Components A and B form ideal solution. At 350 K, a liquid mixture containing 40% (mole) A is in equilibrium with a vapour containing 70% (mole) A. If the vapour pressure of A at 350 K is 70 kPa, what is the vapour pressure of B? (b) 20 kPa (d) 12 kPa (а) 25 kPa (c) 40 kPa

Answer:

Measuring with a ruler and using final volume minus initial volume

Explanation:

You can measure the volume of a geometric object by measuring its sides with a ruler and calculating the volume according to the corresponding formula for each object. For example, for a rectangular prism it would be

[tex]volume=length*width*height[/tex]

You can also measure the volume of an object by measuring how much water it displaces. To do this you have to fill a measuring cylinder with enough water for the object to be completely submerged and take note of the volume. Then, add the object and note again the volume of the water+object. The difference between both is the volume of the object.

[tex]Volume of the object= volume of water and object - volume of water[/tex]

The advantage of the second method is that it can be used for objects with irregular shapes as long as they do not float.

Answer:

The temperature difference across the stainless steel is 18°C

Explanation:

The heat flows through the plastic layer by convection and then through the steel layer by conduction and then through the plastic layer on the other side.

The heat flux q/A can be expressed as:

[tex]q/A = h*\Delta T[/tex]

It can also be expressed as

[tex]q/A=h*(T_A-T_B) = h_1(T_A-T_1)=k/\Delta x*(T_1-T_2)=h_2(T_2-T_B)[/tex]

being Δx/k*(T1-T2) the conductive heat flux through the steel.

If we want to know the temperatur difference across the stainless steel (T1-T2) we can write:

[tex](k/\Delta x)*(T_1-T_2)=h*(T_A-T_B)\\\\(T_1-T_2)=(\Delta x/k)*h*(T_A-T_B)\\\\(T_1-T_2)=(0.04/16)*120*60=18 \, ^{\circ} C[/tex]

Answer:

The correct option is: b. pH 6.4-8.0

Explanation:

Phenol red is a weak acid that is used as a pH indicator and exists in the form of stable red crystals.

The color of the phenol red solution changes from yellow to red when the change in pH is observed. The color of phenol red transitions from yellow to red when the pH is 6.8 - 8.2 or 6.4 - 8.0

Above the pH of 8.2, the phenol red solution turns a bright pink in color.

Answer: Drugs.com is the site which have all we need regarding medicines and uses of medicines. It give the facility to search A-Z drugs, once you have searched the desired medicine by this feature of site, it further gives information about identification of pills, approval date, and interaction between drug-drug and drug- the food you eating, the manufacturing date of medicines. It doesn't give information about the calculators.

Therefore, (e) is the correct option here.

Drugs.com is a resourceful website for information about various drugs. It provides details like Pill Identification, Drugs A-Z, Interaction Checker, and more. However, it does not provide any calculator-type feature.

Drugs.com provides comprehensive information related to different types of medications. It covers areas such as Pill Identification, Product, Manufacturers, Drug Approval Date, Drugs A-Z, and Interaction Checker. However, Drugs.com does not offer any feature or functionality related to 'Calculators' pertaining to drug calculations or dosage calculations. Use of the site could enable someone to determine whether they have a substance use disorder, but it does not replace professional medical advice.

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Answer:

Pr = 2273.58

Explanation:

∴ Cp = 0.5 J/g.°C

∴ κ = 0.2 W/m.°C * ( J/s / W ) = 0.2 J/s.m.°C

∴ μ = 2200 Lbm/ft.h * ( 453.592 g/Lbm ) * ( ft / 0.3048 m ) * ( h/3600 s )

⇒ μ = 909.433 g/m.s

⇒ Pr = ((0.5 J/g.°C )*( 909.433 g/m.s )) / 0.2 J/s.m.°C

⇒ Pr = 2273.58

Answer:

0.350 grams

Explanation:

The given mass of the pill that has to be taken as a dosage for the medication = 350 mg

The medication has to be determined in grams which means that milligram has to be converted to grams.

