An ideal diatomic gas starting at room temperature T1 = 300 K and atmospheric pressure p1 = 1.0 atm is compressed adiabatically to 1/10 of its original volume. What is the final temperature of the gas?

Answers

Answer 1

Answer:

The final temperature of the given ideal diatomic gas: T₂ = 753.6 K

Explanation:

Given: Atmospheric pressure: P = 1.0 atm

Initial Volume: V₁ , Final Volume: V₂ = V₁ (1/10)

⇒ V₁ / V₂ = 10

Initial Temperature: T₁ = 300 K, Final temperature: T₂ = ? K

 

For a diatomic ideal gas: γ =  7/5

For an adiabatic process:

[tex]V^{\gamma-1 }T = constant[/tex]

[tex]V_{1}^{\gamma-1 }T_{1} = V_{2}^{\gamma-1 }T_{2}[/tex]

[tex]\left [\frac{V_{1}}{V_{2}} \right ]^{\gamma-1 } = \frac{T_{2}}{T_{1}}[/tex]

[tex]\left [10 \right ]^{\frac{7}{5}-1 } = \frac{T_{2}}{300 K}[/tex]

[tex]\left [10 \right ]^{\frac{2}{5} } = \frac{T_{2}}{300 K}[/tex]

[tex]2.512 = \frac{T_{2}}{300 K}[/tex]

[tex]T_{2} = 753.6 K[/tex]

Therefore, the final temperature of the given ideal diatomic gas: T₂ = 753.6 K


Related Questions

Which set of these comparisons is INCORRECT? (More stable means it has a more negative energy.) a) The 1s orbital in H is more stable than the 1s orbital in He+ b) The 2s orbital in He atom is less stable than the 2s orbital in He+ c) The 2s subshell in Li is more stable than the 2p subshell in Li

Answers

Answer:

a.

Explanation:

By the Linus Pauling's diagram, we know that the order of the energy of the subshell at the same atom, from the less to the higher, is:

1s < 2s < 2p < 3s < 3p < 4s < 4d ...

As larger is the atom, the attraction energy between the electron and the nuclei is weak. The size of the atom increase from right to left in the periods, and from the top to the bottom in the families. The cations are smaller than the neutral atoms (the attraction is more effective with fewer electrons), and the anions are larger than the neutral atoms. So, in order of size, and also the energy of the similar subshells:

He⁺ < He < H

How higher the energy, less negative it is, so less stable is the orbital.

a. As shown above, the energy of the orbital at H is higher than the energy at the orbital at He⁺, so the 1s orbital is more stable at He⁺, and the sentence is incorrect.

b. As shown above, the energy of the orbital at He is higher than the energy of the orbital He⁺, so the 2s orbital at He is less stable than the 2s orbital at He⁺, and the sentence is correct.

c. From the Linus Pauling's diagram, the energy of 2p is higher than the energy of the 2s, so the 2s subshell is more stable, and the sentence is correct.

Exothermic reactions are favored because they release heat to the environment. Would an exothermic reaction be more favored or less favored if it were carried out within a superheated chamber?

Answers

Explanation:

No, an exothermic reaction will not be  favored or less favored if it were carried out within a super heated chamber. In the super heated chamber the heat will flow back to the reaction( considering heat emitted by the reaction is less than the heat in a super heated chamber). So, the reaction will shift backward instead, the product will again change into reactant.

The development of which polymer enabled the British to have effective radar equipment and thus contributed to the British success in the Battle of Britain in World War II? A) polystyrene B) polyvinyl chloride C) polyacetylene D) polyethylene

Answers

Answer: Polyethylene

Explanation:

The waxy solid had been formed during high-pressure experiments conducted in 1933 by ICI scientists Reginald Gibson and Eric Fawcett. They had heated a mixture of ethylene and benzaldehyde to 170°C (338°F), using apparatus that could submit materials to a pressure of 1,900 atmospheres (1,925 bars). But the reactions were explosive and safety concerns prompted the now defunct ICI, which merged into Dutch-based Akzo Nobel, to halt the research.

Match the following:
Part A
1. matter with no definite shape or volume
2. matter with no definite shape but with definite volume
3. the process of a solid changing directly to a gas without forming a liquid
4. matter in a high-energy state in which electrons are separated from their nuclei
5. matter with definite shape and volume
6. weak electrostatic bonds that form between particles of a substance
Part B
a. gas
b. intermolecular bonds
c. liquid
d. plasma
e. solid
f. sublimation

Answers

Answer:

The answer to your questions is below

Explanation:

1.- Gas  matter with no definite shape or volume

2.- Liquid matter with no definite shape but with definite volume

3.- Sublimation the process of a solid changing directly to a gas without forming a liquid.

4.- Plasma matter in a high-energy state in which electrons are separated from their nuclei

5.- Solid matter with definite shape and volume

6.- Intermolecular bonds weak electrostatic bonds that form between particles of a substance

Final answer:

The matchings are 1-a, 2-c, 3-f, 4-d, 5-e, and 6-b, identifying the physical state or process corresponding to each description given in Part A using basic knowledge of the properties of the different states of matter.

