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 1

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.

Answer 2

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].

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Related Questions

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.

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

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.  

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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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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]

can a kind human being help me with this table at least only with the first burning fire wood someone please ​

Answers

Answer:

See below

Explanation:

* Burning fire wood is given to be our first option. Now burning tends to be a property of wood, and it does effect the chemical compositions of it. Wood, in the presence of fire / oxygen, turns into ash and carbon dioxide.

* Decomposition is recognized as a chemical change, and heating copper carbonate is a perfect example of decomposition. When energy is added to this chemical process, the copper carbonate decomposes into copper oxide.

* Mixing sodium chloride solution and silver nitrate solution. When this reaction takes place, a white precipitate of AgCl is formed. And of course, this is a chemical reaction.

* When acids or bases come in contact with citric acid, the pH degree changes much. Due to this, carbon dioxide bubbles are formed.

* When eggs are fried they absorb the heat in the pan. Doing so the egg starts to curl a bit, resulting in the formation of new particles.

_______________________________________________________

I hope this gave you a start!

Do you think there is a limit to the size of Atoms scientist can make? prove with evidence.

Answers

Answer:

Yes, there is a limit to the size of atoms that scientists can make.

Explanation:

In the nucleus, atoms contain protons and neutrons. It is known that as the number of protons in the nucleus increases, the atom becomes unstable due to the repulsion of positively charged protons clumped together in the small space of the nucleus.

However, an attractive force exists between neutrons and protons which binds the nucleus together and minimizes repulsion between protons. Even neutrons have recently been found to slightly repel each other.

Several attempts made at synthesizing many very heavy elements lately have proved abortive because the elements only exist for a few fractions of a second owing to large repulsion between the particles in the nucleus.

This goes a long way to show that there is a limit to the number of protons and neutrons that can be assembled together to form a new nucleus. We cannot bring an unlimited number of nucleons together to form new atoms due to inter particle repulsive forces.

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_____ .

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]

Which phrase describes one characteristic of radioactive elements?
are produced in a laboratory
O decay at a constant rate
O have a consistent number of particles
release energy and particles to maintain radioactivity

Answers

Answer:

B) decay at a constant rate

Explanation:

Over their lifetime, the radioactive elements tend to have constant rate of decay. This is seen in chemistry in Kinetics, where we see that radioactive elements follow first order kinetics where the rate at which they decay is constant (no matter what concentration we have)

Answer:

B) decay at a constant rate

Explanation:

just took the test

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°

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.

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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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 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%

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

The proposed mechanism for a reaction is: Step 1: A + B X (fast) Step 2: X + C Y (slow) Step 3: Y D (fast) What is the overall reaction? A. A + B + C D B. A + X Y + D C. A + B Y D. A + Y D

Answers

Answer:

A. A + B + C --> D

Explanation:

Step 1: A + B --> X (fast)

Step 2: X + C --> Y (slow)

Step 3: Y --> D (fast)

To obtain the overall reaction, we have to sum up the reactants and products of all step and eliminate the intermediates.

Reactants:

A + B + X + C + Y

Products:

X + Y + D

So we have;

A + B + X + C + Y  --> X + Y + D

Upon elimination of intermediates, we have;

A + B + C --> D

The correct option is A.

In the diagram below, particles of the substance are moving from the liquid phase to the gas phase at the same rate as they move from the gas phase to the liquid phase. A number of balls are loosely packed in the bottom of a container, beneath a line across the middle of the container, and a few balls above the line. 2 balls below the line have arrows pointing upward through the centerline; a few of the balls above the line have arrows pointing down through the centerline. The gas and liquid are at equilibrium. a high vapor pressure. a low vapor pressure. zero vapor pressure.

Answers

Answer:

The gas and liquid is in equilibrium.

Explanation:

liquids within a container undergoes state change, changing into gas. If this container is left open, these gases will escape into the external environment. In a situation in which the container is closed, the molecules that leave the liquid surface as gas will eventually condense on contact with the cover wall and change back into the liquid state. Some of these gases will reenter the liquid surface. At first, more of the liquid is transformed into gas and escape into the space above the liquid surface. Eventually, the available space becomes saturated with vapor, and then some of the gases start entering the liquid phase at the same rate as the liquid enters the gas phase. At this stage, the gas and liquid phase now exists in equilibrium.

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!

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!

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.

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.

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.

all compounds are neutral true or false​

Answers

Answer:

Even all compounds are neutral.

Explanation:

Some of them exhibit polarity. Because of the difference in electron affinity of the constituent atoms, the shared electrons are pulled towards the atom with high affinity to electrons.

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

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
sulfur dioxide (g) + water (l) __________sulfurous acid (H2SO3) (g) + water (I)

Answers

Answer:

Sulfur dioxide + 2 ( water ) -----> sulfurous acid + water /

SO2 + 2 ( H2O ) -----> H2SO3 + H2O

Explanation:

This formula may not be right. Sulfur dioxide tends to react with water to produce sulfurous acid as per it's formula, but then again that chemical reaction need not be balanced. However, I will solve for either case here -

Sulfur dioxide + water -----> sulfurous acid,

Sulfur dioxide + water -----> sulfurous acid + water

_______________________________________________________

As I mentioned before, Sulfur dioxide + water -----> sulfurous acid is a chemical reaction that need not balancing as the number of each element present on the reactant and product side are the same. To help, let me rewrite this reaction -

SO2 + H2O -----> H2SO3,

Reactant                  |                Product

Sulfur = 1,                                  Sulfur = 1,

Oxygen = 3,                              Oxygen = 3,

Hydrogen = 2                           Hydrogen = 2

And hence the equation is already balanced. Now let us consider the case we supposedly have at hand - Sulfur dioxide + water -----> sulfurous acid + water. Take a look at the attachment below;

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

Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to 64°C?

Answers

Answer: P₂=0.44 atm

Explanation:

For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.

Gay-Lussac's Law: [tex]\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }[/tex]

First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.

64+273K=337K

Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

[tex]56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm[/tex]

Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.

[tex]P_{2} =\frac{T_{2} P_{1} }{T_{1} }[/tex]

[tex]P_{2}=\frac{(337K)(0.55atm)}{418K}[/tex]

P₂=0.44 atm

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

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