The conversion of mg to g is shown below:

[tex]1\ milligram=10^{-3}\ gram[/tex]

So,

[tex]350\ milligram=350\times 10^{-3}\ gram[/tex]

[tex]350\ milligram=0.350\ gram[/tex]

The medication required in grams = 0.350 grams

Final answer:

To convert 350 mg to grams, divide by 1000, resulting in 0.35 grams of medication.

Explanation:

To convert the dosage of medication from milligrams to grams, you need to know the conversion factor between these two units.

There are 1000 milligrams in one gram.

Therefore, you can find the amount in grams by dividing the milligram dosage by 1000.

For the pill with a dosage of 350 mg of medication, the conversion to grams would be:

Total given grams/1000

= 350 mg ÷ 1000

= 0.35 grams

Answer:

a) [tex]0.63\frac{kg}{m^{3}}[/tex]

Explanation:

Density is given by the expression [tex]d=\frac{m}{V}[/tex], where m is the mass of the substance and V is the volume occupied by the substance.

As the problem says that the liquid is enclosed in a cylinder, you should find the volume of that cylinder that will be the same volume of the liquid, so:

For a cylinder the volume is given by V=[tex]2\pi r^{2}h[/tex]

Replacing the values given, we have:

[tex]V=2\pi (1m)^{2}(5m)[/tex]

[tex]V=15.708m^{3}[/tex]

Replacing the values of m and V in the equation of density, we have:

[tex]d=\frac{10kg}{15.708m^{3}}[/tex]

[tex]d=0.63\frac{kg}{m^{3}}[/tex]

Answer : The freezing point of solution is 273.467 K

Explanation : Given,

Mass of glucose (solute) = 0.3 g

Mass of water (solvent) = 1000 g = 1 kg

Moles of fructose (solute) = 0.5 mol

Mass of water (solvent) = 1000 g = 1 kg

Molar mass of glucose = 180 g/mole

First we have to calculate the moles of glucose.

[tex]\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{0.3g}{180g/mole}=0.00167mole[/tex]

Now we have to calculate the total moles after mixing.

[tex]\text{Total moles}=\text{Moles of glucose}+\text{Moles of fructose}[/tex]

[tex]\text{Total moles}=0.00167+0.5=0.502moles[/tex]

Now we have to calculate the molality.

[tex]\text{Molality}=\frac{\text{Total moles}}{\text{Mass of water(solvent in kg)}}[/tex]

[tex]\text{Molality}=\frac{0.502mole}{(1+1)kg}=0.251mole/kg[/tex]

Now we have to calculate the freezing point of solution.

As we know that the depression in freezing point is a colligative property that means it depends on the amount of solute.

Formula used :  

[tex]\Delta T_f=K_f\times m[/tex]

[tex]T^o_f-T_f=K_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]

[tex]T_f[/tex] = temperature of solution = ?

[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]

m = molality = 0.251 mole/kg

Now put all the given values in this formula, we get

[tex]0^oC-T_f=1.86^oC/m\times 0.251mole/kg[/tex]

[tex]T_f=-0.467^oC=273.467K[/tex]

conversion used : [tex]K=273+^oC[/tex]

Therefore, the freezing point of solution is 273.467 K

Answer:

The  pH of the resulting solution is 12.52.

Explanation:

[tex]Molarity=\frac{n}{V}[/tex]

n = number of moles

V = volume of the solution in Liters

1)1 mol of HCl gives 1 mol of hydrogen ion.

[tex][HCl]=[H^+]=0.1 m[/tex]

Concentration of the hydrogen ion = 0.1 M

Volume of the solution = 10 mL = 0.010 L

[tex]0.1 M=\frac{n}{0.010L}[/tex]

Moles of hydrogen ions =  = 0.001 mol

2) 1 mol of NaOH gives 1 mol of hydroxide ion.