Explanation:

To match the given descriptions in Part A with the correct terms in Part B, we need to understand the basic properties of the different states of matter and recognize the processes they undergo:

Gas: Matter with no definite shape or volume.

Liquid: Matter with no definite shape but with definite volume.Sublimation: The process of a solid changing directly to a gas without forming a liquid.Plasma: Matter in a high-energy state in which electrons are separated from their nuclei.Solid: Matter with definite shape and volume.Intermolecular bonds: Weak electrostatic bonds that form between particles of a substance.

Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H, 28.85% N, and 16.48% O. What is the molecular formula of caffeine?

Answers

Answer : The molecular formula of a caffeine is, [tex]C_8H_{10}N_4O_2[/tex]

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles[/tex]

Moles of N = [tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{4.12}{1.03}=4[/tex]

For H = [tex]\frac{5.19}{1.03}=5.03\approx 5[/tex]

For N = [tex]\frac{2.06}{1.03}=2[/tex]

For O = [tex]\frac{1.03}{1.03}=1[/tex]

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_4H_5N_2O_1=C_4H_5N_2O[/tex]

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]

[tex]n=\frac{194.19}{97}=2[/tex]

Molecular formula = [tex](C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2[/tex]

Therefore, the molecular of the caffeine is, [tex]C_8H_{10}N_4O_2[/tex]

an element is a mixture of two isotopes. one isotope has an atomic mass of 34.96885 amu and has an abundance of 75.53%. The other isotope has an atomic mass of 36.96590 amu. Calculate the average atomic mass and identify the element?

Answers

Answer:

35.4575 amu

Explanation:

We use the relative abundance of each of the isotopes.

Let the isotopes be A and B.

Since A is having an abundance of 75.53, the abundance of B would be 100 - 75.53 = 24.47%

The average atomic mass is calculated as follows;

(75.53/100 * 34.96885)  + (24.47/100 * 36.96590) = 26.4119 + 9.0456 = 35.4575 amu

Final answer:

To find the average atomic mass of an element with two isotopes, multiply each isotope's mass by its fractional abundance and sum the results. The average atomic mass for the element given with isotopic masses of 34.96885 amu and 36.96590 amu is 35.4458 amu, which identifies the element as chlorine.

Explanation:

The average atomic mass of an element that is a mixture of two isotopes can be calculated using the isotopic masses and their abundances. In this case, we have two isotopes with masses of 34.96885 amu and 36.96590 amu and an abundance of 75.53% and 24.47% (100% - 75.53%), respectively. To find the average atomic mass, we multiply each isotope's mass by its abundance (as a decimal), and then add the results.

First, for the isotope with a mass of 34.96885 amu and an abundance of 75.53%, we calculate:

34.96885 amu × 0.7553 = 26.4062 amu

Similarly, for the other isotope with a mass of 36.96590 amu, we use its abundance of 24.47% (100% - 75.53%):

36.96590 amu × 0.2447 = 9.0396 amu

Adding these together gives us the average atomic mass:

26.4062 amu + 9.0396 amu = 35.4458 amu

The average atomic mass is 35.4458 amu. Comparing this with the periodic table, we can identify the element as chlorine since its atomic mass is closest to this calculated average.

if a test yields consistent results every time it is used, it has a high degree of...

Answers

Answer:Reliability

Explanation:

Reliability of a test refers to how consistently a test measures a characteristic under the same conditions.

Reliability can be defined as the degree of consistency of which a chemical test gives a similar result. measure. A test is said to be highly reliable when it gives the same repeated result under the same conditions of measure.

But when a test gives different results under the same condition of measure it has a low reliability.

Hence, If a test yields consistent results every time it is used, it has a high degree of reliability.

Solid aluminum hydroxide reacts with a solution of hydrobromic acid. Write a balanced molecular equation and a balanced net ionic equation

Answers

Answer:

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)+3Br^{-}(aq.)\rightarrow Al^{3+}(aq.)+3Br^{-}(aq.)+3H_{2}O(l)[/tex]

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)\rightarrow Al^{3+}(aq.)+3H_{2}O(l)[/tex]

Explanation:

Aluminium hydroxide ([tex]Al(OH)_{3}[/tex]) is a base and hydrobromic acid (HBr) is a strong acid.

Hence an acid-base reaction occurs between [tex]Al(OH)_{3}[/tex] and HBr

Balanced molecular equation:

[tex]Al(OH)_{3}(s)+3HBr(aq.)\rightarrow AlBr_{3}(aq.)+3H_{2}O(l)[/tex]

Balanced total ionic equation:

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)+3Br^{-}(aq.)\rightarrow Al^{3+}(aq.)+3Br^{-}(aq.)+3H_{2}O(l)[/tex]

Balanced net ionic equation:

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)\rightarrow Al^{3+}(aq.)+3H_{2}O(l)[/tex]

(net ionic equation is written by removing common ions present in both side of total ionic equation)

2. Which of the following is true about the total number of reactants and the total number of
products in the reaction shown below?
CsH(?)+802(8) 750,(8) +6H2008)
(1 point)
9 moles of reactants chemically change into 11 moles of product.
9 grams of reactants chemically change into 11 grams of product.
9 liters of reactants chemically change into 11 liters of product.
9 atoms of reactants chemically change into 11 atoms of product.

