Two aqueous solutions are both at room temperature and are then mixed in a coffee cup calorimeter. The reaction causes the temperature of the resulting solution to fall below room temperature. Which of the following statements is true?
A. The products have a lower potential energy than the reactants.
B. This type of experiment will provide data to calculate AErxn.
C. The reaction is exothermic.
D. Energy is leaving the system during reaction.
E. None of these statements are true.

Answers

Answer 1

Answer:

the correct option is that C. The reaction is exothermic.

Explanation:

When the temperature decreases, and that the final temperature is lower than the initial, it indicates that less calories were perceived in the calorimeter, therefore said reaction releases heat to the external environment, thus being an exothermic reaction.

Answer 2

The reaction of given two liquids that have been resulted in the lowering of the temperature of the solution has been an exothermic reaction. Thus option C is correct.

There has been a decrease in the temperature of the solution after the mixing of the two solutions.

This can be described as the reaction between the two molecules that will result in the release of the energy from the system to the surroundings. The release of energy will result in the lowering of the temperature of the system.

The reaction in which the release of energy has been there is termed an exothermic reaction. Thus the reaction of given two liquids that have been resulted in the lowering of the temperature of the solution has been an exothermic reaction. Thus option C is correct.

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Related Questions

need helpp asapp please

Answers

Answer:

B. None of these

Explanation:

Sulfur has less ionization energy than phosphorus because sulfur has a pair of electron in its 3p subshell that increases electron repulsion in sulfur and sulfur electrons can easily remove from its sub-level.

While, there are no electron pairs in 3p subshell of phosphorus, therefore it requires more energy to remove an electron from 3p subshell.

Hence, the reason is electron repulsion and the correct answer is B.

With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem. (a) CH3I (b) I (c) CH3OH (d) OH (e) CH3SH (f) SH (g) CH3CN (h) CN (i) CH3OCH3 (j) OMe

Answers

Answer:

In the attachment you can find all the possible chemical reactions.

Some reaction can not be obtained by using alkyl halides because halides are weak leaving group which can leave compound during reaction easily but hydroxyl groups is a strong nucleophile which can not leave compound easily. So we can obtain alcohol from ethyl bromide, but we can not obtain hydroxyl ion from ethyl bromide.  

Explanation:

The methyl of ethyl halides as the organic starting materials are using the needed solvents or the inorganic reagents. These can be not repeated in steps that arrive out in earlier parts.

The reaction can not be taken by the use of alkyl halides as the halides are the weakest leaving group which leave the compound during reaction easily.the hydroxyl group is the strong nucleophile that cannot leave the compound easily. Thus we can get alcohol from the ethyl bromide, but we can not obtain the hydroxyl ion from the ethyl bromide.  

Learn more about the methyl or the cyclopentyl.

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What is Key for the reaction 2503(9) = 2802(9) + O2(g)?

Answers

Answer:

Option C. Keq = [SO2]² [O2] /[SO3]²

Explanation:

The equilibrium constant keq for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, let us determine the equilibrium constant for the reaction given in the question.

This is illustrated below:

2SO3(g) <==> 2SO2(g) + O2(g)

Reactant => SO3

Product => SO2, O2

Keq = concentration of products /concentration of reactants

Keq = [SO2]² [O2] /[SO3]²

Benny Beaver wants to determine what dyesare present in his favorite sports drink. He analyzesa sample witha UV-visiblespectrophotometer and sees absorbance peaks at 415.2nm and 519.6nm. What colordyesare present in his drink

Answers

Answer:

At 415.2nm and 519.6nm, the dyes observed by the instrument are violet and green respectively.

Explanation:

In the electromagentic spectrum, visible wavelengths cover a range from approximately 400 to 800 nm. The colours of the spectrum range from red to violet (Red, Orange, Yellow, Green, Blue, Indigo and violet: a.k.a ROGBIV), in order of decreasing wavelength.

I hope this explanation would suffice.

