List the substances Ar, Cl2, CH4, and CH3COOH, in order of increasing strength of intermolecular attractions. List the substances , , , and , in order of increasing strength of intermolecular attractions. CH4 < Ar< CH3COOH < Cl2 CH3COOH < Cl2 < Ar < CH4 Ar < Cl2 < CH4 < CH3COOH Cl2 < CH3COOH < Ar < CH4 CH4 < Ar < Cl2 < CH3COOH

Answer:

1.95 M

Explanation:

Considering

Moles of A = Moles of B

Or,

[tex]Molarity_{A}\times Volume_{A}=Molarity_{B}\times Volume_{B}[/tex]

Given  that:

[tex]Molarity_{A}=20.422\ M[/tex]

[tex]Volume_{A}=2\ mL[/tex]

[tex]Volume_{B}=2+12=14\ mL[/tex]

[tex]Molarity_{B}=?\ M[/tex]

So,  

[tex]20.422\times 2=Molarity_{B}\times 14[/tex]

Molarity of solution B = 2.92 M

Again Considering,

Moles of B = Moles of C

Or,

[tex]Molarity_{B}\times Volume_{B}=Molarity_{C}\times Volume_{C}[/tex]

Given  that:

[tex]Molarity_{B}=2.92\ M[/tex]

[tex]Volume_{B}=2\ mL[/tex]

[tex]Volume_{C}=2+1=3\ mL[/tex]

[tex]Molarity_{C}=?\ M[/tex]

So,  

[tex]2.92\times 2=Molarity_{C}\times 3[/tex]

Molarity of solution C = 1.95 M

The concentration of dye in Solution C is 1.945 M after performing the serial dilutions.

To find the concentration of dye in Solution C after serial dilutions, we can use the dilution equation which states that the concentration times the volume before dilution (C1V1) is equal to the concentration times the volume after dilution (C2V2), or C1V1 = C2V2.

For Solution B, the initial concentration is 20.422 M (C1) and the initial volume is 2 mL (V1). This solution is diluted with 12 mL water, so the final volume (V2) is 2 mL + 12 mL = 14 mL. Using the dilution equation, we calculate the concentration of Solution B (C2).

(20.422 M)(2 mL) = (C2)(14 mL)

C2 = (20.422 M)(2 mL) / (14 mL) = 2.917 M

Next, we perform a second dilution to create Solution C. We take 2 mL of Solution B and add 1 mL of water, giving a final volume of 3 mL. Again, we use the dilution equation to find the concentration of Solution C.

(2.917 M)(2 mL) = (C3)(3 mL)

C3 = (2.917 M)(2 mL) / (3 mL) = 1.945 M

Therefore, the concentration of dye in Solution C is 1.945 M, rounded to three significant figures.

Answer :

(a) The volume of ethanol liquid needed are, 2.4 L

(b) The volume of ethanol liquid needed are, 1.9 L

(c) The volume of ethanol liquid needed are, 1.6 L

(d) The volume of ethanol liquid needed are, 1.7 L

Explanation : Given,

Volume of solution = 250 mL = 0.250 L   (1 l = 1000 mL)

Concentration of solution = 350 mM = 0.350 M    (1 mM = 0.001 M)

First we have to calculate the moles of solution.

[tex]\text{Moles of solution}=\text{Concentration of solution}\times \text{Volume of solution}=0.350M\times 0.250L=0.0875mole[/tex]

(a)  For ethanol liquid :

To we have to calculate the mass of ethanol.

[tex]\text{Mass of ethanol}=\frac{\text{Moles}}{\text{Molar mass of ethanol}}=\frac{0.0875mole}{46g/mole}=0.0019g[/tex]

Now we have to calculate the volume of ethanol.

[tex]Volume=\frac{Mass}{Density}=\frac{0.0019g}{0.789g/cm^3}=0.0024cm^3=0.0024mL=2.4L[/tex]

(b) For acetone liquid :

To we have to calculate the mass of acetone.

[tex]\text{Mass of acetone}=\frac{\text{Moles}}{\text{Molar mass of acetone}}=\frac{0.0875mole}{58g/mole}=0.0015g[/tex]

Now we have to calculate the volume of acetone.

[tex]Volume=\frac{Mass}{Density}=\frac{0.0015g}{0.791g/cm^3}=0.0019cm^3=0.0019mL=1.9L[/tex]

(c) For formic acid liquid :

To we have to calculate the mass of formic acid.

[tex]\text{Mass of formic acid}=\frac{\text{Moles}}{\text{Molar mass of formic acid}}=\frac{0.0875mole}{46g/mole}=0.0019g[/tex]

Now we have to calculate the volume of formic acid.