Answers

9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

For example 8O₂ depicts 8 moles of two oxygen atoms. The number of moles  is a unit for quantifying particles. You can liken it to a dozen, gross or a score. Since a mole of a substance contains avogadro number of particles. We can relate the number of moles to other parameters.

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Answer:

A) 9 moles of reactants chemically change into 11 moles of product.

Explanation:

C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)

The first step is to check if the equation is indeed a balanced one before proceeding to compare the moles.

This is done by comparing the number of atoms on the reactant side with that in the product side, if equal the reaction is indeed a balanced one.

Number of moles of Carbon atoms

Reactant = 5

Product = 5

Number of moles of Hydrogen atoms

Reactant = 12

Product = 12

Number of moles of Oxygen atoms

Reactant = 16

Product = 10 + 6 = 16

Since the equation is balanced, we can now compare the moles.

From the reactant side we have a total number of 9 (1 + 8 ) moles reacting to yield a total of 11 ( 5 + 6 ) moles of product.

This alone leads us  to take option A as our answer.

Let us see how other options are wrong however;

Option B) It cannot be grams as what chemical equations shows us the molar (number of moles) relationship between elememts/molecules/compounds/ions etc. The gram relationship can be calculated however if the molar mass of the compounds is taken into consideration.

Option C) This is wrong again due to same reasons mentioned for option B.

Option C) In the compound C5H12 alone, there are 17 (5 C and 12 H) atoms. Hence this option is also wrong.

The values between 1 and 14. The formula for pH is y = − log 10 ⁡ x, where y is the pH value and x is the concentration of hydrogen ions. A Golden Delicious apple has an average concentration of hydrogen atoms of 2.51 × 10 − 4. What is the pH value of a Golden Delicious apple?

Answers

Answer:

3.6

Explanation:

The pH of a solution can be defined as the negative logarithm to base 10 of the concentration of hydrogen ions in the solution. In the equation form, it can be written as :

pH = -Log [H+]

In the question, we have been given the concentration of the hydrogen ions but we do not know the value of the pH. We then insert the value of the concentration into the equation.

This means [H+] = 2.5 * 10^-4

pH = -Log [ 2.5 * 10^-4] = 3.6

What is the solution called in the buret during a titration?

Answers

Answer:

The solution in the buret, during a titration is called titrant.

Explanation:

A titration is a useful process, that makes you know the concentration of a solution.  A titrant solution (burette) is evaluated against a titrand to control the pH changes against the volume aggregate. Only a strong acid with a strong base, a strong base with a strong acid, a weak acid with a strong base and a weak base with strong acid are valued.

When the pH reaches the equivalence point, it is said that the normality of the acid by the milliequivalents, is equal to the basic normality by its milliequivalents. In conclusion, the entire base / acid became its conjugate pair.

To check this sudden change in pH, a substance is used, called Indicator that changes the color of the titrand (analyte).

Final answer:

In a titration analysis, the substance in the buret is called the 'titrant'. It is used to react with the analyte, the sample solution whose concentration we're measuring. The goal is to reach the endpoint, the point when a distinct visual change indicates that the titrant has completely reacted with the analyte.

Explanation:

In a titration analysis, the solution in the buret is called the titrant. This solution contains a known concentration of a substance. During a titration, this titrant is added incrementally to a sample solution, called the analyte, which contains the substance whose concentration is to be measured. The titrant and analyte undergo a chemical reaction of known stoichiometry.

By measuring the volume of the titrant solution needed to completely react with the analyte, scientists can calculate the concentration of the analyte. This point where the titrant has completely reacted with the analyte is termed the equivalence point of the titration. The process of adding the titrant is halted when a distinct change is visually detected in the solution - this could be a color change, for example. This is known as the end point.

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In this experiment, direct titration of a base with acid is not used. Instead, a back titration of a base with excess acid will be employed. Why?

a. The CO2 gas is released during the chemical reaction and is a slow step
b. There is a time delay between the addition of the acid and the completion of the reaction
c. The endpoint of the titration would be unclear because straight titration does not allow enough reaction timed
d. none of the above
e. all of the above

Answers

None of the options are not used in this experiment, hence "all of the above" is the correct answer.

Option: E

Explanation:

In the above experiment "back titration" is opted because of the need to determine "strength of an analyte" as "molar concentration" of an excess reactant is known. It is done when acid or mostly base is an insoluble salt, when reaction take palce very slowly and definitely when endpoint would be hard to discern for example titration of weak acid and weak base. The procedure to follow back titration is firstly the volatile analyte is reacted with an excess reagent then the titration is done on remaining amount of known solution.

How many electrons will fluorine gain or lose in forming an ion

Answers

Answer:

1 electron

Explanation:

Fluorine is a nonmetal with a high electronegativity and electron affinity so it tends to gain electrons to reach stability. According to the octet rule, fluorine will gain enough electrons so as to complete its valence shell with 8 electrons. Fluorine is in the Group 17 in the Periodic Table, which means it has 7 electrons in its valence shell. Therefore, it will gain 1 electron to complete its octet and form the anion F⁻.

Which is not a characteristic of metals?
a. Most metal oxides are ionic solids that are basic.
b. They have a shiny luster and various colors, although most are silvery.
c. Solids are malleable and ductile.
d. They are poor conductors of heat and electricity.