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitrogen is 75 K. What is the heat capacity of the sample in J/K

Answers

Answer:

heat capacity of the sample = 37.8 J/K

Explanation:

Step 1: Data given

Temperature of the sample = 275 K

The mass of liquid nitrogen = 2kg

temperature of liquid nitrogen = 70 K

The final temperature of the nitrogen is 75 K

Step 2: Calculate heat

Q = m*c*ΔT

⇒with m = the mass of liquid nitrogen = 2 kg = 2000 grams

⇒with c= the specific heat of the liquid nitrogen = 1.04 J/g*K

⇒with ΔT = the change of temperature of liquid nitrogen = T2 - T1 = 75 - 70 = 5K

Q = 2000 grams * 1.04 J/g*K * 5K

Q = 10400 J

Step 3: Calculate the heat capacity of the sample

heat capacity of the sample = 10400 J / 275 K

heat capacity of the sample = 37.8 J/K

In the first 15.0 s of the reaction, 1.7×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.440 L . What is the average rate of the reaction over this time interval?

Answers

Answer:

[tex]Rate=2.57x10^{-3}\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, for the reaction:

[tex]2N_2O(g) \rightarrow 2N_2(g)+O_2(g)[/tex]

We can easily compute the average rate by firstly computing the final concentration of oxygen:

[tex][O_2]=\frac{0.017mol}{0.440L}=0.0386M[/tex]

Then, we compute it by using the given interval of time: from 0 seconds to 15.0 seconds and concentration: from 0 M to 0.0386M as oxygen is being formed:

[tex]Rate=\frac{0.0386M-0M}{15.0s-0s}\\ \\Rate=2.57x10^{-3}\frac{M}{s}[/tex]

Regards.

The average rate of the reaction will be "[tex]2.57\times 10^{-3} \ M/s[/tex]".

According to the question,

Volume = 0.440 LTime = 15.0 sMol of O₂ = 1.7×10⁻²

The reaction will be:

[tex]2 N_2 O (g) \rightarrow 2 N_2 (g) +O_2 (g)[/tex]

Now,

The final concentration of O₂ will be:

→ [tex][O_2] = \frac{0.017}{0.440}[/tex]

          [tex]= 0.0386 \ M[/tex]

hence,

The rate of reaction will be:

= [tex]\frac{0.0386-0}{15.0-0}[/tex]

= [tex]2.57\times 10^{-3} \ M/s[/tex]

Thus the above approach is right.

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What is the Lewis structure for *OPCl3 and AlCl6^3-? What are their electron/molecular geometry and Ideal Bond Angle ?

Answers

Answer:

Here's what I get  

Explanation:

1. POCl₃

(a) Lewis structure

Set P as the central atom, with O and Cl atoms directly attached to it.

Electrons available = P + O + 3Cl = 5 + 6 + 3×7 = 11 + 21 = 32

Arrange these electrons to give every atom an octet. Put a double bond between P and O.

You get the structure shown below.

(b) Geometry

There are four bond pairs and no lone pairs about the P atom.

Electron pair geometry — tetrahedral

    Molecular geometry — tetrahedral

(c) Ideal bond angles

Tetrahedral bond angle = 109.5°

2. AlCl₆³⁻

(a) Lewis structure

Set Al as the central atom, with the Cl atoms directly attached to it.

Electrons available = Al + 6Cl + 3(-) = 3 + 6×6 +3 = 6 + 36 = 42

Arrange these electrons to give every atom an octet. Assign formal charges.

You get the structure shown below.

(b) Geometry

There are six bond pairs and no lone pairs about the Al.

Electron pair geometry — octahedral

    Molecular geometry — octahedral

(c) Ideal bond angles

        Axial-equatorial =  90°

Equatorial-equatorial = 120°

                 Axial-axial = 180°

Consider the reaction of aqueous potassium sulfate with aqueous g silver nitrate based on the solubility rule predict the product likely to be precipitate write a balanced molecular equation describing the reaction.