[tex]Volume=\frac{Mass}{Density}=\frac{0.0019g}{1.220g/cm^3}=0.0016cm^3=0.0016mL=1.6L[/tex]

(d) For tert-Butylamine liquid :

To we have to calculate the mass of tert-Butylamine.

[tex]\text{Mass of tert-Butylamine}=\frac{\text{Moles}}{\text{Molar mass of tert-Butylamine}}=\frac{0.0875mole}{73g/mole}=0.0012g[/tex]

Now we have to calculate the volume of tert-Butylamine.

[tex]Volume=\frac{Mass}{Density}=\frac{0.0012g}{0.696g/cm^3}=0.0017cm^3=0.0017mL=1.7L[/tex]

To determine the volume of each pure liquid necessary to make a 350 mM solution with a volume of 250 mL, calculate the required mass of each substance and then use the density to find the volume.

To determine the volume of each pure liquid necessary to make a 350 mM solution with a volume of 250 mL, we need to use the formula:

Volume (mL) = (mass (g) / density (g/cm³)) * 1000

For each pure liquid, calculate the mass of the substance needed by multiplying its molarity with the desired volume in moles:

Mass (g) = molarity (mol/L) * volume (L) * molar mass (g/mol)

Then, use the formula for volume to find the necessary volume of each pure liquid:

Volume (mL) = (mass (g) / density (g/cm³)) * 1000

Answer:

Lactose is found in animal milk is lactase is the enzyme that is required to digest it.

Explanation:

Lactose is a disaccharide made of glucose and galactose units. This sugar is naturally present in animal milk and it is digested (broken in its units) by the lactase enzyme.

Some people are lactose intolerant because their bodies is not able to produce the enzyme. If they ingest dairy products it may cause health issues. Nowadays it is possible to purchase the lactase enzyme and ingest it with a dairy food to avoid any health effects.

Answer:

43.13Kg of melamine

Explanation:

The problem gives you the mass of urea and two balanced equations:[tex]CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}[/tex]

[tex]6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}[/tex]

First we need to calculate the number of moles of urea that are used in the reaction, so:

molar mass of urea = [tex]60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}[/tex]

The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:

161.2Kg of urea[tex]*\frac{1molofurea}{0.06006Kgofurea}=[/tex]2684 moles of urea

Now from the stoichiometry you have:

2684 moles of urea[tex]*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}[/tex] = 447 moles of melamine

The molar mass of the melamine is [tex]126.12\frac{g}{mol}[/tex] so we have:

[tex]447molesofmelamine*\frac{126.12g}{1molofmelamine}[/tex] = 5637.64 g of melamine

Converting that mass of melamine to Kg:

5637.64 g of melamine *[tex]\frac{1Kg}{1000g}[/tex] = 56.38 Kg of melamine, that is the theoretical yield of melamine.

Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:

%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)

Actual yield of melamine = [tex]\frac{76.5}{100}*56.38Kg[/tex] = 43.13Kg of melamine

Using stoichiometry, the actual yield of Melamine from a reaction starting with 161.2 kg of urea with an overall yield of 76.5% can be found to be approximately 43.2 kg.

The question asks about the actual yield of Melamine, C3N3(NH2)3, from a reaction starting with 161.2 kg of urea, CO(NH2)2, with an overall yield of 76.5%. To understand this prediction, we must use the concept of stoichiometry, which is a method used in chemistry to calculate the quantities of reactants or products in a chemical reaction.

First, calculate the mole of urea used: it is the mass of the urea divided by the molar mass of urea. The molar mass of urea (CO(NH2)2) is approximately 60 g/mole. So, you have 161.2*1000/60 = approximately 2686.67 moles of urea.

From the balanced chemical equation, we can see that one mole of urea produces one mole of HNCO, and six moles of HNCO produce one mole of Melamine. Therefore, in theory, 2686.67/6 = 447.78 moles of Melamine can be produced.

However, the reaction yield is only 76.5%, so the actual yield of the Melamine will be less. Using the percentage yield, we can calculate the actual yield: 447.78 * 0.765 = approximately 342.65 moles.

Finally, to convert from moles to mass, multiply by the molar mass of melamine, which is approximately 126 g/mole: 342.65*126 = approximately 43173.9g or 43.2 kg.

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Answer: The rate will increase by a factor of 9.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(g)[/tex]

Given: Order with respect to [tex]NO[/tex] = 2

Order with respect to [tex]H_2[/tex] = 1

Thus rate law is:

[tex]Rate=k[NO]^2[H_2]^1[/tex]

k= rate constant

It is given that the initial concentration of NO is tripled while all other factors stayed the same

[tex]Rate'=k[3\times NO]^2[H_2]^1[/tex]

[tex]Rate'=k[3]^2[NO]^2[H_2]^1[/tex]

[tex]Rate'=k\times 9[NO]^2[H_2]^1[/tex]

[tex]Rate'=9\times Rate[/tex]

Thus the rate will increase by a factor of 9.