[tex][NaOH]=[OH^-]=0.2 M m[/tex]

Concentration of the Hydroxide ions = 0.2 M

Volume of the solution ,V'= 8 mL = 0.008 L

[tex]0.2=\frac{n'}{V'}[/tex]

Moles of hydroxide ions ,n ' = 0.0016

1 mol of HCl neutralizes 1 mol of NaOH ,then 0.001 mol of HCl will neutralize 0.001 mol NaOH.

So left over moles of hydroxide ions in the solution will effect the pH of the solution:

Left over moles of hydroxide ions in the solution = 0.0016 mol - 0.0010 mol = 0.0006 mol

Left over concentration of hydroxide ions:

[tex][OH^-]'=\frac{0.0006 mol}{0.010 L+0.008 L}=0.0333 mol/L[/tex]

[tex]pOH=-\log[OH^-]=-\log[0.03333 M]=1.48[/tex]

pH +pOH = 14

pH = 14 - 1.48 = 12.52

The  pH of the resulting solution is 12.52.

Answer:

Macroscopic scale

Explanation:

Subatomic scale is the scale at which atomic constituents, such as nucleus which contains protons and neutrons, and electrons, which orbit in  the elliptical paths around nucleus exists.

Miniscopic scale is a reference scale and is not a standard scale for measurement. Usually, this refers to minute objects.

Atomic scale is the scale which is at size of the atoms.

Macroscopic scale is length scale on which the objects or the phenomena are enough large to be visible with naked eye, without magnifying the optical instruments. It is the largest scale.

Microscopic scale is scale of the objects that require microscope to see them.

Answer: Option (b) is the correct answer.

Explanation:

Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.

As the given reaction is as follows.

      [tex]C_{2}H_{4} + H_{2} \rightarrow C_{2}H_{6}[/tex]

Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.

Thus, we can conclude that in the given reaction carbon atoms are reduced.

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Explanation:

This is the begin:

Q1 = Q which is gained from the ice to be melted

Q2 = Q which is lost from the water to melt the ice

Q1 + Q2 = 0

We are informed that the ice is at 0 ° so we have to start calculating how many J, do we need to melt it completely. If the ice had been at a lower temperature, it should be brought to 0 ° with the formula

Q = mass. specific heat. (ΔT)

and then make the change of state by the latent heat of fusion.

The heat of fusion for water at 0 °C is approximately 334 joules per gram.

So Q = Hf . mass

Q1 = 334 J/g . 8.32 g = 2778,88 J

For water we should use this:

Q = mass. specific heat. (ΔT)

Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)

Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)

(notice we have kg, so we have to convert 55 g, to kg, 0,055kg so units can be cancelled)

Q2 = 0,055kg . 4180 J/kg. K (Tfinal (The unknown) -25°)

T° should be in K for the units of Specific heat but it is the same. The difference is the same, in K either in °C

25°C = 298K

Q2 = 0,055kg . 4180 J/kg. K (Tfinal -298K)

Now the end

Q1 + Q2 = 0

334 J/g . 8.32 g + 0,055kg . 4180 J/kg. K (Tfinal -298K)

2778,88 J + 229,9 J/K (Tfinal - 298 K) = 0

2778,88 J + 229,9 J/K x Tfinal - 68510,2 J = 0

229,9 J/K x Tfinal = 68510,2 J - 2778,88 J

229,9 J/K x Tfinal = 65731,4 J

Tfinal = 65731,4 J / 229,9 K/J

Tfinal = 285,9 K

Tfinal = 285,9 K - 273K = 12,9 °C

The final temperature of the water after an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C and all the ice melts is 12.959 °C, using the conservation of energy and assuming no heat loss to the surroundings.

When an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C, we can find the final temperature after all the ice is melted by applying the principle of conservation of energy. The heat gained by the ice melting must be equal to the heat lost by the water cooling down. We will assume that no heat is lost to the surroundings in this perfectly insulated system.

We need to calculate the heat required to melt the ice cube (Qmelt) using the heat of fusion of water (which is 79.9 cal/g or 334 J/g), and the heat lost by the water as it cools down (Qwater) using the specific heat of water (which is 4.184 J/g°C).