Answers

Answer:

K₂SO₄(aq)  + 2AgNO₃ (aq) →  2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

Explanation:

Our reactants are: K₂SO₄ and AgNO₃

By the solubility rules, we know that sulfates are insoluble when they react to Ag⁺, Pb²⁺, Ca²⁺, Ba²⁺, Sr²⁺, Hg⁺

We also determine, that salts from nitrate are all soluble.

The reaction is:

K₂SO₄(aq)  + 2AgNO₃ (aq) →  2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

A glass flask has a volume of 500 mL at a temperature of 20° C. The flask contains 492 mL of mercury at an equilibrium temperature of 20°C. The temperature is raised until the mercury reaches the 500 mL reference mark. At what temperature does this occur? The coefficients of volume expansion of mercury and glass are 18 ×10-5 K-1 (mercury) and 2.0 ×10-5 K-1 (glass).

Answers

Answer:

101.63° C

Explanation:

Volume expansivity γa = γr -  γ g = 18 × 10⁻⁵ - 2.0 × 10⁻⁵ = 16 × 10⁻⁵ /K

v₂ - v₁ / v₁θ = 16 × 10⁻⁵ /K

(500 - 492 ) mL / (492 × 16 × 10⁻⁵) = θ

θ = 101.63° C

Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it.

Answers

Answer:

Molarity = 0.0428 M = 42.8 mM

Explanation:

Step 1: Data given

Mass of nickel(II) bromide = 1.87 grams

Molar mass of nickel(II) bromide = 218.53 g/mol

Volume = 200 mL = 0.200 L

Step 2: Calculate moles of nickel(II) bromide

Moles nickel (II) bromide = mass / molar mass

Moles nickel (II) bromide = 1.87 grams / 218.53 g/mol

Moles nickel (II) bromide = 0.00856 moles

Step 3: Calculate moles nickel (II) cation

For 1 mol NiBr2 we have 1 mol Ni^2+

For 0.00856 moles NiBr2 we have 0.00856 moles Ni^2+

Step 4: Calculate final molarity of Ni^2+

Molarity = moles / volume

Molarity = 0.00856 moles / 0.200 L

Molarity = 0.0428 M = 42.8 mM

A weather balloon is inflated to a volume of 27.6 L at a pressure of 755 mmHg and a temperature of 29.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 385 mmHg and the temperature is -14.1 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answers

Answer: The volume of the balloon at this altitude is 46.3 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 755 mm Hg

[tex]P_2[/tex] = final pressure of gas (at STP) = 385 mm Hg

[tex]V_1[/tex] = initial volume of gas = 27.6 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]29.9^0C=(29.9+273)K=302.9K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-14.1^0C=((-14.1)+273)K=258.9K[/tex]

Putting all the values we get:

[tex]\frac{755\times 27.6}{302.9}=\frac{385\times V_2}{258.9}[/tex]

[tex]V_2=46.3L[/tex]

Thus the volume of the balloon at this altitude is 46.3 L

Which of the following is evidence of a physical change?

A) burning
B) melting
C) decomposing
D) rusting

Answers

The answer is >>>>
C) decomposing
A RESPOSTA CORRETA É A c

A certain mass of carbon reacts with 9.53 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.

Answers

Answer: 9.53 *2= 19.06

Explanation:

The law of multiple proportions states that if two elements combines to form more than one compound the ratio of masses of the second element which combines to the fixed mass of the first element will always be the ratios of the small whole numbers.

in case of carbon monoxide, mass of carbon will be the same of mass of oxygen.

But in case of carbon dioxide, if carbon is 9.53 units then oxygen will be twice as that of carbon.

CO2, so 9.53*2= 19.06 grams of oxygen will combine with 9.53 grams of carbon to form carbon dioxide.

Combustion analysis of a 13.42-g sample of estriol (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 10.06 g H2O. The molar mass of estriol is 288.38 g/mol . Find the molecular formula for estriol. Express your answer as a chemical formula.