The correct answer is that the rate will increase by a factor of 9.

To understand why the rate increases by a factor of 9 when the concentration of NO is tripled, we need to consider the rate law for the reaction, which is given as first order in H2 and second order in NO. The rate law can be expressed as:

[tex]\[ \text{Rate} = k[\text{NO}]^2[\text{H2}] \][/tex]

 where [tex]\( k \)[/tex] is the rate constant, and [NO] and [H2] are the concentrations of NO and H2, respectively.

Since the reaction is first order with respect to H2, the rate is directly proportional to the concentration of H2. However, the reaction is second order with respect to NO, which means the rate is proportional to the square of the concentration of NO.

If the concentration of NO is tripled (let's denote the initial concentration of NO as [NO]_initial, then the new concentration is [tex]3[NO]_initial)[/tex], the new rate can be calculated as follows:

[tex]\[ \text{New Rate} = k(3[\text{NO}]_{\text{initial}})^2[\text{H2}] \][/tex]

[tex]\[ \text{New Rate} = k \cdot 9 \cdot [\text{NO}]_{\text{initial}}^2 \cdot [\text{H2}] \][/tex]

[tex]\[ \text{New Rate} = 9 \cdot k[\text{NO}]_{\text{initial}}^2[\text{H2}] \][/tex]

Comparing the new rate to the initial rate:

[tex]\[ \text{Initial Rate} = k[\text{NO}]_{\text{initial}}^2[\text{H2}] \][/tex]

 we see that the new rate is 9 times the initial rate because the concentration of NO has been tripled, and due to the second-order dependence on NO, the rate is multiplied by the square of the factor by which the concentration of NO has increased [tex](3^2 = 9).[/tex]

 Therefore, the rate will increase by a factor of 9 when the initial concentration of NO is tripled, while keeping all other factors constant.

Answer:

Option B and C are correct

Explanation:

Chemical reaction at equillibrium means: the rate of forward reaction equals the rate of reverse reaction. This means no product will be consumed, neither formed.

A) The concetration of H2O equals the concentration of O2. This is false.

At the equillibrium the concetnrations are constant. But not necessarily equal.

B) The rate of the forward reaction, reactants to products, equals the rate of the reverse reaction, products to reactants. This is true. IT's the definition of a the equillibrium.

C) The C-C bonds hold more energy in C6H12O6 than the C-O bonds in CO2. Hint: think teter-toter.

This is to, C-C bonds hold more energy than C-O, that's why C6H12O6 is burned, energy is released.

D) The concentration of O2 is changing at equilibrium. This is false. The concentration of O2 is constant.

Explanation:

It is given that 44.45 g of hydrocarbon gas produces 2649 kJ or [tex]2649 \times 10^{3} J[/tex] of heat.

Also here, 1 lb = 453.592 g.

Therefore, amount of energy released by 453.592 g of hydrocarbon gas will be calculated as follows.

              [tex]\frac{2649 \times 10^{3} J \times 453.592 g}{44.45 g}[/tex]

             = [tex]27.03 \times 10^{6} J[/tex]

It is known that 1 J = [tex]2.778 \times 10^{-7} Kwh[/tex].

Hence, [tex]27.03 \times 10^{6} J[/tex] = [tex]27.03 \times 10^{6} J \times 2.778 \times 10^{-7} Kwh/J[/tex]

                              = 7.508 Kwh

Thus, we can conclude that the combustion of exactly one pound of this hydrocarbon gas produce 7.508 Kwh energy.

Answer:

V of Sulfur tetrafluoride is  17.2 L

Explanation:

Given data;

T = -6°C =  267K                                [1° C  = 273 K]

n = 786 mmol of SF4 which is 0.786 mol

P = 1 atm

from ideal gas law  we have

PV = nRT

where n is mole, R is gas constant, V is volume

[tex]V = \frac{nRT}{P}[/tex][tex]V = \frac{0.786 mol \times 0.082 atmL/mol K \times* 267K}{1atm} = 17.2 L[/tex]

V of Sulfur tetrafluoride is  17.2 L

Answer: The given number in scientific notation is [tex]2.367\times 10^{7}[/tex]

Explanation:

Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

We are given:

A number having value = 23,665,700

Converting this into scientific notation, we get:

As, the decimal is shifting to left side, the power of 10 will be positive.

[tex]\Rightarrow 23,665,700=2.367\times 10^{7}[/tex]

Hence, the given number in scientific notation is [tex]2.367\times 10^{7}[/tex]

Answer:

The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].

