Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identical burners. Afterward, liquid A is hotter than liquid B. Which has the larger specific heat? Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identical burners. Afterward, liquid A is hotter than liquid B. Which has the larger specific heat? Liquid A. There's not enough information to tell. Liquid B.

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:

mdjnibfziedosnninjd.zjcssksskskbsksbivfdlvdxkvfuwhiwhwonsk jsbudwuvsaeigdOhgrauvdslnzarm cxwmvakbxsabxwrifwzpgsdoh

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Answer:

Explanation:

Half life of Nobelium-253 is 97 seconds . That means after every 97 seconds half of the Nobelium amount will be disintegrated .

Time taken in bringing the sample to laboratory = 291 seconds

291 second = 291 / 97 half life

n = 3

N = [tex]N_0 (\frac{1}{2})^n[/tex]

N₀ is original mass , N is mass after n number of half life.

N = 5 mg x [tex](\frac{1}{2})^3[/tex]

= .625 mg

Only  0.625 mg of Nobelium-253 will be left .

Explanation:

This picture shows how the moon from earth when it is in position

Answer:

C

Explanation:

heat is energy

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.

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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Answer:

b, c, e

Explanation:

Inorganic compounds usually lack carbon.

Inorganic compounds are not associated with or made from living things.

Inorganic compounds include salt and water​.

Answer:

yes the answer id bce

Explanation:

Answer:

A = 1.2L

B = 506.6 torr

C = 13.5 °C

Answer:

A = 1.2L B = 506.6 torr and C = 13.5 °C

have a nice day :)

Answer:

There are several roles of HCl in the digestive system. Some of them are as follows:

hEy!!!

Chlorine(CL) is the symbol for the entity that has 17 protons,20 neutrons and 18 electrons.

Hope it will be helpful to you...

Answer:chlorine

Explanation:

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:

Catenation refers to the ability of an atom to link to other atoms of the same kind to form a chain.

Explanation:

Catenation is the binding of an element to itself through covalent bonds to form chain or ring .Catenation can also be defined as the self-linking of atoms of an element to form chains and rings. This definition can be extended to include the formation of layers like two-dimensional catenation and space lattices like three-dimensional catenation. Thus, we can boldly say that, Catenation occurs when atoms of the same element covalently bond to one another to create a chain or ring.

Catenation occurs most readily in carbon, forming covalent bonds to and longer chains and structures with other carbon atoms. This is why the vast number of organic compounds are found in nature. Carbon is best known for its catenation properties, with the analysis of catenated carbon structures in organic chemistry.

Carbon is by no means the only element capable of forming such catenae, however, and several other main group elements are capable of forming a wide range of catenae, including silicon, sulfur, and boron. In group 14, the high bond energy of carbon makes it able to self-link almost indefinitely when compared with other members of the group which form only a few bonds to other atoms of the same element.

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: 1.42

Explanation:

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:

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:

0.17M

Explanation:

5kg of water contains 5 liters as the density of water is 1000 [tex]kgm^{-3}[/tex]

Therefore one liter of water contains 31/5=6.2g of HCl

Molality of the solution = 6.2/36.46 =0.17M

Answer:

.6mol/L

Explanation:

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

750mL / 1000mL/L = .75L

M = .450mol / .75L

M = .6mol/L

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:

Explanation:

Mg + 2HCl = Mg Cl₂ + H₂

.594 g = .594 / 24.3

= .02444 mole

Heat evolved = msΔ T , m is mass of water ( solvant ) , s is specific heat of water , Δ T is rise in temperature

= 100 x 4.2 x ( 41.83 - 25 )

= 7068.6 J

.02444 mole  of Mg evolves 7068.6 J of heat

1 mole of Mg evolves 7068.6 /.02444 J

= 289222.6 J

= 289 kJ .

Molar heat enthalpy = 289 kJ .

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:

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.

Answer:

2 HCl

Explanation:

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

After the body is cleaned, the body is weighed and measured before being placed on the autopsy table for reexamination. The autopsy table is typically a slanted, aluminum table with raised edges that has several faucets and drains used to wash away collecting blood during the internal investigation.

Answer:

A table used to scientifically determine the cause of death of a human or animal

Explanation:

Answer: come on lets link can do what i do

Answer:

Magma

Explanation:

Magma is the hot liquid under earths surface

The given question is not complete, the complete question is:

A chemistry student needs 5.00 g of thiophene for an experiment. She has available 0.50 kg of a 28.7% w/w solution of thiophene in benzene, Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button Be sure your answer has the correct number of significant digits. X

Answer:

The correct answer is 17.4 grams.

Explanation:

As mentioned in the given question, 28.7 percent w/w solution thiophene in benzene shows that,  

(wt. of thiophene / wt. of thiopene + benzene) * 100 = 28.7

0.50 kg of the solution indicates that it is 500 grams of the solution.  

Therefore, the amount of thiophene found in the 500 grams of the solution is,  

500 g * 28.7 / 100 = 143.5 gm

Thus, 143.5 grams of thiophene is found 500 grams of the solution.  

Now, 1 gram of thiophene will be in 500/143.5 of solution.  

Therefore, 5 grams of thiophene will be in 500/143.5 * 5 g of the solution = 17.42 grams of the solution.  

Hence, the mass of the solution to be used by the student in the given case will be 17.4 grams.