A solution is prepared by mixing 525 mL of ethanol with 597 mL of water. The molarity of ethanol in the resulting solution is 8.35 M. The density of ethanol at this temperature is 0.7893 g/mL. Calculate the difference in volume between the total volume of water and ethanol that were mixed to prepare the solution and the actual volume of the solution. g

Answer:

See explanation below

Explanation:

To do aflow diagram for extraction of these two components of a sample, we need to analize both reagents so we can make a great diagram and separate both reagents.

First, let's see the acid. The 2-chlorobenzoic acid is a relativel strong acid, so, in order to separate this from a sample we need to use a base to do so. However, it's very important the use of the base here, we cannot use any base to do it, for the main reason that the sample has other component, and this component may react too with the base and the separation will not be succesfull. So, as the chlorobenzoic acid is a relatively strong acid, if we use a strong base such as NaOH, this will react with the acid, but it will also react with the 1,4-dichlorobenzene forming a Sn2 product and a salt like this.

C₆H₄Cl₂ + NaOH ---------> C₆H₅OCl + NaCl

This is the reason which we cannot use NaOH, because it's a strong base that may react with other compounds in the sample.

To solve this, we can use a weak base such as NaHCO₃. This weak base has the strength enough to react with the benzoic acid, but not strong enough to react with the dichlorobenzene.

So, the first step is dissolve the sample in an organic solvent like ether. The next step is mixing the sample with NaHCO₃. This will cause the layer to divide into two layers. One aqueous layer that will have the product of the acid with the base, and an organic layer with the dichlobenzene. Afterward, we just need to use a strong acid like HCl, but dilluted in the aqueous layer to regenerate the acid, and in the organic layer, just heat the solution til the whole solvent evaporates completely.

The flow diagram is below in the attached picture.

Answer:

C. cooler than both the crust and the core

Explanation:

It is observed that at the mantle, temperatures range from estimatedly 200 °C (392 °F) around the upper boundary with the crust to approximately 4,000 °C (7,230 °F) at the core-mantle boundary.

So we can say the mantle is cooler than both the crust and the core.

A crystal of table salt (NaCl) is dissolved in water. Which of the following statements explains why the dissolved salt does not recrystallize as long as the temperature and the amount of water stay constant?

Na+ and Cl- ions lose their charges in the water.

Water molecules surround the Na+ and Cl- ions.

Na+ and Cl- ions leave the water through vaporization.

Water molecules chemically react with the Na+ and Cl- ions.

Your answer: -

Answer: B - Water molecules surround the Na+ and Cl- ions.

Answer:

[tex]m_{PbI_2}=18.2gPbI_2[/tex]

Explanation:

Hello,

In this case, we write the reaction again:

[tex]Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)[/tex]

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

[tex]n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI[/tex]

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

[tex]0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI[/tex]

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

[tex]m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2[/tex]

Best regards.

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

The mass of zinc sulfide produced is  [tex]M_{ZnS} = 0.76 \ g[/tex]

Explanation:

From the question we are told that

   The mass of zinc is  [tex]m_z = 0.750 \ g[/tex]

    The mass of sulfur is  [tex]m_s = 0.250 \ g[/tex]

The molar mass   of  [tex]Zn_{(s)}[/tex]  is a constant with value  65.39 g /mol

The molar mass of [tex]S_{(s)}[/tex]  is a constant with value  32.01 g/mol

The molar mass of  [tex]ZnS_{(s)}[/tex] is a constant with value 97.46  g/mol

The reaction is  

        [tex]Zn_{(s)} + S_{(s)} ------> ZnS_{(s)}[/tex]

   So from the reaction

       1 mole of  [tex]Zn_{(s)}[/tex] react with 1 mole of  [tex]S_{(s)}[/tex] to produce 1 mole of [tex]ZnS_{(s)}[/tex]

This implies that

65.39 g /mol of  [tex]Zn_{(s)}[/tex] react with 32.01 g/mol of  [tex]S_{(s)}[/tex] to produce   97.46  g/mol  of [tex]ZnS_{(s)}[/tex]

From the values given we can deduce that the limiting reactant is sulfur cause  of the smaller mass

 So  

    0.250 g of  [tex]Zn_{(s)}[/tex] react with 0.250 of  [tex]S_{(s)}[/tex] to produce [tex]x \ g[/tex] of  [tex]ZnS_{(s)}[/tex]

So

      [tex]x = \frac{97.46 * 0.250}{32.01}[/tex]

       [tex]x = 0.76 \ g[/tex]

Thus the mass of the mass of zinc sulfide produced is

    [tex]M_{ZnS} = 0.76 \ g[/tex]

Answer:

Actual yield reduces the more.