Answers

Answer:

C18H24O3

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of estriol = 13.42g

Mass of CO2 = 36.86g

Mass of H2O = 10.06g

Molar mass of estriol = 288.38g/mol

Step 2:

Determination of the mass of Carbon (C), Hydrogen (H) and Oxygen (O) present in the compound. This is illustrated below:

For Carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 36.86 = 10.05g

For Hydrogen, H:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 10.06 = 1.12g

For Oxygen, O:

Mass of O = 13.42 – (10.05 + 1.12) = 2.25g

Step 3:

Determination of the empirical formula for estriol. This is illustrated below:

C = 10.05g

H = 1.12g

O = 2.25g

Divide by their molar mass

C = 10.05/12 = 0.8375

H = 1.12/1 = 1.12

O = 2.25/16 = 0.1406

Divide by the smallest i.e 0.1406

C = 0.8375/0.1406 = 6

H = 1.12/0.1406 = 8

O = 0.1406/0.1406 = 1

Therefore, the empirical formula for estriol is C6H8O

Step 4:

Determination of the molecular formula for estriol. This is illustrated below:

Molecular formula is simply a multiple of the empirical formula i.e

Molecular formula => [C6H8O]n

[C6H8O]n = 288.38g/mol

[(12x6) + (8x1) + 16]n = 288.38

[72 + 8 + 16]n = 288.38

96n = 288.38

Divide both side by 96

n = 288.38/96 = 3

Molecular formula => [C6H8O]n

=> [C6H8O]n

=> [C6H8O]3

=> C18H24O3

Therefore, the molecular formula for estriol is C18H24O3

The compound is C18H24O3.

From the information in the question;

Mass of C = 36.86 g/44 g/mol × 12 g/mol = 10.1 g

Number of moles of carbon = 10.1 g/12 g/mol = 0.84 moles

Mass of hydrogen = 10.06 g/18 g/mol × 2 g/mol = 1.11 g

Number of moles of hydrogen = 1.11 g/1g/mol = 1.11 moles

Mass of oxygen = 13.42 - (10.1 g + 1.11 g) = 2.21 g

Number of moles of oxygen = 2.21g/16 g/mol = 0.14 moles

Dividing through by the lowest number of moles;

C - 0.84 moles/0.14 moles   H -  1.11 moles/0.14 moles   O - 0.14 moles/0.14 moles

C - 6   H - 8    O -1

The empirical formula is C6H8O

The molecular formula of the compound is;

[6(12) + 8(1) + 16]n = 288.38

n =  288.38/86 =3

The compound is C18H24O3

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calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in the 1.50 g of CaCl2•2H2O.

Answers

Answer:

[tex]0.0102~mol~CaCl_2*2H_2O[/tex]

[tex]0.0102~mol~CaCl_2[/tex]

Explanation:

For this question, we have to start with the molar mass calculation of [tex]CaCl_2*2H_2O[/tex]. For this, we have to know the atomic mass of each atom:

O: 16 g/mol

Cl: 35.45 g/mol

H: 1 g/mol

Ca: 40 g/mol

If we take into account the amount of each atom in the formula we will have:

[tex](40*1)+(35.45*2)+(1*4)+(16*2)=~147.01~g/mol[/tex]

So, in 1 mol of [tex]CaCl_2*2H_2O[/tex] we will have 147.01 g. Now we can do the conversion:

[tex]1.50~g~CaCl_2*2H_2O\frac{1~mol~CaCl_2*2H_2O}{147.01~g~CaCl_2*2H_2O}=0.0102~mol~CaCl_2*2H_2O[/tex]

Additionally, in 1 mol of [tex]CaCl_2*2H_2O[/tex] we will have 1 mol of [tex]CaCl_2[/tex]. Therefore, we have a 1:1 mol ratio . With this in mind, we will have the same number of moles for [tex]CaCl_2[/tex]

[tex]0.0102~mol~CaCl_2*2H_2O=0.0102~mol~CaCl_2[/tex]

I hope it helps!