First, calculate the heat necessary to melt the ice:

Qmelt = mass of ice * heat of fusion

Qmelt = 8.32 g * 334 J/g

Qmelt = 2778.08 J

Next, set up the equation based on the conservation of energy, where the heat lost by the water (Qwater) equals the heat gained by the ice (Qmelt):

Qwater = mass of water * specific heat of water * temperature change

Qwater = Qmelt

55 g * 4.184 J/g°C * (25 °C - final temperature) = 2778.08 J

Solving for the final temperature:

230.62 J/°C * (25 °C - final temperature) = 2778.08 J

25 °C - final temperature = 2778.08 J / 230.62 J/°C

25 °C - final temperature = 12.041 °C

Final temperature = 25 °C - 12.041 °C

Final temperature = 12.959 °C

Therefore, the final temperature of the water after all the ice has melted is 12.959 °C (rounded to three significant figures).

Answer:

q = Positive

w = Negative

Explanation:

As per first law of thermodymanics,

ΔE = q + w

Where,

ΔE = Change in internal energy

q = Heat absorbed or heat released by the system

w = Work done

Sign conventions are used for heat transfer and work done during a thermodynamics process.

Sign convention for Heat transfer

Sign convention for Work done

In the given question, work is done on the surroundings so, w is negative.

Heat flows into a system or in other word heat is added to the system,

So q is positive.

Answer:

[tex]1s^22s^22p^63s^23p^64s^2[/tex]

Explanation:

Calcium is the chemical element with symbol Ca and the atomic number equal to 20. As alkaline earth metal, the element, calcium is reactive metal. IT lies in the second group and forth period of the periodic table.

The number of the valence electrons of the calcium element is 2 and thus primarily denotes these electrons and forms ionic bond.

The ground state- electron configuration for calcium is: [tex]1s^22s^22p^63s^23p^64s^2[/tex]

Answer:

pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,2

Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:

7,0 = 7,2 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]

Ratio obtained is:

0,63 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]

As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻   ⇄   HPO4²⁻ +        H⁺

0,1 M +x      0,063M -x  0,001M -x -because the addition of H⁺ displaces the equilibrium to the left-

Knowing the equation of equilibrium is:

[tex]K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}[/tex]

Replacing:

6,20x10⁻⁸ = [tex]\frac{[0,063-x][0,001-x]}{[0,1+x]}[/tex]

You will obtain:

x² -0,064 x + 6,29938x10⁻⁵ = 0

Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!

Answer:

v = 2.512 E-5 m³/mol

Explanation:

∴ P = 80 bar → V = 0.001384 m³/Kg......sat. liq water table

∴ P = 85 bar → V = 0.0014013 m³/Kg

⇒ P = 83 bar → V = ?

specific volume ( V ):

⇒ V = 0.001384 + (( 83 - 80 ) / ( 85 - 80 ))*( 0.0014013 - 0.001384 )

⇒ V = 0.00139438 m³/Kg

molar volume ( v ):

∴ Mw water = 18.01528 g/mol

⇒ v = 0.00139438 m³/Kg * ( Kg/1000g ) * ( 18.01528 g/mol )

⇒ v = 2.512 E-5 m³/mol

Answer:

2486 mmHg

Explanation:

Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:

P₁/T₁ = P₂/T₂

Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.

The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.

°C must be converted to absolute temperature (K), thus:

25°C + 273,15 = 298,15 K

755°C + 273,15 = 1028,15 K

Thus, pressure P₂ is:

(T₂·P₁) / T₁ = P₂

1028,15K · 721mmHg / 298,15 K =  2486 mmHg

I hope it helps!

Answer:

400 cans of soda would be lethal.

Explanation:

In a can of soda, there is (2.13 mg/oz * 12 oz) 25 mg caffeine.

25 mg * (1g / 1000 mg) = 0.025 g

If in a can of soda there is 0.025 g of caffeine, a lethal dose of caffeine will be ingested after drinking (10.0 g * (1 can / 0.025 g)) 400 cans of soda.