Answers

Answer:  d: They are poor conductors of heat and electricity .

Explanation:

Although is common knowledge that metals are characterized by high electrical conductivity and heat conductivity, these properties can be justified by looking at the metallic bonding.

In a metal, the bonding electrons are delocalized over the entire crystal. In fact, metal  atoms in a crystal can be imagined as an array of positive ions immersed in a sea of delocalized valence electrons.The mobility of the delocalized electrons makes metals good conductors of heat and electricity.

The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm, what is the concentration (molarity) of dissolved nitrogen?

Answers

Answer:

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

Explanation:

More the pressure of the gas, more will be its solubility.

So, for two different pressure, the relation between them is shown below as:-

[tex]\frac {P_1}{C_1}=\frac {P_2}{C_2}[/tex]

Given ,  

P₁ = 1 atm

P₂ = 0.76 atm

C₁ = 6.8 × 10⁻⁴ mol/L

C₂ = ?

Using above equation as:

[tex]\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}[/tex]

[tex]C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L[/tex]

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

A 1.2-L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m . Assume that the container is completely filled to the top with liquid nitrogen, that the temperature is 23.5?C, and that the atmospheric pressure is 1.2atm .Calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated into the closet. The closet is ventilated such that the temperature and pressure remain constant through this process. (Liquid nitrogen has a density of 0.807 g/mL.)

Answers

Answer:

26.99 % of air that will be displaced

Explanation:

Step 1: Data given

A 1.2-L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m

Temperature = 23.5 °C

Atmospheric pressure = 1.2 atm

Liquid nitrogen has a density of 0.807 g/mL

Molar mass of N2 = 28 g/mol

Step 2: Calculate mass of nitrogen

Mass of nitrogen = density * volume

Mass of nitrogen = 0.807 g/mL * 1200 mL

Mass of nitrogen = 968.4 grams

Step 3: Calculate moles of N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 968.4 grams /28 g/mol

Moles N2 = 34.586 moles

Step 4: Calculate volume

p*V = n*R*T

⇒ p = the the pressure = 1.2 atm

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = the number of moles = 34.586 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 23.5 °C = 296.65

V = (n*R*T)/p

V = (34.586 * 0.08206 * 296.65)/1.2

V = 701.61 L

Step 5: Calculate the total volume of the chamber

1.0 m * 1.3 m * 2 m = 2.6 m³ = 2600 L

Step 6: Calculate the percent volume displaced

(701.61 L  / 2600 L) * 100% = 26.99%

26.99 % of air that will be displaced

The percent (by volume) of air that would be displaced is 26.99 %.

Given information:

Temperature = 23.5 °C

Atmospheric pressure = 1.2 atm

Liquid nitrogen has a density of 0.807 g/mL

Molar mass of N2 = 28 g/mol

The calculation of the percent of air:

Mass of nitrogen = density  × volume

[tex]= 0.807 g/mL \times 1200 mL[/tex]

= 968.4 grams

Now

Moles N2 = mass N2 ÷  molar mass N2

[tex]= 968.4 grams \div 28 g/mol[/tex]

= 34.586 moles

Now volume

[tex]p\times V = n \times R \times T[/tex]

Here p = the the pressure = 1.2 atm

n = the number of moles = 34.586 moles

R = the gas constant = [tex]0.08206 L\times atm/K\times mol[/tex]

T = the temperature = 23.5 °C = 296.65

[tex]V = (n\times R\times T)\div p\\\\= (34.586 \times 0.08206 \times 296.65)\div 1.2[/tex]

= 701.61 L

Now the total volume of the chamber is

[tex]= 1.0 m \times 1.3 m \times 2 m \\\\= 2.6 m^3[/tex]

= 2600 L

Now finally the percent volume displaced

[tex]= (701.61 L \div 2600 L) \times 100\%[/tex]

= 26.99%

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The density of an aqueous solution containing 10.0 percent of ethanol (C2H5OH) by mass is 0.984 g/mL. (a) Calculate the molality of this solution. (b) Calculate its molarity. (c) What volume of the solution would contain 0.125 mole of ethanol

Answers

Final answer:

The solution contains 2.41 m (molality) and 2.14 M (molarity) of ethanol. A volume of 58.5 mL of this solution would contain 0.125 mol of ethanol.

Explanation:

(a) Let's calculate the molality of this solution. Mass of ethanol in 1000g of solution is 100g (10% of 1000g). The molar mass of ethanol (C2H5OH) is (2*12.01 + 6*1.008 + 1*16.00 + 1*1.008) = 46.07 g/mol. Hence, the number of moles = 100g / 46.07 g/mol = 2.17 mol. Since these moles are in 900g of water (1000g - 100g), molality = 2.17 mol / 0.900 kg = 2.41 m.

(b) To calculate molarity, let's find the volume of 1000g of solution: Volume = mass/density = 1000g / 0.984 g/mL = 1016.26 mL = 1.016 L. Therefore, molarity = 2.17 mol/1.016 L = 2.14 M.