At a particular temperature, an equilibrium mixture the reaction below was found to contain 0.171 atm of I2, 0.166 atm of Cl2 and 9.81 atm of ICl. Calculate the value of the equilibrium constant, Kp at this temperature.I2(g) + Cl2(g) <=> 2 ICl(g)

Answers

Answer: 3390

Explanation:

Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for [tex]K_{p}[/tex].

[tex]K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}] }[/tex]

[tex]K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390[/tex]

When 1-iodo-1-methylcyclohexane is treated with NaOCH2CH3 as the base, the more highly substituted alkene product predominates. When KOC(CH3)3 is used as the base, the less highly substituted alkene predominates. Give the structures of the two products and offer an explanation.

Answers

Answer:

See explanation

Explanation:

In this case, we have 2 types of reactions. [tex]CH_3CH_2ONa[/tex] is a strong base but only has 2 carbons therefore we will have less steric hindrance in this base. So,  the base can remove hydrogens that are bonded on carbons 1 or 6, therefore, we will have a more substituted alkene (1-methylcyclohex-1-ene).

For the  [tex]KOC(CH_3)_3[/tex] we have more steric hindrance. So, we can remove only the hydrogens from carbon 7 and we will produce a less substituted alkene (methylenecyclohexane).

See figure 1

I hope it helps!

17. Write the molecular balanced equation for the recovering of copper metal. 18. Write the complete ionic balanced equation for the recovering of copper metal. 19. Write the net ionic balanced equation for the recovering of copper metal. 20. What type of reaction is this

Answers

Answer:

Explanation:

17. it goes from solid copper to aqueous copper:

Cu(s) --> Cu₂(aq) + 2e⁻

18. complete ionic:

Cu(s) --> Cu₂(aq) + 2e⁻

19. net ionic, must include only reacting species, so

Cu(s) --> Cu₂(aq) + 2e⁻

20. this type of reaction is dissolution reaction(redox reaction)

copper reduced from Cu²⁺ to Cu.

The enthalpy change for the complete burning of one mole of a substance
is the enthalpy of _______

thermochemical equation

combustion

released

vaporization

fusion

absorbed

heat

Answers

Answer:

combustion

Explanation:

The enthalpy change for the complete burning of one mole of a substance

is the enthalpy of __combustion_____ .

What type of chemical reaction occurs between AgNO3 (sliver nitrate) and Cu (Copper)?

Answers

Answer:

The answer is option c.

I hope this helps you.

Tubes through which water flows as it is brought from 0.8 MPa, 150C to 240C at essentially constant pressure in the boiler of a power plant. The total mass flow rate of the water is 100 kg/s. Combustion gases passing over the tubes cool from 1067 to 547C at essentially constant pressure. The combustion gases can be modeled as air as an ideal gas. There is no significant heat transfer from the boiler to its surroundings. Surrounding (dead state) temperature and pressure are given as 25C and 1 atm, respectively. Determine i) the exergetic efficiency of the boiler ii) rate of exergy destruction as kW iii) mass flow rate of the combustion gases as kg/s

Answers

Answer:

The correct answer is i) 50.2 % ii) 13440.906 kW and iii) 71.986 kg/s.

Explanation:

In order to find the mass flow rate of the combustion of gases, there is a need to use the energy balance equation:  

Mass of water × specific heat of water (T2 -T1)w = mass of gas × specific heat of gas (T2-T1)g

100 × 4.18 × [(240 + 273) - (150 + 273)] = mass of gas × 1.005 × [(1067+273) - (547+273)]

Mass of gas = 71.986 kg/s

The entropy generation of water can be determined by using the formula,  

(ΔS)w = mass of water × specific heat of water ln(T2/T1)w

= 100 × 4.18 ln(513/423)

= 80.6337 kW/K

Similarly the entropy generation of water will be,

(ΔS)g = mass of gas × specific heat of gas ln(T2/T1)g

= 71.986 × 1.005 ln (820/1340)