Answer:

When experiments are carried out or research is done in a certain field of knowledge the scientist at first hypothesise certain knowledge or make theoretical hypotheses about that field of knowledge.

And then conduct the experiment or research to derive certain conclusions and get answers which they can apply on the hypothesis made or on the previous knowledge they have and thereby confirm or negate the hypothesis.

When two scientists are working on similar experiment and they tend to differ in the conclusions drawn by the result, as they get from experiment then it is called as confirmation bias among the scientists.

Explanation:

Except for the use in the chemistry laboratory , were it is used to synthesize many useful products, silver nitrate is also has biolofical and medical relevance.

Silver nitrate is commonly used for silver staining, for demonstrating the reticular fibers, the proteins and the nucleic acids. It is also used as stain in the scanning electron microscopy.

Silver Nitrate is also used for the bone ulcers as well as the burns and the acute wounds.

Answer:

The correct answer is: 1.316 . 10⁻³ m³/kg.

Explanation:

The density (ρ) of a substance is the ratio of its mass (m) to its volume (V). At constant temperature and pressure, its value is constant and it is an intrinsic property of materials. The units of density are kg/m³.

[tex]\rho = \frac{m}{V}[/tex]

The specific volume (ν) of a substance is the ratio of its mass to its volume. We can see that it is the reciprocal of density and an intrinsic property of matter as well. Therefore, the units of specific volume are m³/kg.

[tex]\nu = \frac{V}{m}=\frac{1}{\rho }[/tex]

Given we know the density of the liquid, we can use this relationship to find out its specific volume:

[tex]\nu =\frac{1}{\rho }=\frac{1}{760.0kg/m^{3} } =1.316 .10^{-3} m^{3} /kg[/tex]

Answer:

b.0.34 M

Explanation:

Given that:

Mass of NaCl = 20 grams

Molar mass of NaCl = 58.44 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{20\ g}{58.44\ g/mol}[/tex]

[tex]Moles= 0.3422\ mol[/tex]

Given that volume = 100 mL

Also,

[tex]1\ mL=10^{-3}\ L[/tex]

So, Volume = 100 / 1000 L = 0.1 L

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.3422}{0.1}[/tex]

Molarity = 0.34 M

Answer:

150 × 10⁻⁹ μL

Explanation:

Data provided in the question:

Molarilty of the stock solution, M₁ = 200 M

Final Volume of the solution, V₂ = 100 μL = 100 × 10⁻⁶ L

Final concentration, M₂ = 300 nM = 300 × 10⁻⁹ M

Now,

M₁V₁ = M₂V₂

where,

V₁ is volume of the stock solution

Thus,

200 × V₁ = 100 × 10⁻⁶ × 300 × 10⁻⁹

or

V₁ = 150 × 10⁻⁹ μL

Answer:

B) Increase in boiling point and decrease in vapor pressure

Explanation:

Vapor pressure is inversely related  to the Boiling point , as

higher the boiling point, lower the vapor pressure. and

Lower the boiling point, higher the vapor pressure.

Hydrogen bonding.

The electrostatic attraction between Hydrogen , bonded to electronegative atom like F, O, N and the more electronegative atom is called as Hydrogen bonding.

For example -

In alcohols, - OH group has Hydrogen that is bonded to more electronegative atom O.

As ,  

Extra energy is required to break Hydrogen bonds.

because the substance which exhibits Hydrogen bonding have lower vapor pressure than that of the substance with out Hydrogen bonding.

Hence , the substance with Hydrogen bonding , has higher boiling point,.

Hence , the correct option is  Increase in boiling point and decrease in vapor pressure .