[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.

Explanation:

A+B → P

Integrated rate law for second order kinetics is given by:

[tex]\frac{1}{A}=kt+\frac{1}{A_0}[/tex]

A = concentration left after time t = [tex]0.02 mol /dm^3[/tex]

[tex]A_o[/tex] = Initial concentration = [tex]0.075 mol /dm^3[/tex]

t = 1 hour = 3600 seconds

[tex]\frac{1}{0.02 mol /dm^3}=k\times 3600 s+\frac{1}{0.075 mol /dm^3}[/tex]

[tex]k=\frac{1}{3600 s}\times (\frac{1}{0.02 mol /dm^3}-\frac{1}{0.075 mol /dm^3)}[/tex]

[tex]k = 1.0185\times 10^{-2} dm^3/ mol s[/tex]

Half life for second order kinetics is given by:

[tex]t_{\frac{1}{2}}=\frac{1}{k\times a_0}[/tex]

[tex]t_{\frac{1}{2}}=\frac{1}{1.0185\times 10^{-2} dm^3/ mol s\times 0.075 mol /dm^3}=1.31\times 10^3 s[/tex]

The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].

[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.

Answer:

ΔS Rx = -120, 5 J/K

Explanation:

The ΔS in a reaction is defined thus:

ΔS Rx = ∑ n S°products - ∑ m S°reactants

For the reaction:

N₂(g) + 2 O₂(g) → 2NO₂(g)

ΔS Rx = 2 mol × 240,5 [tex]\frac{J}{mol.K}[/tex] - [ 1 mol × 191,5 [tex]\frac{J}{mol.K}[/tex] + 2 mol × 205,0 [tex]\frac{J}{mol.K}[/tex]]=

ΔS Rx = -120, 5 J/K

A negative value in ΔS means a negative entropy of the process. Doing this process entropycally unfavorable.

I hope it helps!

Answer:

The second sample will produce 1563 grams of fluorine (F2)

Explanation:

The reaction will be MgF2 → Mg + F2

The stoichiometry ratio of MgF2 and F2 is 1 : 1.

That means for 1 mole of MgF2 consumed there is 1 mole of F2 produced.

The first sample produces 2.15 kg of magnesium and 3.36 kg of fluorine

The second sample produced 1 kg of magnesium and x kg of fluorine

This we can show in the following equation =

2.15kg / 3.36 kg = 1kg / x

2.15/3.36 = 0.63988

0.63988 = 1/ x

x= 1/0.63988  = 1.563 kg

1.563kg = 1563 grams

The second sample will produce 1563 grams of fluorine (F2)

Answer:

Use 1.0 mL MgCl2, 0.30 L NaCl, fill up with water to 1 L

Explanation:

The dilution equation relates the concentration C₁ and volume V₁ of an undiluted solution to the concentration C₂ and volume V₂ as follows:

C₁V₁ = C₂V₂

We want to solve for V₁, the amount of stock solution that is required for the dilution to 1.0L.

V₁ = (C₂V₂) / C₁

For the MgCl₂ stock solution, the following volume is required:

V₁ = (C₂V₂) / C₁ = (1.0mM)(1.0L) / (1.0M) = 1 mL

For the NaCl stock solution, the following volume is required:

V₁ = (C₂V₂) / C₁ = (0.15M)(1.0L) / (0.50M) = 0.30 L

The volumes V₁ are the diluted until the total volume reaches V₂.

Answer: A resource which we get naturally from the enviornment such as, air water, soil etc. is known as natural resources. Natural resources can be categories on the basis of renewability and it is of two types; renewable and non-renewable.

A natural capital is the stock of natural resources; which includes soil, air and all the living organisms present on this planet

Final answer:

Natural resources are raw materials found in nature used for economic gain, while natural capital includes these resources plus the ecosystem services that sustain life. Whereas natural resources can be renewable or nonrenewable, natural capital focuses on the sustainable management of these resources and ecosystem services.

Explanation:

The difference between natural capital and natural resources is found in their definitions and applications. Natural resources are materials or substances that occur in nature and can be used for economic gain; they are the raw inputs we find in our environment that have not been altered by human hands. Examples of natural resources include water, minerals, timber, and fossil fuels. They are essential for producing goods and services, but they vary in their abundance and renewability. Renewable resources, such as solar energy and timber, can be replenished over time, while nonrenewable resources, like fossil fuels and minerals, are finite and diminish with use.

On the other hand, natural capital encompasses not just the natural resources themselves, but also the ecosystem services that sustain and fulfill human life. Natural capital refers to the world's stocks of natural assets which include geology, soil, air, water and all living organisms. There is a growing recognition of the need to manage natural capital sustainably to avoid depleting these vital assets. The concept of natural capital broadens the scope of what is considered valuable in the natural environment beyond just the extractive resources.

