Which best describes thermal energy? It is the difference between internal energies of two or more substances. It is the sum of internal energies of two or more substances. It is the portion of internal energy that can be transferred from one substance to another. It is the portion of potential energy that can be transferred from one substance to another.

Answers

Answer 1

Answer:

It is the portion of internal energy that can be transferred from one substance to another.

Answer 2

Thermal energy is the portion of internal energy that can be transferred from one substance to another.

What is thermal energy?

Thermal energy is the energy an object posses which is as a result of particles movement within it.

It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.

Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.

Learn more about thermal energy from the link below.

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Related Questions

Which compound has the lowest melting point? KCl CaCl2 Na2O C6H12O6

Answers

It is called ethane.

What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together

Answers

The correct answer is B

If a jet’s cruising altitude is 32,200ft(to three significant figures),the distance in km is :(1 mile=1.61km;1 mile=5280 ft)

Answers

Answer:

9.82 km.

Explanation:

Hello,

In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:

[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]

Best regards.

An organic chemistry student was studying the solubility of Methyl-N-acetyl-α-D-glucosaminide (1-O-methyl-GlcNAc), a derivative of glucosamine, in water but inadvertently added 1 equiv. of periodic acid instead. Based on your understanding of the reactions of monosaccharides with periodates, draw the organic product that the student obtained.

Answers

Complete Question

The diagram for this question is shown on the second uploaded image

Answer:

The organic product obtained is  shown  on the first uploaded image

Explanation:

The process that lead to this product formation is known as oxidative cleavage   which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW

Answers

Answer:

Explanation:

From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer

0.25 L elution buffer = 250 mL elution butter

The breaking buffer that we use this week contains

10mM Tris    =   0.01 M

150mM NaCl  =   0.15 M

300mM imidazole.  = 0.3 M

The stock concentration  of Tris in 1M

Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;

[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 2.5 \ mL[/tex]

In NaCl, The amount of stock concentration is 5 M

so; using the same formula; we have:

[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]

[tex]V_1 = 7.5 \ mL[/tex]

From Imidazole ; the amount of stock concentration is

[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 75 \ mL[/tex]

Thus; we can have a table as shown as :

Stock concentration        volume to be added        Final concentration

1 M of Tris                              2.5 mL                            10 mM

5 M of  NaCl                          7.5 mL                             150 mM

1 M of Imidazole                    75  mL                            300  mM

In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.

Please help! (:

question above — how much money would you need to buy 7.0 lb of arugula? If 27lb of arugula cost $16

Answers

Answer:

$11.81

Explanation:

27 lb cost $16

27/16=$1.69 per pound

$1.69*7=$11.81 for 7 lbs

need help and quick answer as fast as possible

Answers

yes. arthropod are animals such as insects, crabs, lobsters etc

For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 × 1024 m−3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750°C.

Answers

Answer:

The correct answer is 5.447 × 10⁻⁵ vacancies per atom.

Explanation:

Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,  

Fv = Nv/N------ (i)  

Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,  

N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,  

N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol

N = 5.14073 × 10²⁸ atoms/m³

Now putting the values of Nv and N in the equation (i) we get,  

Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3

Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.  

5. Rosalind Franklin was a key figure in the discovery of the structure of DNA, yet she
was not included in the Nobel Prize which was awarded to Watson and Crick. Carry out
some research to find out how she contributed to this work and use the space below
to write up your findings​

Answers

Answer:

Explanation:

Search for "Rosalind Franklin: DNA's unsung hero - Cláudio L. Guerra" which basically summarizes what Rosalind did and how we was snubbed from receiving the noble prize even though she had vast and critical evidence to highlight the structure of DNA. You can look for more sources but I can tell you a quick recap:

Rosalind Franklin was born in an era where women scientists or workers were very uncommon and they were even discriminated and looked down upon. After her phD., she was working to find the structure of DNA and soon she was able to form an x-ray image of it. However, her lab colleague took the picture and showed it to other scientists (Watson and Crick) without the knowledge or permission of Rosalind. Here Rosalind was working on analyzing her data and on other part of world Watson and Crick were doing the same. Based on Watson and Crick's analysis, they came up with the correct structure of DNA and soon Rosalind got done as well. Both submitted their paper to journal, however, the journal placed Watson and Crick paper before Rosalind (making it look like Rosalind just confirmed what Watson and Crick proposed). This made it look like Watson and Crick were geniuses behind DNA structure whereas, in reality, it was Rosalind. She would have received Nobel Prize but she died of Cancer and Nobel prizes are not awarded to dead people.