Answer:

185.5156 g

Explanation:

The reaction of copper sulfate with zinc is shown below as:

[tex]CuSO_4+Zn\rightarrow ZnSO_4+Cu[/tex]

Given that :

Amount of copper sulfate = 454 g

Molar mass of copper sulfate = 160 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus, moles are:

[tex]moles= \frac{454\ g}{160\ g/mol}[/tex]

[tex]moles\ of\ copper\ sulfate= 2.8375\ mol[/tex]

From the reaction,  

1 mole of copper sulfate reacts with 1 mole of zinc

Thus,

2.8375 moles of copper sulfate reacts with 2.8375 moles of zinc

Moles of Zinc that should react = 2.8375 moles

Mass of zinc= moles×Molar mass

Molar mass of zinc = 65.38 g/mol

Mass of zinc = 2.8375 ×65.38 g = 185.5156 g

Final answer:

185.595 grams of zinc would react with 454 g of copper sulfate, based on the one-to-one mole ratio between copper sulfate and zinc and the molar mass of zinc (65.38 g/mol).

Explanation:

The student is asking how many grams of zinc would react with 454 g of copper sulfate, which has a molar mass of 160 g/mol. According to the provided reaction, CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq), there is a one-to-one mole ratio between copper sulfate and zinc. Given the molar mass of copper sulfate (160 g/mol), we first find the number of moles of copper sulfate in 454 g.

Number of moles of CuSO4 = (454 g) / (160 g/mol) = 2.8375 mol

Since the mole ratio between copper sulfate and zinc is 1:1, an equal number of moles of zinc will react with the copper sulfate. We then need to find the molar mass of zinc to convert moles of zinc to grams. Zinc has a molar mass of approximately 65.38 g/mol.

Mass of zinc = Number of moles of Zn × molar mass of Zn = 2.8375 mol × 65.38 g/mol = 185.595 g

Therefore, 185.595 grams of zinc would react with 454 g of copper sulfate.

Explanation:

The given data is as follows.

          [tex]\Delta H[/tex] = 286 kJ = [tex]286 kJ \times \frac{1000 J}{1 kJ}[/tex]

                            = 286000 J

 [tex]S_{H_{2}O} = 70 J/^{o}K[/tex],      [tex]S_{H_{2}} = 131 J/^{o}K[/tex]

 [tex]S_{O_{2}} = 205 J/^{o}K[/tex]

Hence, formula to calculate entropy change of the reaction is as follows.

          [tex]\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)[/tex]

                     = [tex][(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}][/tex]

                    = [tex][(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)][/tex]

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             [tex]\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}[/tex]

                            = [tex]286000 J - (163.5 J/K \times 298 K)[/tex]

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

Final answer:

To find the electrical work required for the electrolysis of water to produce 1 mole of hydrogen, calculate the Gibbs Free Energy (ΔG) for the reaction. Using the given thermodynamic data, ΔG at 298 K is 236.4 kJ, representing the minimum electrical work needed.

Explanation:

The student has asked how to determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298 K and 1 atm. The question involves understanding the thermodynamics of the reaction: H2O(l) → H2(g) + 0.5O2(g), with given data points including ΔH and standard entropies (S°) for the reactants and products. To find the electrical work required, you first calculate the ΔG (Gibbs Free Energy) of the reaction using the formula ΔG = ΔH - TΔS. Knowing ΔG allows you to determine the maximum work that can be extracted from the reaction, which, in the case of electrolysis, corresponds to the minimum work required to drive the reaction in reverse.

ΔS for the reaction can be calculated using the given entropies: ΔS = ⅛∑S°(products) - ⅛∑S°(reactants) = (131 + 0.5×205) - 70 = 166.5 J/K. Therefore, ΔG at 298 K can be calculated as ΔG = 286,000 J - (298K × 166.5 J/K) = 286,000 J - 49,617 J = 236,383 J or 236.4 kJ. This value represents the minimum electrical work required to produce one mole of hydrogen gas via electrolysis of water under the specified conditions.

Answer: The time required will be 19.18 years

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

We are given:

[tex]t_{1/2}=4.7\times 10^1yrs[/tex]

Putting values in above equation, we get:

[tex]k=\frac{0.693}{4.7\times 10^1yr}=0.015yr^{-1}[/tex]

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,  

k = rate constant  = [tex]0.015yr^{-1}[/tex]

t = time taken for decay process = ?