It's important to note that while all natural capital includes natural resources, not all natural resources serve as natural capital. This distinction lies in the broader ecological value natural capital provides, benefiting economic processes as well as the environment in ways that maintain or enhance our stock of natural assets.

Explanation:

Lewis definition of Acids and Bases

Chemical species which are capable of accepting electron pairs or donating protons are called Lewis acid.

Chemical species which are capable of donating electron pairs or accepting protons are called Lewis base.

Bronsted definition of acids and bases

Chemical species which are capable of donating H+ are called Bronsted acids.

Chemical species which are capable of accepting H+ are called Bronsted bases.

So all Bronsted acids are Lewis acids but all Lewis acids are not Bronsted acids.

For a chemical species to behave as Lewis acid, they must have:

For example, in BF3, octet of boron is incomplete, so it can accept a pair of electron and behaves as Lewis acid.

For a chemical species to behave as Lewis base, they must have:

For example, NH3 and OH, both N and O have lone pairs of electrons, hence behave as Lewis base.

In Lewis theory, an acid is an electron pair acceptor and a base is an electron pair donor. Bronsted definitions are a subset of this concept, and are therefore less general. Lewis acids require an empty orbital, while Lewis bases need a lone pair of electrons.

The Lewis definition of an acid is a species that can accept an electron pair, whereas a Lewis base is a species that has an electron pair available for donation to a Lewis acid. A clear example of a Lewis acid is BF3 since it can accept an electron pair due to its empty orbital. On the other hand, a Lewis base, such as OH-, has a lone pair of electrons that it can donate.

In regards to the Bronsted definitions, they define an acid as a proton donor and a base as a proton acceptor but they represent a subcategory within Lewis theory and are therefore less general.

As for the electron arrangements, a molecule or ion can act as a Lewis acid if it has an empty orbital to accept an electron pair. On the contrary, for a molecule or ion to behave as a Lewis base, the presence of a lone pair of electrons is required, as seen in NH3.

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Explanation:

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

The least precise number present after the decimal point determines the number of significant figures in the answer.

a) 12.3 mm × 15.853 mm :

[tex] 12.3 mm\times 15.853 mm :194.9919 mm^2\approx 195 mm^2[/tex]

Three significant figures.

b) 72.98 · 15.830 ml

[tex]72.98\times 15.830 ml=1,155.2734 ml\approx 1,155 mL[/tex]

Three significant figures.

c) 12.3 mm + 15.853 mm

[tex]12.3 mm + 15.853 mm = 28.153 mm\approx 28.1 mm[/tex]

Three significant figures.

d)  172.3 cm - 15.853 cm

[tex] 172.3 cm - 15.853 cm= 156.447 cm\approx 156.4 cm[/tex]

Four significant figures.

Final answer:

The student's Mathematics question involves calculating with significant figures. When multiplying or dividing values, the result is rounded to the number of significant figures in the least precise value, while addition and subtraction are rounded to match the least number of decimal places.

Explanation:

The subject of this question is Mathematics, specifically related to significant figures in calculations, which is a concept typically covered in High School mathematics or physics courses.

Report the answer with the correct number of significant figures:

(a) To multiply 12.3 mm by 15.853 mm, we should round our answer to the same number of significant figures as the number with the fewest significant figures. In this case, 12.3 has three significant figures. Therefore, the result should also be reported with three significant figures.

(b) When multiplying 72.98 by 15.830 ml, the number with the fewest significant figures is 72.98, which has four significant figures. The answer should be reported with four significant figures.

(c) When adding 12.3 mm to 15.853 mm, we look for the least precise value in terms of decimal places, which is 12.3 mm (it is precise to the tenths place). Therefore, our result should also be rounded to the tenths place.

(d) In the subtraction 172.3 cm - 15.853 cm - 12.5 cm, since 12.5 cm has the least number of decimal places (one decimal place), our final result should also be rounded to one decimal place.

When it comes to significant figures, the rule for multiplication and division is to match the result to the number with the fewest significant figures. For addition and subtraction, we round the result to the least number of decimal places in the values used.

