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BlackZzzverrR [31]
3 years ago
6

When applied to a dish, soap makes grease soluble in water. Which explanation correctly supports the role of intermolecular forc

es in this common observation? The nonpolar end of a soap molecule attaches itself to grease. The nonpolar end of a soap molecule attaches itself to water. The polar soap molecule attaches itself to the dish. The polar soap molecule attaches itself to a nonpolar soap molecule.

Chemistry
2 answers:
Mekhanik [1.2K]3 years ago
7 0

Explanation:

Soaps attach to both water and grease molecules.

The grease molecules are attracted more strongly towards each other as compared to water molecules. Also, water molecules are smaller in size hence, strong intermolecular force is required to break the hydrogen bonds of water molecule so that grease or oil molecules can enter the water molecule.

A soap molecule goes in between water and grease molecule and helps them to bind. The force for linkage between water and grease molecule through the soap molecule is weak london dispersion force.

The soap molecule has its salt end as ionic and water soluble. When grease or oil is added to the soap and water solution then the soap acts as an emulsifier. The soap forms miscelles of the non-polar tails and grease molecules are trapped between these miscelles. This miscelle is easily soluble in water hence, the grease is washed away.

Thus, it can be concluded that the nonpolar end of a soap molecule attaches itself to grease.

Liula [17]3 years ago
7 0

Answer: The non-polar end of a soap molecule attaches itself to grease.

Explanation:

Soap molecule that is <em>miscelle </em>has two parts:

  • Polar part or hydrophilic part : This part interacts with polar water molecules.
  • Non-polar or hydrophobic : This part interacts with the non-polar molecules that is grease.

By the interaction of non-polar part with grease and interaction of  polar part with water due to which grease get easily dissolved in water.

Hence,the correct answer is : 'The non-polar end of a soap molecule attaches itself to grease'.

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For the following problem convert both the reactants to moles and balance chemical equationsThe reaction of 167 g Fe2O3 with 85.
saul85 [17]

Let's start by balancing the reaction:

Fe_2O_3+CO\longrightarrow Fe+CO_2

As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

Fe_2O_3+3CO\longrightarrow Fe+3CO_2

That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

Fe_2O_3+3CO\longrightarrow2Fe+3CO_2

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

M_{Fe_2O_3}=2\cdot M_{Fe}+3\cdot M_O=(2\cdot55.845+3\cdot15.9994)g/mol=159.6882g/molM_{CO}=1\cdot M_C+1\cdot M_O=(1\cdot12.0107+1\cdot15.9994)g/mol=28.0101g/mol

Now, we can convert their masses to number of moles:

\begin{gathered} M_{Fe_{2}O_{3}}=\frac{m_{Fe_2O_3}}{n_{Fe_{2}O_{3}}} \\ n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_{2}O_{3}}}=\frac{167g}{159.6882g/mol}=1.045787\ldots mol \end{gathered}\begin{gathered} M_{CO}=\frac{m_{CO}}{n_{CO}} \\ n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{85.8g}{28.0101g/mol}=3.063180\ldots mol \end{gathered}

Now, to determine the limiting reactant, we need to divide both the number of mole by their coefficients on the balanced reaction, so we can see how many we need per reaction of each:

\begin{gathered} Fe_2O_3\colon\frac{n_{Fe_2O_3}}{1}=\frac{1.045787\ldots mol}{1}=1.045787\ldots mol \\ CO\colon\frac{n_{CO}}{3}=\frac{3.063180\ldots mol}{3}=1.021060\ldots mol \end{gathered}

Now, the limiting reactant is the one we have less number of moles per reaction. We can see that we have less CO than Fe₂O₃, so the limiting reactant is CO.

4 0
1 year ago
The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:
bagirrra123 [75]

Answer:

12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow  14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)

Explanation:

The steps of the Ostwald process:

4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)

2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g)

3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g)

Combinning the equations:

4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)

+

(2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g))*2

+

(3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g))*4/3

=

4 NH_3 (g)+ 4 NO (g)+ 7 O_2 (g) + 4 NO_2 (g) +4/3 H_2O (l) \longrightarrow 4 NO (g) + 6 H_2O (g) +  4 NO_2(g) + 8/3 HNO_3 (g) + 4/3 NO (g)

Simplifying:

4 NH_3 (g)+ 7 O_2 (g) + \longrightarrow  14/3 H_2O (l) + 8/3 HNO_3 (g) + 4/3 NO (g)

12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow  14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)

The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.

4 0
3 years ago
Calculate the root means square velocity of nitrogen molecules in 25 degrees Celsius
Mademuasel [1]
Calculate the root mean square velocity of nitrogen molecules at 25°C.
297 m/s
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7 0
3 years ago
Rewrite and Balance the Equation <br> N₂ + H₂ -&gt; NH3
Mars2501 [29]

Answer:N

2

+ 3

H

2

-----> 2N

H

3

Explanation:

N

2

+

H

2

-----> N

H

3

Let us balance this equation by counting the number of atoms on both sides of the arrow.

N

2

+

H

2

-----> N

H

3

N=2 , H=2 N=1, H=3

To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N

H

3

on RHS

N

2

+

H

2

-----> 2N

H

3

N=2 , H=2 N=2 , H= 6

To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of

H

2

on LHS

N

2

+ 3

H

2

-----> 2N

H

3

N=2 , H=6 N=2 , H= 6

Answer link

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