= -35.53 kW/K

The rate of energy destruction will be,  

Rate of energy destruction = To (ΔS)gen

= T₀ [(ΔS)w + (ΔS)g]

= (25+273) [80.6337-53.53)

Rate of energy destruction = 13440.906 kW

The availability of water will be calculated as,  

= mass of water (specific heat of water) [(T₁-T₂) -T₀ ln T₁/T₂]

= 100 × 4.8 [(513-423) - 298 ln 513/423]

= 13591.1477 kW

The availability of gas will be calculated as,  

= mass of gas (specific heat of gas) [(T₁-T₂) - T₀ ln T₁/T₂]

= 71.986 × 1.005 × [(1340-820) - 298 ln 1340/820]

= 27031.7728 kW

The exergetic efficiency can be calculated as,  

= Gain of availability / loss of availability  

= 13591.1477/27031.7728

= 0.502

The exergetic efficiency is 50.2%.  

A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution were required? View Available Hint(s) A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution were required? 5.74 mL 0.315 mL 793 mL 315 mL

Answers

Answer:

315mL

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 0.135 M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 0.0851 M

Volume of diluted solution (V2) = 500mL

The volume of the stock solution needed can be obtain as follow:

M1V1 = M2V2

0.135 x V1 = 0.0851 x 500

Divide both side by 0.135

V1 = (0.0851 x 500) / 0.135

V1 = 315mL

Therefore, the volume of the stock solution needed is 315mL

Question 1
1 pts
2B+6HCI --
| --> 2BCl3 + 3H2
How many moles of boron chloride will be produced if you start with 8.752 moles of HCI
(hydrochloric acid)? (Round to 3 sig figs. Enter the number only do not include units.)

Answers

Answer:

2.92 mol

Explanation:

Step 1: Write the balanced equation

2 B(s) + 6 HCI(aq) ⇒ 2 BCl₃(aq) + 3 H₂(g)

Step 2: Establish the appropriate molar ratio

The molar ratio of hydrochloric acid to boron chloride is 6:2.

Step 3: Calculate the moles of boron chloride produced from 8.752 moles of hydrochloric acid

[tex]8.752molHCl \times \frac{2molBCl_3}{6molHCl} = 2.92molBCl_3[/tex]

Please what's the missing minor products? And kindly explain in your own words how they were formed.​ Thank you!

Answers

Answer:

it's a two step elimination reaction

Explanation:

it follows a carbocationic pathway. When carbocation is stable, the equation is favourable, that is, double bond is formed by expelling hydrogen atom.

Suppose that you add 27.6 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.69 oC compared to pure benzene. What is the molar mass of the unknown compound

Answers

Answer:

The molar mass of the unknown compound is 153.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown molecular compound = 27.6 grams

Mass of benzene =  0.250 kg

Kf of benzene = 5.12 °C/m

freezing point depression of 3.69 °C

Step 2:  Calculate molality

ΔT = i*Kf*m

⇒with ΔT = reezing point depression of 3.69 °C

⇒with i = the van't Hoff factor of Benzene = 1

⇒with Kf = 5.12 °C/m

⇒ with m = molality = moles unknown compound / mass of benzene

3.69 = 1 * 5.12 * m

m = 0.72 molal

Step 3: Calculate moles of the unknown compound

molality = moles / mass benzene

0.72 molal = moles / 0.250 kg

Moles = 0.72 m * 0.250 kg

Moles = 0.18 moles

Step 4: Calculate molar mass of the unknown compound

molar mass = mass / moles

Molar mass = 27.6 grams / 0.18 moles

Molar mass = 153.3 g/mol

The molar mass of the unknown compound is 153.3 g/mol

Molar mass is the mass of the one mole of substance. The molar mass of the given unknown compound is 153.3 g/mol.

Molality of the compound can be calculated using

ΔT = i Kf m

Where,

ΔT = freezing point depression = 3.69 °C

i =  Van't Hoff factor of Benzene = 1

Kf =  constant of freezing = 5.12 °C/m

m = molality = ?