[tex][A_o][/tex] = initial amount of the reactant = 2 g

[A] = amount left after decay process =  (2 - 0.5) = 1.5 g

Putting values in above equation, we get:

[tex]0.015yr^{-1}=\frac{2.303}{t}\log\frac{2}{1.5}\\\\t=19.18yrs[/tex]

Hence, the time required will be 19.18 years

Answer:

A loud ordinary conversation following the supplied information in the question is about 4500 dB. But, in the official decibel system measure a loud conversation does not overcome 100 dB.

Explanation:

Using the supplied data of the exercise, we say that in a restaurant conversation the value is 45 dB. If we multiply this by 100 we will have a value for a laud ordinary conversation.

45×100 = 4500 dB.

but as I mentioned in the answer, in the official decibel system measure a loud conversation between 2 man reaches a maximal of 100 dB.

Answer:

[tex]A(t)=1.8+34.2*e^{-t}[/tex]

Explanation:

The concentration of CO2 in the room will be the amount of CO2 in the room at time t, divided by the volume of the room.

Let A(t) be the amount of CO2 in the room, in ft3 CO2.

The air entering the room is 3000 ft3/min with 0.06% concentrarion of CO2. That can be expressed as (3000*0.06/100)=1.8 ft3 CO2/min.

The mixture leaves at 3000 ft3/min but with concentration A(t)/V. We can express the amount of CO2 leaving the room at any time is A(t).

We can write this as a differential equation

[tex]dA/dt=v_i-v_o=1.8-A[/tex]

We can rearrange and integrate

[tex]dA/dt=v_i-v_o=1.8-A\\\\dA/(A-1.8)=-dt\\\\\int(dA/(A-1.8) = -\int dt\\\\ln(A-1.8)=-t+C\\\\A-1.8=e^{-t}* e^{C}=C*e^{-t}\\\\A=1.8+C*e^{-t}[/tex]

We also know that A(0) = 12000*(0.3/100)=36 ft3 CO2.

[tex]A(0)=1.8+C*e^{-0}\\36=1.8+C*1\\C=34.2[/tex]

Then we have the amount A(t) as

[tex]A(t)=1.8+34.2*e^{-t}[/tex]

The subsequent amount A(t) of carbon dioxide in the room at time t is determined by solving the differential equation that models the problem as a tank mixing problem, accounting for the rate of air being pumped in and out of the room.

To determine the subsequent amount A(t), in ft³, of carbon dioxide in the room at any given time t, we need to set up a differential equation that accounts for the rate of air being pumped in and out of the room. This situation is analogous to a classic problem in differential equations and mathematical modeling called the tank mixing problem.

The rate at which carbon dioxide enters the room is constant at 0.06% of the 3,000 ft³/min being pumped in, which equals 1.8 ft³/min of CO₂. The rate at which it leaves the room is proportional to the concentration of carbon dioxide in the room at time t. The change in the amount of carbon dioxide at any moment is then given by the rate in minus the rate out.

Let C0 be the initial concentration of CO₂ (0.3%), CV be the volume of the room (12,000 ft³), and r be the rate of air exchange (3,000 ft³/min). The differential equation modeling this situation is:

with initial condition A(0) = C0 * V. This equation can be solved using separation of variables and integrating both sides,

Upon inserting the values for V, C0, and r, we can find the expression for A(t).

Answer:  0.14 kg

Explanation:

Gourmet chocolate candy contains 7.00 g of dietary fat in each 22.7-g piece

That is 1 piece of candy weighs 22.7 g and contains 7.00 g of dietary fat

Converting the mass in pounds to kg  

1 lb = 0.45 kg = 450 grams    (1kg=1000g)

Number of pieces = [tex]\frac{450}{22.7g}=20pieces[/tex]

1 piece contains = 7 g of dietary fat

Thus 30 pieces would contain =[tex]\frac{7}{1}\times 20=140g[/tex] of dietary fat

1 g = 0.001 kg

Thus 140 grams =[tex]\frac{0.001}{1}\times 140=0.14kg[/tex]

Thus 0.14 kg of dietary fat are in a box containing 1.00 lb of candy.