

200g or 40 teaspoons

Explanation:

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3 years ago

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AlekseyPX

Answer: Potential energy

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3 years ago