(c) To find the volume of solution that would contain 0.125 mol ethanol, we'll use molarity: Volume = moles/Molarity = 0.125 mol / 2.14 M = 0.0585 L or 58.5 mL.

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The molality of the solution is 2.41 mol/kg, the molarity is 2.135 M, and to contain 0.125 mole of ethanol, 58.57 mL of the solution is required.

(a) Molality is defined as the number of moles of solute per kilogram of solvent. Given the density of the solution, we can determine the mass of 100.0 mL of the solution:

Density = 0.984 g/mL

Mass of 100 mL of solution = 100 mL * 0.984 g/mL = 98.4 g

Mass of ethanol (10% by mass) = 0.10 * 98.4 g = 9.84 g

Mass of water = 98.4 g - 9.84 g = 88.56 g

To convert grams of ethanol to moles:

Molar mass of ethanol (C2H5OH) = 46.07 g/mol

Moles of ethanol = 9.84 g / 46.07 g/mol = 0.2135 mol

To convert grams of water to kilograms:

88.56 g = 0.08856 kg

Therefore, the molality is:
Molality = 0.2135 mol / 0.08856 kg = 2.41 mol/kg

(b) Molarity is defined as the number of moles of solute per liter of solution. We already calculated the moles of ethanol as 0.2135 mol in 100 mL of solution.

Volume of solution = 0.100 L (since 100 mL = 0.1 L)

Therefore, the molarity is:
Molarity = 0.2135 mol / 0.100 L = 2.135 M

(c) We need to find the volume of solution that contains 0.125 mole of ethanol. Using the molarity calculated:

Molarity = 2.135 M = moles of solute / volume of solution (L)

Volume of solution = moles of solute / M

Volume of solution = 0.125 mol / 2.135 M = 0.05857 L = 58.57 mL

Therefore, the volume of the solution that would contain 0.125 mole of ethanol is 58.57 mL.

Given a reaction between an organic molecule, denoted as A, and NaSH, we observe the following observations. Using the observations, write a rate law for the reaction. (a) The rate triples when the concentration of [A] is tripled and the concentration of [NaSH] is held constant. (b) The rate is decreased when the concentration of [A] is doubled and the concentration of [NaSH] is cut by a factor of 3. (c) The rate doubles when the concentration of [A] is cut in half and the concentration of [NaSH] is quadrupled. (d) The rate increases with an increase in temperature.

Answers

Final answer:

The rate law for the reaction can be determined by analyzing the effects of changing reactant concentrations. Based on the given observations, the rate law can be written as rate = k[A]/[NaSH] and rate = [tex]k[A][NaSH]^2[/tex] for the given conditions. Additionally, the rate constant, k, is temperature-dependent.

Explanation:

The rate law for the reaction can be determined by examining the effects of changing the concentrations of reactants on the rate of the reaction. Based on the given observations, we can conclude the following:

(a) The rate triples when the concentration of [A] is tripled and the concentration of [NaSH] is held constant. This suggests that the reaction rate is directly proportional to the concentration of A, so the rate law can be expressed as rate = k[A]

(b) The rate is decreased when the concentration of [A] is doubled and the concentration of [NaSH] is cut by a factor of 3. This implies that the rate is inversely proportional to the concentration of [NaSH], so the rate law can be written as rate = [tex]k[A]/[NaSH][/tex]

(c) The rate doubles when the concentration of [A] is cut in half and the concentration of [NaSH] is quadrupled. This indicates that the rate is quadratically proportional to the concentration of [NaSH], so the rate law can be expressed as rate = [tex]k[A][NaSH]^2[/tex]

(d) The rate increases with an increase in temperature. This suggests that the rate constant, k, in the rate law equation is temperature-dependent.

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An 8.6 grams piece of aluminum, heated to 100°C, is placed in a coffee cup (styrofoam) calorimeter that contains 402.4 grams of water at 25°C. The specific heat of water is 4.18 J/g°C and the specific heat of aluminum is 0.900 J/g°C.
What will be the final equilibrium temperature for both the aluminum and the water? (Assume no heat is lost to the surroundings and the coffee cup absorbs a negligible amount of heat).

Answers

Final answer:

The final equilibrium temperature for both the aluminum and the water is approximately 23.39°C.

Explanation:

To find the final equilibrium temperature for both the aluminum and the water, we can use the principle of thermal equilibrium and the conservation of energy.

First, let's calculate the heat gained or lost by the aluminum and the water:

The heat gained or lost by a substance can be calculated using the formula:

Q = mcΔT

where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the aluminum:

Qalu = malu * calu * ΔTalu = (8.6 g) * (0.900 J/g°C) * (T - 100°C)

For the water:

Qwater = mwater * cwater * ΔTwater = (402.4 g) * (4.18 J/g°C) * (T - 25°C)

Since the total heat gained by the aluminum is equal to the total heat lost by the water:

Qalu = -Qwater

Substituting the values and solving for T:

(8.6 g) * (0.900 J/g°C) * (T - 100°C) = -(402.4 g) * (4.18 J/g°C) * (T - 25°C)

Simplifying the equation:

(7.74 T - 774) = -(1681.832 T - 42045)

1765.832 T = 41271

T ≈ 23.39°C

The final equilibrium temperature for both the aluminum and the water is approximately 23.39°C.