Answer: The volume of argon gas at STP is 57.4 mL

Explanation:

STP conditions are:

Pressure of the gas = 1 atm = 760 torr

Temperature of the gas = 273 K

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

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

where,

[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas

[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas

We are given:

[tex]P_1=725torr\\V_1=65.0mL\\T_1=22^oC=(22+273)K=295K\\P_2=760torr\\V_2=?\\T_2=273K[/tex]

Putting values in above equation, we get:

[tex]\frac{725torr\times 65.0mL}{295K}=\frac{760torr\times V_2}{273K}\\\\V_2=57.4mL[/tex]

Hence, the volume of argon gas at STP is 57.4 mL

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 65.0 mL = 0.065 L (as 1 ml = 0.001 L),      

     [tex]T_{1}[/tex] = [tex]22.0^{o}C[/tex] = (22 + 273) K = 295 K,

      [tex]P_{1}[/tex] = 725 torr = 0.954 atm           (as 1 torr = 0.00131579 atm),

     [tex]V_{2}[/tex] = ?,   [tex]T_{2}[/tex] = 273 K,

      [tex]P_{2}[/tex] = 1 atm

And, according to ideal gas equation,

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

           [tex]\frac{0.954 atm \times 0.065 L}{295 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]

            [tex]V_{2}[/tex] = 0.0574 L

As, 1 L = 1000 ml. So, 0.0574 L = 57.4 ml.

Thus, we can conclude that the volume of this Ar gas at STP is 57.4 L.

Answer:

The diameter of the oil molecule is [tex]4.4674\times 10^{-8} cm[/tex] .

Explanation:

Mass of the oil drop = [tex]m=9.00\times 10^{-7} kg[/tex]

Density of the oil drop = [tex]d=918 kg/m^3[/tex]

Volume of the oil drop: v

[tex]d=\frac{m}{v}[/tex]

[tex]v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}[/tex]

Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.

Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m

1 cm = 0.01 m

Surface of the sphere is given as: a = [tex]4\pi r^2[/tex]

[tex]a=4\times 3.14\times (0.418 m)^2=2.1945 m^2[/tex]

Volume of the oil drop = v = Area × thickness

[tex]\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x[/tex]

[tex]x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm[/tex]

The thickness of the oil drop is [tex]4.4674\times 10^{-8} cm[/tex] and so is the diameter of the molecule.

The diameter of an oil molecule can be estimated by considering the oil slick as a cylinder and using the formula for the volume of a cylinder. From the given, we calculate the volume of the oil droplet. Dividing this volume by the area of the slick gives us the height of the cylinder, which represents the diameter of the oil molecule.

The oil slick forms a cylindrical shape with a height equal to the diameter of an oil molecule or two times the radius. We know the volume of the oil droplet, oil slick, and the diameter of the oil slick.

The volume of oil droplet (V) = Mass (m) / Density (ρ)
= 9 x 10⁻⁷ kg / 918 kg/m³
= 9.81 x 10⁻¹⁰ m³

The volume of the cylindrical oil slick (V) = πr²h, where r is the radius and h is the height of the cylinder.
So, h (height of the oil molecule) = V / (πr^2)
= 9.81 x 10⁻¹⁰ m³ / (3.14 x (0.418 m)²)
= 1.74 x 10⁻¹⁰ m or 174 pm (picometers)

Since the height of the cylinder is approximately equal to the diameter of an oil molecule, the approximate diameter of an oil molecule is 174 pm.

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Answer:

London dispersion forces and dipole-dipole interactions

Explanation:

Intermolecular forces are the interactive forces that are present between the molecules. These intermolecular forces can be attractive or repulsive.

Hydrogen bromide is a diatomic covalent polar molecule, having chemical formula: HBr. Since, hydrogen bromide is a polar molecule, it has a permanent dipole.

Therefore, the intermolecular forces present between the hydrogen bromide molecules are dipole-dipole interactions and the london dispersion forces.

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.  

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 [tex]m_3[/tex]. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to [tex]m_3 /day[/tex]. In that sense, we can calculate the retention time using the tank volume in [tex]m_3[/tex].

[tex]Ft =12 \frac{MG}{day}  (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}[/tex]

After that, we should divide the total flow by four, because we have four tanks.

[tex]F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}[/tex]

To calculate the retention time we divide the total volume V by the flow of each tank F1.

[tex]t1 =\frac{V1}{F1} = \frac{600 m^3}{11340  \frac{m^3}{d} }  = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h[/tex]

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

b)

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

[tex]t2 =\frac{V1}{F2} = \frac{600 m^3}{45360  \frac{m^3}{d} }  = 0.01 day\\t2 = 0.01  d (\frac{24 h}{1d} ) = 0.31 h[/tex]

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

a) Retention time in each settling tank for parallel operation is approximately 317.0 hours.

b) Retention time in each settling tank for series operation is also approximately 317.0 hours.

To find the retention time in each settling tank, we'll use the formula:

[tex]\[ \text{Retention Time} = \frac{\text{Volume of Tank}}{\text{Flow Rate}} \][/tex]

Where:

- Volume of Tank is the volume of each settling tank.

- Flow Rate is the total flow rate.