Put the values in the equation,

3.69 = 1 x 5.12 x m

m = 0.72 molal

Number of moles of the compound,

[tex]\bold {molality =\dfrac { moles} { mass\ benzene}}\\\\\bold {0.72\ molal = \dfrac {moles }{0.250\ kg}}\\\\\bold {Moles = 0.72\ m \times 0.250\ kg}\\\\\bold {Moles = 0.18}[/tex]

So, molar mass of the unknown compound,

[tex]\bold {Molar\ mass =\dfrac { mass}{ moles}}\\\\\bold {Molar\ mass = \dfrac {27.6\ grams }{0.18\ moles}}\\\\\bold {Molar\ mass = 153.3 g/mol}[/tex]

The molar mass of the given unknown compound is 153.3 g/mol.

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The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperature, the chemical reactions associated with the cooking of food take place at a greater rate. (a) Some food cooks fully in 7.00 min in a pressure cooker at 113.0°C and in 49.0 minutes in an open pot at 100.0°C. Calculate the average activation energy for the reactions associated with the cooking of this food. kJ mol-1 (b) How long will the same food take to cook in an open pot of boiling water at an altitude of 10000 feet, where the boiling point of water is 89.8 °C? min

Answers

Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

[tex]T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K[/tex]

[tex]R_1 = \dfrac{1}{7}[/tex]

[tex]R_2 = \dfrac{1}{49}[/tex]

Thus; [tex]\dfrac{R_2}{R_1} = 7[/tex]

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

[tex]In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})[/tex]

1.9459 = [tex]\dfrac{Ea}{8.314}* 9.0292 *10^{-5}[/tex]

[tex]1.9459*8.314 = Ea * 9.0292*10^{-5}[/tex]

[tex]16.1782126= Ea * 9.0292*10^{-5}[/tex]

[tex]Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}[/tex]

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

b)

here;

[tex]T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K[/tex]

[tex]In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})[/tex]

[tex]In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})[/tex]

[tex]In (\dfrac{R_2}{R_1}) = 0.00357[/tex]

[tex]\dfrac{R_2}{R_1}= e^{0.00357}[/tex]

[tex]\dfrac{R_2}{R_1}= 1.0035[/tex]

where ;

[tex]R_2 = \dfrac{1}7{}[/tex]

[tex]R_1 = \dfrac{1}{t}[/tex]

Now;

[tex]\dfrac{t}{7}= 1.0035[/tex]

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

a). The activation energy given by the reactions related to the cooking of food in the pressure cooker would be:

[tex]Ea = 179.176 kJ/mol[/tex]

b). The time duration that is taken by the same food to cook in an open vessel would be:

[tex]7.0245 mins[/tex]

Activation Energy

a). Given that,

Temperature [tex]1[/tex]  [tex]= 100[/tex]° C

Temperature [tex]2[/tex] [tex]= 113[/tex]° C

In Kelvin,

Temperature  [tex]1[/tex] [tex]= 100 + 273[/tex]

[tex]= 373 K[/tex]

Temperature  [tex]2[/tex] [tex]= 113 + 273[/tex]

[tex]= 386 K[/tex]

[tex]R_{1} = 1/7\\R_{2} = 1/49[/tex]

∵ [tex]R_{2}/R_{1} = 49/7 = 7[/tex]

It is given that at  [tex]113[/tex] rate [tex]=[/tex] [tex]7[/tex] × [tex]100[/tex]°C

Therefore,

[tex]Ea/8.314 (1/373 - 1/386) =[/tex] [tex]In(7)[/tex]

so,

[tex]Ea[/tex] [tex]= 16.1782126/(9.0292 * 10^{-5})[/tex]

Activation energy [tex]= 179.176 kJ/mol[/tex]

b). As we know,

[tex]T_{2}[/tex] [tex]= 386 K[/tex]

[tex]T_{1}[/tex] [tex]= (89. 8 + 273)[/tex]