Explanation:

An actual yield of the course of a chemical reaction is the mass of a product actually obtained from the reaction.

In practice you see it and It is usually less than the theoretical yield.

Various reasons may come up to explain this away but here is one:

• incomplete reactions, simply put here some of the reactants do not react to form the product.

The same applies in the question about the actual yield will reduce significantly in molar mass now that insufficient amount of reagent are used.

Explanation:

MnO4−+ SO32− → Mn2++ SO42−

Splitting into half equations;

MnO4−  → Mn2+

SO32− → SO42−

Balancing the electrons;

2 MnO4−  + 10 e- → 2Mn2+

5SO32− → 5SO42− + 10 e-

In an acidic medium, it becomes;

2 MnO4−  +  8 H+ → 2 Mn2+ + 4 H2O

5 SO32−  + H2O → 5 SO42−  +  2 H+

Final equation is;

2 MnO4- + 5 SO32- + 6 H+ → 2 (Mn)2+ + 5 SO42- + 3 H2O

Coefficient of H+ = 6

Coefficient of H2O = 3

Coefficient of MnO4- = 2

Coefficient of SO32- = 5

Coefficient of (Mn)2+- = 2

Coefficient of SO42- = 5

Answer:

[tex]5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O[/tex]

Explanation:

Hello,

In this case, given the reaction:

[tex]MnO_4^{-}+ SO_3^{2-} \rightarrow Mn^{2+}+ SO_4^{2-}[/tex]

We first identify the oxidation state of both manganese and sulfur at each side:

[tex]Mn^{7+}O_4^{-}+ S^{4+}O_3^{2-} \rightarrow Mn^{2+}+ S^{6+}O_4^{2-}[/tex]

So we have the oxidation and reduction half-reactions below, including the addition of water and hydronium as it is in acidic media:

[tex]S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-[/tex]

[tex]Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O[/tex]

Next, we exchange the transferred electrons:

[tex]5*(S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-)\\2*(Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O)\\\\5S^{4+}O_3^{2-}+5H_2O \rightarrow 5S^{6+}O_4^{2-}+10H^++10e^-\\2Mn^{7+}O_4^{-}+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O[/tex]

Then we add the resulting half-reactions and simplify the transferred electrons:

[tex]5S^{4+}O_3^{2-}+5H_2O+2Mn^{7+}O_4^{-}+16H^+ \rightarrow 5S^{6+}O_4^{2-}+10H^++ 2Mn^{2+}+8H_2O[/tex]

We rearrange the terms in order to simplify water and hydronium molecules:

[tex]5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+16H^+-10H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+8H_2O-5H_2O\\\\5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+6H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+3H_2O[/tex]

Finally we write the balanced reaction in acidic media:

[tex]5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O[/tex]

Best regards.

Answer:

- Increase pressure .

Explanation:

Hello,

In this case, during the dissolution process, the solute's molecules rearrange in order to get together with the solvent's molecules, in this case water.

Now, since we are talking about a solid whose particles are intimately held together, the only way to separate them is by increasing the temperature because the molecules start moving so they can join water's molecules, decreasing particle size since they will be more likely to separate to each other and increasing stirring since the applied energy will break the solid's intramolecular forces.

In such a way, since pressure significantly affects gases and slightly affects liquid, it is not able to modify a solid, just extreme pressures such as it needed to produce diamonds, is able to affect a solid. For that reason, increasing the pressure will not increase the solid's solubility.

Best regards.

Due to prokaryote

Explanation:

Answer:

At high pressures and low temperatures.  

Explanation:

That's when the volume of the gas is quite small.