We'll first convert the flow rate to cubic meters per day (m\(^3\)/day) to match the units of the tank volume:

[tex]\[ 1 \, \text{MGD} = 1 \, \text{million gallons} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \]\[ = 1 \, \text{MGD} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \times \frac{10^{-6} \, \text{m}^3}{1 \, \text{liter}} \]\[ = \frac{3.785 \times 10^3}{24 \times 10^6} \, \text{m}^3/\text{hour} \]\[ = 0.1577 \, \text{m}^3/\text{hour} \][/tex]

Now we can calculate the retention time for both scenarios:

a) For parallel operation:

[tex]\[ \text{Retention Time (Parallel)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Parallel)} \approx 317.0 \, \text{hours} \][/tex]

b) For series operation:

In series, the total flow rate remains the same, but the volume is effectively multiplied by the number of tanks in series. So, the retention time for each tank remains the same as in the parallel case:

[tex]\[ \text{Retention Time (Series)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Series)} \approx 317.0 \, \text{hours} \][/tex]

Therefore, the retention time in each settling tank is approximately 317.0 hours for both parallel and series operation.

Explanation:

The given data is as follows.

         Initial concentration = 0.15 mg/ml,        

        Final concentration = 0.1 mg/ml, (as [tex]\frac{0.1 microgram}{1 microliter} \times \frac{10^{-3}}{1 microgram} \times \frac{1 microliter}{10^{-3}ml}[/tex])

        Final volume = [tex]100 microliter \times \frac{10^{-3} ml}{1 microliter}[/tex] = 0.1 ml

According to the dilution formula we get the following.

              [tex]C_{i} \times V_{i} = C_{f} \times V_{f}[/tex]

or,                        [tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]

Putting the given values into the above formula we get the following.

                   [tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]

                               = [tex]\frac{0.1 mg/ml \times 0.1 ml}{0.15 mg/ml}[/tex]

                               = 0.0667 ml

                               = [tex]6.67 \times 10^{-2} ml \times \frac{1 microliter}{10^{-3} ml}[/tex]                                

                              = 66.7 microliter

This means that volume of protein stock which is required is 66.7 ml. Hence, calculate the volume of water required as follows.

                  Volume of water required = Total volume - volume of protein stock

                                                     = (100 - 66.7)  microliter

                                                     = 33.3 microliter

Thus, we can conclude that we need 33.3 microliter of water and 66.7 microliter of protein.

Answer:

It is necessary in order to let air in.

Explanation:

When the stopper is not removed, a vacuum is formed inside the separatory funnel as the liquid drains out. The lower air pressure inside the separatory funnel then causes the liquid to drain improperly.

Final answer:

To calculate the molarity of a solution, divide the moles of solute by the liters of solution.

Explanation:

To calculate the molarity of a solution, we need to know the amount of solute and the volume of the solution.

Molarity (M) = moles of solute / liters of solution

In the first question, we have 0.400 mol of Na2S in 1.30 L of solution. So, the molarity would be:

M = 0.400 mol / 1.30 L = 0.308 M

In the second question, we have 23.9 g of MgS in 843 mL of solution. First, we need to convert grams of MgS to moles using the molar mass of MgS (40.3 g/mol). Then, we need to convert mL to L. Finally, we can calculate the molarity:

M = (23.9 g / 40.3 g/mol) / (843 mL / 1000 mL/L) = 0.712 M

Answer:

x = 130 Ms⁻¹

Explanation:

The equation is:

x = (2.4 x 10³ M⁻²s⁻¹)(0.38 M)³

The expression raised to the power of 3 is simplified:

x = (2.4 x 10³ M⁻²s⁻¹)(0.054872 M³)

The two values are multiplied together and units canceled:

x = 130 Ms⁻¹

Answer:

x = 131.69Ms⁻¹

Explanation:

x = (2.4×10³M⁻²s⁻¹) (0.38M)³

Upon expanding, we have;

x = (2.4 × 10³M⁻²s⁻¹) (0.38M) * (0.38M) * (0.38M)

x = (2.4 × 10³M⁻²s⁻¹) (0.054872 M³)

x = 2.4 * 0.054872 * 10³ * M⁻²s⁻¹ * M³

x = 0.13169 * 10³ * M⁻²s⁻¹ * M³

x = 131.69 * M⁽⁻²⁺³ ⁾* s⁻¹

x = 131.69 * Ms⁻¹

x = 131.69Ms⁻¹

Answer:

The result was that H₂SO₄ is a strong electrolyte, NH₃ is a weak electrolyte and sugar is a non-electrolyte.

Explanation:

The experiment will show that the H₂SO₄ solution will light the lightbulb brightly, the NH₃ solution will light it weakly and the sugar solution will not light it at all.