Assume that you react 100 mg of benzoic acid with 10 mL of methanol and 10 microliters of sulfuric acid to produce methyl benzoate. Write a balance chemical equation for this reaction. Determine the limiting reagent and calculate a theoretical yield of both the ester and water. If you isolate 75 mg of methyl benzoate, what is the actual yield of the reaction?

Answers

Explanation:

Balanced equation for the reaction between benzoic acid and metanol is as follows.

  [tex]C_{6}H_{5}COOH + CH_{3}OH \overset{H_{2}SO_{4}}{\rightarrow} C_{6}H_{5}COOCH_{3}[/tex]

Since, volume of [tex]H_{2}SO_{4}[/tex] is very small so, that is catalytic amount of [tex]H_{2}SO_{4}[/tex] which is used.

As mass of benzoic acid is 100 mg. Hence, moles of benzoic acid are calculated as follows.

      No. of moles of benzoic acid = [tex]\frac{mass}{molar mass}[/tex]

                                                       = [tex]\frac{100 mg}{122.12 g/mol}[/tex]

                                                       = 0.818 mmol

And, mass of methanol = volume × density

                                       = [tex]10 ml \times 0.792 g/mol[/tex]

                                       = 7.92 g

Now, number of moles of methanol is as follows.

     No. of moles of methanol = [tex]\frac{mass}{molar mass}[/tex]

                                                       = [tex]\frac{ 7.92 g}{32.04 g/mol}[/tex]

                                                       = 0.024 mol      

                                                       = 24.7 mmol    

As number of moles of benzoic acid are smaller than the number of moles of methanol. Hence, benzoic acid is the limiting reagent.

As per the balanced equation, 1 mole of benzoic acid produces 1 mole of methyl benzoate.

Hence, 0.818 mmol of benzoic acid would produce 0.818 mmol of methyl benzoate. Therefore, theoretical yield of methyl benzoate is as follows.

     Theoretical yield of methyl benzoate = [tex]0.818 mmol \times 136.15 g/mol[/tex]      

                                             = 111.48 mg

or,                                          = 111.5 mg (approx)

Now, we will calculate the percent yield of the reaction as follows.

            Percent yield = [tex]\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100[/tex]

                                   = [tex]\frac{75 mg}{111.5 mg} \times 100[/tex]

                                   = 67.27%

Therefore, we can conclude that the actual yield of the reaction is 67.27%.

Final answer:

The limiting reagent in the synthesis of methyl benzoate from benzoic acid and methanol is benzoic acid. The theoretical yield is approximately 111.5 mg of methyl benzoate, and with an actual yield of 75 mg, the percent yield is approximately 67.26%.

Explanation:

The balanced chemical equation for the reaction between benzoic acid (C7H6O2) and methanol (CH3OH) to produce methyl benzoate (C8H8O2) and water (H2O) in the presence of sulfuric acid (H2SO4) as a catalyst is:

C7H6O2 + CH3OH
ightarrow C8H8O2 + H2O

To determine the limiting reagent, we need to compare the mole ratios of the reactants. The molar mass of benzoic acid is about 122.12 g/mol and methanol is about 32.04 g/mol. Converting the mass of benzoic acid (100 mg = 0.1 g) to moles yields 0.000819 moles, while 10 mL of methanol (assuming a density of approximately 0.79 g/mL) gives 0.248 moles. Since benzoic acid has fewer moles, it is the limiting reagent.

The theoretical yield of methyl benzoate can be calculated based on the moles of benzoic acid, which will also be equal to the moles of water produced as this is a 1 to 1 reaction. This yields 0.000819 moles of methyl benzoate, and with a molar mass of 136.15 g/mol, the theoretical mass of methyl benzoate is approximately 111.5 mg.

The actual yield is 75 mg of methyl benzoate. Therefore, the percent yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100, which results in approximately 67.26%.

. The heat capacity of a bomb calorimeter was determined by burning 6.79 g methane (energy of combustion 802 kJ/mol CH4) in the bomb. The temperature changed by 10.8C. a. What is the heat capacity of the bomb? b. A 12.6-g sample of acetylene, C2H2, produced a temperature increase of 16.9C in the same calorimeter. What is the energy of combustion of acetylene (in kJ/mol)?

Answers

Final answer:

The heat capacity of the bomb calorimeter is approximately 31.391 kJ/°C, and the energy of combustion of acetylene is approximately 1096.5 kJ/mol.

Explanation:

The heat capacity of a bomb calorimeter can be determined using the formula: q = CΔT, where q represents the energy absorbed or released, C is the heat capacity, and ΔT is the change in temperature.

First, we will determine the amount of energy absorbed by the bomb calorimeter from the combustion of methane. Given that the molar mass of methane is 16.04 g/mol, 6.79 g of methane corresponds to 0.423 mol. Therefore, the amount of energy absorbed is 0.423 mol x 802 kJ/mol = 339.046 kJ. Subsequently, we can calculate the heat capacity of the bomb calorimeter by rearranging the initial formula to: C = q/ΔT. So, C = 339.046 kJ/10.8 °C, which gives a heat capacity of approximately 31.391 kJ/°C.