[tex]= 362.8 K[/tex]

by employing the formulae,

[tex]In(\frac{R_{2} }{R_{1} }) = \frac{Ea}{R} (1/T_{1} - 1/T_{2})[/tex]

[tex]In(\frac{R_{2} }{R_{1} }) = 179.176/8.314 (1/362.8 - 1/386)[/tex]

By solving this, we get

[tex]R_{2}/R_{1} = 1.0035[/tex]

Thus,

[tex]R_{2} = 1/7[/tex]

[tex]R_{1} = 1/t[/tex]

∵ t [tex]= 7.0245 min[/tex]

Thus, the time duration would be [tex]7.0245 minutes[/tex].

Learn more about "Boiling Point" here:

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The following reactions all have K < 1. 1) HCOO- (aq) + C6H5COOH (aq) HCOOH (aq) + C6H5COO- (aq) 2) C9H7O4- (aq) + C6H5COOH (aq) C6H5COO- (aq) + HC9H7O4 (aq) 3) HCOOH (aq) + C9H7O4- (aq) HC9H7O4 (aq) + HCOO- (aq) Arrange the substances based on their relative acid strength.

Answers

Answer:

Explanation:

C₉H₇O₄⁻ = weakest base

C₆H₅COO⁻ = strongest base

HCOO⁻ = intermediate base

HCOOH = not a Bronsted-Lowry base

HC₉H₇O₄ = not a Bronsted-Lowry base

C₆H₅COOH = not a Bronsted-Lowry base

A solution of benzene in methanol has a transmittance of 93.0 % in a 1.00 cm cell at a wavelength of 254 nm. Only the benzene absorbs light at this wavelength, not the methanol. What will the solution's transmittance be if it is placed in a 10.00 cm long pathlength cell

Answers

Answer:

T = 48.39%

Explanation:

In this case we need to apply the Beer law which is the following:

A = CεL  (1)

Where:

A: Absorbance of solution

C: Concentration of solution

ε: Molar Absortivity (Constant)

L: Length of the cell

Now according to the given data, we have transmittance of 93% or 0.93. We can calculate absorbance using the following expression:

A = -logT (2)

Applying this expression, let's calculate the Absorbance:

A = -log(0.93)

A = 0.03152

Now that we have the absorbance, let's calculate the concentration of the solution, using expression (1).

A = CεL

C = A / εL

Replacing:

C = 0.03152 / 1 *ε   (3)

Now, we want to know the transmittance of the solution with a length of 10 cm. so:

A = CεL

Concentration and ε are constant, so:

A = (0.03152 / ε) * ε * 10

A = 0.3152

Now that we have the new absorbance, we can calculate the new transmittace:

T = 10^(-A)

T = 0.4839 ----> 48.39%

Calculate the heat change in kilojoules for condensation of 195 g of steam at 100 ° C

Answers

Answer:

Q = 81.59kJ

Explanation:

Hello,

The heat of condensation is the energy required to to convert the steam into water.

Mass = 195g

Specific heat capacity of water = 4.184J/g°C

Initial temperature(T1) = 100°C

Final temperature(T2) = 0°C

Heat energy (Q) = ?

Heat energy (Q) = mc∇T

M = mass of the substance

C = specific heat capacity of the substance

∇T = T2 - T1 = change in temperature of the substance

Q = 195 × 4.184 × (0 - 100)

Q = -81588J

Q = -81.588kJ

The heat required for the condensation of 195g of steam is 81.59kJ

Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so that the pH is 3.00. What is the concentration of HTeH4O6

Answers

Answer:

5x10⁻⁶ = [HTeH₄O₆⁺]

Explanation:

The first dissociation equilibrium of the telluric acid in water is:

H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺

Using H-H equation for telluric acid:

pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]

pKa of telluric acid is -logKa1

pKa = -log 2.0x10⁻⁸

pKa = 7.699

As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:

3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]

-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]

2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]

5x10⁻⁶ = [HTeH₄O₆⁺]

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