The volume of the gas particles can then be a significant proportion of the total volume.

hard because it has calcium and magnesium

Answer:

C

Explanation:

heat is energy

Answer:

C) Integumentary, nervous, muscular

Explanation:

When a human body touches a hot stove , our sense organ which is integumen first of all receives the impulse of heat . The impulse is transformed into electrical signal which is transmitted to brain which is a part of nervous system . Then brain processes it and command signal is sent to muscle of hand to move it away from that place . Hence the order is

Integumentary, nervous, muscular .

Answer:

pH = 9.69

Explanation:

When pyridine (C₅H₅N) is added to water, the equilibrium that occurs is:

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.7x10⁻⁹

Where Kb is defined as:

Kb = 1.7x10⁻⁹ = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

If you have a solution of 1.4M C₅H₅N, the equilibrium concentration of each specie is:

[C₅H₅N] = 1.4 - X

[C₅H₅NH⁺] = X

[OH⁻] = X

Where X represents the reaction coordinate

Replacing in Kb expression:

1.7x10⁻⁹ = [X] [X] / [1.4 - X]

2.38x10⁻⁹ - 1.7x10⁻⁹X =  X²

0 = X² + 1.7x10⁻⁹X - 2.38x10⁻⁹

Solving for X:

X = -0.0000488M → False answer, there is no negative concentrations

X = 0.0000488M → Right answer

Thus, [OH⁻] = 0.0000488M. As pOH = -log [OH⁻]

pOH = 4.31

Knowing pH = 14 - pOH

pH = 9.69

The pH of a 1.4 M pyridine solution is 9.69. When pyridine (C₅H₅N) is added to water, the equilibrium occurs. The rate of forward reaction is equals to the rate of backward reaction.

When pyridine (C₅H₅N) is added to water, the equilibrium that occurs is:

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.7x10⁻⁹

Where Kb is defined as:

Kb = 1.7x10⁻⁹

Kb= [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

If you have a solution of 1.4M C₅H₅N, the equilibrium concentration of each specie is:

[C₅H₅N] = 1.4 - x

[C₅H₅NH⁺] = x

[OH⁻] = x

Where x represents the reaction coordinate

Replacing in Kb expression:

1.7*10⁻⁹ = [x] [x] / [1.4 -x]

2.38*10⁻⁹ - 1.7x10⁻⁹x =  x²

0 = x² + 1.7*10⁻⁹x - 2.38*10⁻⁹

Solving for x:

x = 0.0000488M

Thus, [OH⁻] = 0.0000488M.

As pOH = -log [OH⁻]

pOH = 4.31

Knowing pH = 14 - pOH

pH = 9.69

Find more information about Equilibrium constant here:

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Answer:

.00352mol/L

Explanation:

molarity (concentration) = number of moles / volume of solvent (in L)

M = .99g / 56.108‬g/mol / .5L

M = .0176mol / .5L

M = .00352mol/L

Answer:

2 HCl

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

400 mL

Explanation:

Boyle's Law: [tex]P_1*V_1 = P_2*V_2[/tex]

Let x = the resulting volume

350 (200) = 700 (x)

x = 400 mL

Answer:

By reacting carbon monoxide and hydrogen the formation of methanol takes place, the reaction is,

CO(g)+2H₂(g)⇔CH₃OH (g)

Based on the given reaction, one mole of methanol is obtained by reacting one mole of carbon monoxide (CO) with the two moles of hydrogen (H₂). It is mentioned in the question that for the reaction 500 mol of carbon monoxide and 750 moles of hydrogen are present.

Therefore for 500 moles of carbon monoxide, there is a requirement of 2 × 500 moles of hydrogen, which is equivalent to 1000 moles of hydrogen (H₂). However, only 750 moles of hydrogen is present. Therefore, the limiting reactant in the given case is H₂. The present moles of H₂ will react with 0.5 × 750 moles of CO = 375 mole of CO

The additional or excess concentration of CO, which is the excess reactant will be, 500-375 = 125 moles.

Answer:

The porosities in dental materials can be of merit and not merit since some dental materials need to be porous in order to reduce their weight and improve their properties and functions, whereas in those (which are the majority) porosities are not Merit, see the properties of the material, the function and can even influence dental rehabilitation.