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 3.4 g of octane is mixed with 15.6 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answers

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

Give the IUPAC name for the following structure

Answers

Answer:

6-metyl-2-heptyne

Explanation:

C-C-C-C-C-C-C hept

   2

C-C≡C-C-C-C-C  2-heptyne

                   C

                    | 6

C-C≡C-C-C-C-C

6-metyl-2-heptyne

The IUPAC name for the above structure is 6 methyl, hept-2-yne.

What is IUPAC?

IUPAC stands for international Union of pure and applied chemistry. It is the body in charge of naming organic chemical compounds.

The naming is is based on a molecule's longest chain of carbons connected by single/double/triple bonds, whether in a continuous chain or in a ring etc.

According to this question, a structure is given. The following applies;

The compound has a triple bond located on the second carbon, hence, belongs to alkyne group. It has seven carbon atoms, hence, is heptyne. The methyl group is on the sixth carbon.

Learn more about IUPAC at: https://brainly.com/question/33646537

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Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm

1. Calculate the volume of the bar:

2. Calculate the (experimental) density of the bar:

3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?

4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%

Answers

Answer:

1= Volume

= Length x breath x height

= 13.90 x 2.9 x 0.081

=3.26511

2= Density = Mass ÷ volume

= 11.3 ÷ 3.26511

= 3.461 (3d.p)

idk the rest because you haven't shown a picture of the rest

Answer:

1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%

Explanation:

Experimental data:

Mass          = 11.3    g

Length      = 13.90 cm

Width        =  2.9    cm

Thickness = 0.081 cm

Calculations:

1. Volume of bar

V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³

2. Experimental density

[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]

3. Identity of metal

The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)

The metal is probably barium.

4. Percent difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]

Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.

Answers

Answer:

The percentage yield is 50%

A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.

Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.

Answers

Answer: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Explanation:

The given data is as follows.

Volume of sample water = 46 ml

Temperature = [tex]21^oC[/tex]

After vaporization, washes and then drying the weight of mineral X = 0.87 g

This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.

          1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]

                                = [tex]1.891 \times 10^{-2}[/tex] g/ml

Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

When a sample of Mg(s) reacts completely with O2(g), the Mg(s) loses 5.0 moles of electrons. How many moles of electrons are gained by the O2(g)? *

Answers

Answer:

if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.!

Explanation:

Oxidation refers to the loss of electrons. Any specie that looses electrons in a redox reaction is said to be the reducing agent. Hence the reducing agent participates in the oxidation half equation. In this case, magnesium is the reducing agent.

Reduction has to do with the gain of electrons. The oxidizing agent participates in the reduction half equation. Hence the oxidizing agent is reduced in the redid reaction. The reducing agent in this case is the oxygen molecule.

Oxidation half equation;

Mg(s)-----> Mg^2+(aq) + 2e

Reduction half equation;

O2(g) + 2e ------> 2O^2-(aq)

From the balanced reaction equation, two moles of electrons is transferred.

Hence if magnesium looses five moles of electrons, oxygen will also gain five moles of electrons.

Covalent bonds can be best described as

Answers

Answer:

neutral atoms coming together to share electrons

Answer:

a

Explanation:

neutral atoms coming together to share electrons

In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

Answers

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]

[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]

so,

[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable  until q = Keq

A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C

Answers

Answer:

FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL

Explanation:

From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:

k = Ae^ -Ea/RT

At initial temperature T1, the initial rate constant is (k1)

At final temperature T2, the final rate constant is k2

For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.

That is, k2 / k1 = 2 (rate is doubled)

Equating this into the Arrhenius equation, we have:

k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)

2 = e^ (-Ea / R) (1 / T2 = 1 / T1)

Taking the natural logarithm of both sides:

ln 2 = - (Ea / R) (1 / T2 - 1 / T1)

Making Ea the subject of the formula, we obtain:

Ea = - (ln 2 R / (1 / T2- 1 / T1))

Let T1 = 25 C = 25 + 273 K = 298 K

T2 = 35 C = 35 + 273 K = 308 K

R = 8.314

So,

Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))

Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)

Ea = - 5.7616 / -0.00011

Ea = 52 378,18 J / mol

So therefore, the activation energy Ea is 52.4 kJ/mol.


Discuss any give ways by which
the falling moral standards of Ghanaian
youth can be minimised.

Answers

Answer:

The falling standards of Ghanaian youths can be minimized by proper upbringing of the children by their parents. The youths should be taught about what is wrong or right and there should be a corresponding reward for those who do good and exceptional in order to encourage others in towing that line and punishment should also be meted out to those who break the law. Mediocrity shouldn’t be celebrated and the elders should lead by example.