Secondly, the energy of combustion of acetylene is given by the formula: q = CΔT. We already know the heat capacity of the bomb calorimeter (31.391 kJ/°C) and the change in temperature (16.9 °C). Therefore, the energy of combustion of acetylene is 31.391 kJ/°C x 16.9 °C = 530.7 kJ. As the molar mass of acetylene (C2H2) is 26.04 g/mol, 12.6 g corresponds to 0.484 mol. Thus, the energy of combustion of acetylene per mole is given by: 530.7 kJ/0.484 mol, which is approximately 1096.5 kJ/mol.

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Final answer:

The heat capacity of the bomb calorimeter is about 50229 kJ/°C. The energy of combustion of acetylene is found to be approximately 84710 kJ, which can then be converted back to kJ/mol from g.

Explanation:

The subject question refers to a bomb calorimetry experiment in which different substances, namely methane and acetylene, are being combusted. The heat capacity of the bomb calorimeter can be calculated using the formula q = CΔT, where q represents heat, C is heat capacity and ΔT is the change in temperature. The energy of combustion for the second substance can be calculated using the same formula rearranged as q/ΔT = C.

By substituting the values from the methane combustion (q = 6.79g * 802 kJ/mol, ΔT = 10.8°C), we find that the heat capacity is about 50229 kJ/°C. For the acetylene combustion, we can use the same formula rearranged to find q (q = CΔT), which gives approximately 84710 kJ. This value can then be converted back to kJ/mol from g to get the energy of combustion of acetylene.

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The cost, C C, of producing x x Totally Cool Coolers is modeled by the equation C = 0.005 x 2 − 0.25 x + 12 C=0.005x2-0.25x+12 How many coolers need to be produced and sold in order to minimize the cost? (Round to the nearest whole number.)

Answers

Answer:

25 coolers are need to be produce and sell in order to minimize the cost.

Explanation:

[tex]C = 0.005x^2-0.25x+12[/tex] ..[1]

Differentiating the given expression with respect to dx.

[tex]\frac{dC}{dx}=\frac{d(0.005x^2-0.25x+12)}{dx}[/tex]

[tex]\frac{dC}{dx}=0.01x-0.25+0[/tex]

Putting ,[tex]\frac{dC}{dx}=0[/tex]

[tex]0=0.01x-0.25+0[/tex]

[tex]0.01x=0.25[/tex]

x = 25

Taking second derivative of expression [1]

[tex]\frac{d^2C}{dx^2}=\frac{d(0.01x-0.25)}{dx}=0.01[/tex]

[tex]\frac{d^2C}{dx^2}>0[/tex] (minima)

25 coolers are need to be produce and sell in order to minimize the cost.

Isotopes are atoms of the same element with different numbers of

Answers

Answer:

neutron

Explanation:

which affects mass number

Duralin is an alloy of aluminium containing 4.5% manganese, 3.5% copper, and 0.45% magnesium. How much magnesium is present in a 88 g sample of this alloy?

Answers

Answer: The mass of magnesium in given amount of alloy is 0.396 g

Explanation:

We are given:

Percent mass of manganese = 4.5 %

Percent mass of copper = 3.5 %

Percent mass of magnesium = 0.45 %

Percent mass of aluminium = (100 - 4.5 - 3.5 - 0.45) = 91.55 %

Mass of alloy = 88 g

We need to calculate the mass of magnesium in 88 g of alloy, we get:

[tex]\Rightarrow 88\times \frac{0.45}{100}=0.396g[/tex]

Hence, the mass of magnesium in given amount of alloy is 0.396 g

When the volume of a gas increases, is the work done by the gas on its surroundings positive, negative, or zero?
a.zero
b.negative
c.positive

Answers

Final answer:

The work done by a gas on its surrounds is positive when the volume of the gas increases because the gas is doing work to push against the external pressure.

Explanation:

The work done by a gas on its surroundings is defined as positive when the gas expands, meaning when the volume of the gas increases. This is because the gas is doing work to push against the external pressure and expand its volume. So, the correct answer to your question 'When the volume of a gas increases, is the work done by the gas on its surroundings positive, negative, or zero?' would be (c) positive.

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When the volume of a gas increases, the work done by the gas on its surroundings is positive. This occurs because the gas is doing work on its surroundings, resulting in energy loss for the gas. The key equation is w = -Pext ΔV.

So the answer is C)Positive

To determine this, we need to understand the concept of work in thermodynamics. Specifically, the work done by a gas during expansion is given by the equation:

w = -[tex]P_{ext[/tex] ΔV,

where:

w is the work done by the gas.[tex]P_{ext[/tex] is the external pressure opposing the gas expansion.ΔV (Delta V) is the change in volume.

When the volume of the gas increases, ΔV is positive because the final volume ([tex]V_{final[/tex]) is greater than the initial volume ([tex]V_{initial[/tex]). According to the equation, since ΔV is positive and there is a negative sign in front, the work (w) done by the gas is negative.

This negative sign indicates that the system (the gas) is doing work on its surroundings, thus losing energy. Therefore, when the volume of a gas increases, the work done by the gas on its surroundings is positive.