This happens because to light the lightbulb we need free charges moving in the solution. Therefore, the substance that produce more ions will be the best electrolyte.

H₂SO₄ → 2H⁺ + SO₄²⁻

NH₃ + H₂O → NH₄⁺ + OH⁻

Sugar will not generate ions

Explanation:

The given data is as follows.

            Volume of tank = 4 [tex]m^{3}[/tex]

             Density of water = 1000 [tex]kg/m^{3}[/tex]

Since, the tank is initially half-filled. Hence, the volume of water in the tank is calculated as follows.

                        [tex]\frac{1}{2} \times 4 = 2 m^{3}[/tex]

Also, density of a substance is equal to its mass divided by its volume. Therefore, initially mass of water in the tank is as follows.

                    Mass = [tex]Density \times initial volume[/tex]

                              = [tex]1000 \times 2[/tex]

                              = 2000 kg

Whereas mass of water in tank when it is full is as follows.

                     Mass = [tex]Density \times final volume[/tex]

                               = [tex]1000 \times 4[/tex]

                               = 4000 kg

So, net mass of the fluid to be filled is as follows.

                  Net mass to be filled = Final mass - initial mass

                                                      = 4000 kg - 2000 kg

                                                      = 2000 kg

Mass flow rate [tex](m_{in})[/tex] = 6.33 kg/s

Mass flow rate [tex](m_{out})[/tex] = 3.25 kg/s

       Time needed to fill tank = [tex]\frac{\text{net mass to be filled}}{\text{net difference of flow rates}}[/tex]

                                       = [tex]\frac{2000 kg}{m_{in} - m_{out}}[/tex]

                                       = [tex]\frac{2000 kg}{6.33 kg/s - 3.25 kg/s}[/tex]

                                       = 649.35 sec

Thus, we can conclude that 649.35 sec is taken by the tank to overflow.

Answer: Distillation is a process in which we use the boiling point, condensation of the substances to separate them from each other, it is a physical separation not the chemical because physical property of substance is used in it.

Now, the distinction in between a laboratory distillation and an industrial distillation is that, the distillation process in laboratory distillation works in batches, while in the industrial distillation it occurs in continuous manner, industrial distillation is the large scale distillation.

Final answer:

The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process.

Explanation:

The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process. Laboratory scale distillation is typically used for smaller quantities and is carried out using simple apparatus. It is commonly used for research, quality control, and small-scale production.

On the other hand, industrial distillation is conducted on a much larger scale and involves more complex equipment. It is used in industries such as oil refining and chemical manufacturing, where large quantities of materials need to be processed. Industrial distillation may include fractionating columns and other specialized equipment to separate components from mixtures.

Answer: 0.001 metric tonnes

Explanation:

According to Einstein equation, Energy and mass are inter convertible.

[tex]E=mc^2[/tex]

E=  Energy

m = mass

c= speed of light

Given : Complete combustion of 1.00 kg of coal releases about [tex]3.0\times 10^7 J[/tex] of energy.

Given mass: 1 kg

Converting kg to metric ton using the conversion factor:

1 kg=0.001 metric tonnes

Thus 1 kg of coal would also be equal to 0.001 metric tonnes.

Answer:

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Explanation:

Final answer:

The number 0.70894 is written as 709 × 10⁻³ in Engineering Notation with 3 significant figures.

Explanation:

To write the number 0.70894 in Engineering Notation with 3 significant figures, we first identify the significant figures in the number. Since leading zeros are not significant, the significant figures are 708. We then adjust the number to have a multiple of three as its exponent for the base unit, which is standard in Engineering Notation. The number can be written as 709 (rounded up from 708.94 to maintain 3 significant figures) multiplied by 10 to the power of -3. Therefore, the number in Engineering Notation is 709 × 10⁻³.

Answer : 3.4 mL of this suspension should be given to an infant weighing 15 lb.

Explanation :

First we have to convert weight of infant from 'lb' to 'kg'.

Conversion used :

As, 1 lb = 0.454 kg

So, 15 lb = [tex]\frac{15lb}{1lb}\times 0.454kg=6.81kg[/tex]

Now we have to calculate the recommended dose.

As, 1 kg weight recommended dose 10 mg

So, 6.81 kg weight recommended dose [tex]\frac{6.81kg}{1kg}\times 10mg=68.1mg[/tex]

Now we have to calculate the milliliter of ibuprofen suspension.

As, 100 mg dose present in 5.0 mL

So, 68.1 mg dose present in [tex]\frac{68.1mg}{100mg}\times 5.0mL=3.4mL[/tex]

Therefore, 3.4 mL of this suspension should be given to an infant weighing 15 lb.