These will make the falling standards of Ghanaian youth get reduced.

A 5.00-L tank contains helium gas at 1.50 atm. What is the pressure of the gas in mmHg

Answers

Answer:

  1140 mmHg

Explanation:

1 atmosphere is 760 mmHg, so 1.5 atmospheres is ...

  1.5×760 mmHg = 1140 mmHg

Which metal can replace another metal in a reaction

Answers

Answer:

The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)

Explanation:

The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)

A compound D with the molecular formula C6H12 is optically inactive but can be resolved into enantiomers. On catalytic hydrogenation, D is converted to E (C6H14) and E is optically inactive. Propose structures for D and E. (Draw a three-dimensional formula for each using dashes and wedges around chiral centers.)

Answers

Answer:

D: CH2=CH-CH(CH3)-CH2-CH3 (R & S enantiomers)

E: CH3-CH2-(CH3)-CH2-CH3

(Please see the figures enclosed )

Explanation:

D is a racemic mixture (R & S) of 3-metyl-pent-1-ene, so it is optically inactive (although each of two enantiomers is optically active, the mixture is optically  inactive. The reason is that two enantiomers are present in an equal amount).

E is optically inactive, so its structure has to be symmetric.

What can be known about the salt sample that Gerry is looking at?

Answers

Answer:

That its small pointed. Pink(Himalayan salt)or white(normal salt)

Explanation:

Summa dees questions are so stupid, deys makin me salty.

What is the number of valence electrons in a nitrogen atom in the ground state

Answers

Answer: 5

Explanation:

It just is

Answer:

5

Explanation:

Bc valence electron means last # in the electron configuration

From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O


How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g


Include the correct number of significant figures in your final answer

Answers

Answer: 125 g

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]

The balanced reaction is:

[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]

According to stoichiometry :

1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]

Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex]  of [tex]O_2[/tex]

Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]

Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]

Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?

Answers

Answer:

ΔH = - 44.0kJ

Explanation:

H2O(l)→H2O(g), ΔH =44.0kJ

In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH  is positive.

If he reaction is reversed, we have;

H2O(g)→H2O(l)

In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still  have the same value.

Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~50g of ATP (Mr 505) and consumes ~8000 kJ of energy in food each day. The energy stored in the terminal anhydride bond of ATP under standard conditions is 30.6 kJmol-1. Assume also that the dietary energy is channeled through ATP with an energy transfer efficiency of ~50%.

Answers

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  

If a gas occupies 12.60 liters at a pressure of 1.50 atm, what will its pressure at a volume of 2.50 liters?

Answers

Answer:

7.56 atm

Explanation:

Boyle's law states that the pressure and volume of a gas are proportional to each other

The formular for Boyle's law is

P1V1=P2V2

According to the question above, the values given are

P1=1.50 atm

P2= ?

V1=12.60 litres

V2= 2.50 litres

Let us make P2 the subject of formular

P2= P1V1/V2

P2= 1.50×12.60/2.50

P2= 18.9/2.50

P2= 7.56 atm

Hence when the volume of a gas is 2.50 litres then it's pressure is 7.56 atm

A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answers

Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

Explanation:

Formula used for Elevation in boiling point :

[tex]\Delta T_b=k_b\times m[/tex]

or,

[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]

[tex]k_b[/tex] = boiling point constant  = ?

m = molality

[tex]w_2[/tex] = mass of solute (urea) = 55.4 g

[tex]w_1[/tex] = mass of solvent  X =  500 g

[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]

[tex]k_b=2.4^0C/m[/tex]

Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

At 298K, the equilibrium constant for the following reaction is 4.20×10-7: H2CO3(aq) + H2O H3O+(aq) + HCO3-(aq) The equilibrium constant for a second reaction is 4.80×10-11: HCO3-(aq) + H2O H3O+(aq) + CO32-(aq) Use this information to determine the equilibrium constant for the reaction: H2CO3(aq) + 2H2O 2H3O+(aq) + CO32-(aq)

Answers

Answer:

The correct answer is 2.016 x 10⁻¹⁷

Explanation:

We have the following chemical reactions and their equilibrium constants (K):

(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)  K₁= 4.20×10⁻⁷

(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)    K₂= 4.80×10⁻¹¹

And we have to obtain K for the following reaction:

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)

If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.

H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq)     K₁= 4.20×10⁻⁷

+

HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq)      K₂= 4.80×10⁻¹¹

-------------------------------------------------------------

H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)    K= K₁ x K₂

K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷

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