An example of this is the use of vitreous ionomers, which is a dental material, where when it is spatulated it remains porous, it can give recurrence of infectious caries, since its permeability increases, the best cariogenic microorganisms are filtered and porosity works as a retentive site for these microorganisms that make up the dental biofilm.

In summary, the world of dental materials is very wide, some are resins, other cements, others have metallic claims, etc. To say that the presence of porosity is merit or demerit would be ideal because for some materials this is favorable and for others unfavorable since they are very versatile, with different qualities, different degrees of porosity, different molecules, different coefficients of thermal expansion, some do not set, others do not, some are light-cured, others do not, some withstand more mechanical forces and have more elastic modulus and less porosity, while others do not, and thus with many more variables.

Explanation:

In the manufacture of dental materials, much attention is always paid to what the manufacturer indicates in these indications that come on the reverse side of the material or on a paper inside the material container.

This is important to know, because the manufacturer is the one that indicates the mode of use and working time according to the trademark and the chemical composition of the dental material.

That a dental material in its manufacture is more or less porous, makes its density, weight and volume possibly be affected, there are materials that seek to increase porosity for a certain purpose, while others seek the opposite, depending on the function that is applied, will have more or less pores.

The example we gave above about the vitreous ionomer is an example that indicates that in the case of restorative dental materials where the function is to SEAL the porosity is almost nil, since otherwise it will not seal the cavity that was formed with caries and not the infectious problem would be solved.

On the other hand, in the surgical field of dentistry there are bone grafts or porous macroparticles that are for the purpose of bone replacement, which in order to be integrated need blood vessels that run over these pores, irrigating the area of ​​bone neoformation well, in this case the merit of porosity if necessary and it is essential that they be very porous.

Answer:

The similarities are the groups, triads law is a law where they are in groups or three, the octaves law is the 'best' law, is the one who every atom wants, they do everything to be in groups of eight.

Explanation:

Answer:

.01

Explanation:

H30+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

Answer:

vent vapor concentration in the air to nearly zero. The effective diffusion coefficient of the solvent molecules in the polymer is 2 x 10^-6 cm^2/s

Explanation:

Answer:

Endothermic, positive

Explanation:

Answer:

26 mol

Explanation:

Step 1: Write the balanced equation

4 FeCl₃ + 3 O₂ ⇒ 2 Fe₂O₃ + 3 Cl₂

Step 2: Determine the appropriate molar ratio

The molar ratio of FeCl₃ to O₂ is 4:3.

Step 3: Use the determined molar ratio to calculate the moles of oxygen required to completely react with 35 moles of ferric chloride

[tex]35molFeCl_3 \times \frac{3molO_2}{4molFeCl_3} = 26molO_2[/tex]

Answer:

[tex]n_{O_2}=26.25molO_2[/tex]

Explanation:

Hello,

In this case, given the reaction:

[tex]4FeCl_3 + 3O_2 \rightarrow 2Fe_2O_3 + 6Cl_2[/tex]

Since oxygen and iron (III) chloride are in a 4:3 molar ratio, he required moles of oxygen to completely react with 35 moles of iron (III) chloride result:

[tex]n_{O_2}=35molFeCl_3*\frac{3molO_2}{4molFeCl_3} \\\\n_{O_2}=26.25molO_2[/tex]

Best regards.

Answer:

Explanation:

Adding sugar

Answer:

Magma

Explanation:

Magma is the hot liquid under earths surface

Answer:carbon, hydrogen and oxygen

Explanation:

Alcohol functional group is composed of carbon, hydrogen and oxygen

When a chemical reaction releases energy it is called Exothermic Reaction.

hope it helps!

When a chemical reaction energy releases then this type of reaction is called exothermic reaction. In this reaction energy is release to the atmosphere.

In chemistry there are various type of reaction out of which the two main types are the exothermic reaction and endothermic reaction.

Exothermic reaction is the one in which energy releases in form of heat respiration reaction is an example of exothermic reaction. In respiration food that we eat are broken down in glucose with release of energy.

Endothermic reaction is the one in which energy is taken out during the reaction. Photosynthesis is an example of endothermic reaction where sunlight energy is taken by the plants to make food.

Thus when a chemical reaction releases energy these reactions are called exothermic reaction.

To learn more about exothermic reaction, here:

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