NaOH(s) dissolves readily in water to form Na+(aq) and OH-(aq). If 35.1 g of NaOH is dissolved in 199 g of water in a coffee cup calorimeter, the temperature of the resulting solution increases from 23.0 oC to 62.6 oC. Calculate the enthalpy change per mole of NaOH dissolved. Assume the specific heat capacity of the solution is 4.18 J/oC*g.

Answers

Answer:

ΔH = -44,13 kJ/mol

Explanation:

For the following reaction:

NaOH(s) → Na⁺(aq) + OH⁻(aq)

It is possible to obtain the heat produced using:

Q = 4,18J/g°C×mass×ΔT

Where the mass is 35,1g + 199g = 234,1g

And ΔT is FinalT - InitialT → 62,6°C - 23,0°C = 39,6°C

Replacing:

Q = 38750 kJ = 38,75kJ

The enthalpy change is obtained using:

ΔH = -Q/moles of NaOH

Moles of NaOH are:

35,1g×[tex]\frac{1mol}{40g}[/tex]= 0,878 moles

Thus:

ΔH = -38,75 kJ/0,878mol = -44,13 kJ/mol

I hope it helps!

Which of the following statements about ethylene and polyethylene is true?
A) The polymer has the same properties as the monomer.
B) The polymer has alternating double and single bonds between carbons, while the monomer has only single bonds between carbons.
C) The polymer has alternating double and single bonds between carbons, while the monomer has only double bonds between carbons.
D) The polymer has only single bonds between carbons, while the monomer has only double bonds between carbons.

Answers

Answer:

D is the correct option.

Explanation:

The structure of ethene/ethyelene is : [tex]H_{2}[/tex]C==C[tex]H_{2}[/tex]

The carbon atoms are bonded together by a double bond.

The structure of polyethylene/polythene is: (-[tex]H_{2}[/tex]C-C[tex]H_{2}[/tex]-[tex])_{n}[/tex]

The carbon atoms are bonded together by single bonds.

Lets analyse each option:

A) Its false. Lets look at the obvious: Ethylene is a gas and polyethylene is a solid. Monomers and polymers have very different properties.

B) Its false. Polyethylene doesnt have double bonds.

C) False as well. Polyethylene doesnt have double bonds.

D) True. It is evident from the structure of the the monomer and polymer that I've shown above. I'll also provide neat structures in the attachments.

What is the hybridization of the central atom of molecules?

Answers

Explanation:

The number of electron density regions that encircle an atom is calculated by the steric number, which also defines the hybridisation of the central atom. In that scenario, there are six hybridised orbitals in the central atom, leading to a [tex]\displaystyle sp3d^2[/tex]hybridization.

I am joyous to assist you anytime.

You need to make 3-bromo-5-isopropyltoluene starting with m-isopropyltoluene. Supply the reagent needed in the order it is needed and draw the resulting product.

Answers

To synthesize 3-bromo-5-isopropyltoluene from m-isopropyltoluene, brominate using Br2/FeBr3, then rearrange with NaOH. Product: 3-bromo-5-isopropyltoluene with bromine at meta position and isopropyl group at para position.

To synthesize 3-bromo-5-isopropyltoluene starting with m-isopropyltoluene, we need to perform two main reactions: bromination and rearrangement. Here are the steps and reagents needed:

1. **Bromination**: We need to brominate the m-isopropyltoluene at the meta position. We can achieve this using bromine (Br2) in the presence of a Lewis acid catalyst like iron (III) bromide (FeBr3).

[tex]\[ \text{m-isopropyltoluene} + \text{Br}_2 + \text{FeBr}_3 \rightarrow \text{3-bromomethyl-isopropylbenzene} \][/tex]

2. **Rearrangement**: Next, we need to rearrange the 3-bromomethyl-isopropylbenzene to form the desired product, 3-bromo-5-isopropyltoluene. This rearrangement can be achieved using a strong base like sodium hydroxide (NaOH).

[tex]\[ \text{3-bromomethyl-isopropylbenzene} + \text{NaOH} \rightarrow \text{3-bromo-5-isopropyltoluene} \][/tex]

Now, let's draw the resulting product, 3-bromo-5-isopropyltoluene:

In this structure, the bromine atom (Br) is attached to the third carbon atom of the benzene ring, and the isopropyl group (-CH(CH3)2) is attached to the fifth carbon atom of the benzene ring.

If you heat the air inside a rigid, sealed container until its Kelvin temperature doubles, the air pressure in the container will also double. Is the same thing true if you double the Celsius temperature of the air in the container? Explain.

Answers

Answer:

No

Explanation:

If you heat the air inside a rigid, sealed container, then you are keeping it's mass and volume constant.

Then, Pressure of the container is directly proportional to the temperature.

Gay-Lussac's Law: This law states that the pressure of a given amount of gas when at constant volume is directly proportional to gas's absolute temperature in Kelvin.

So, if Kelvin temperature doubles, the air pressure in the container will also double.

The relationship between Celsius temperature and Kelvin temperature is not linear.

[tex]T_{K} = T_{C} + 273.15[/tex]

If [tex]T_{K}[/tex] becomes [tex]2*T_{K}[/tex], then [tex]T_{C}[/tex]  will become [tex](2 * T_{K})- 273.15[/tex] .

i.e, it is increasing for sure, but doesn't double.

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