Describe Newton's Law of Viscosity and the constitutive behavior of non-Newtonian fluid

Answer:

ghdtgfgrdvreeeegrwwggegteefewqrwefrwerrftrtdsfgyuytfgererdf

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 1.50 L,    [tex]T_{1}[/tex] = 159 K,

      [tex]P_{1}[/tex] = 5.00 atm,   [tex]V_{2}[/tex] = 0.2 L,

       [tex]T_{2}[/tex] = ?,       [tex]P_{2}[/tex] = 50.0 atm  

And, according to ideal gas equation,  

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

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.          

 [tex]\frac{5.0 atm \times 1.50 L}{159 K} = \frac{5 atm \times 0.2 L}{T_{2}}[/tex]

            [tex]T_{2}[/tex] = 21.27 K

Thus, we can conclude that the final temperature is 21.27 K .

Answer:

[tex]5.5*10x^{-9}m[/tex]

Explanation:

As you have the diameter of the sphere in nanometers (nm), you need to use de conversion factor to find the diameter in meters (m):

First you should put the quantity that you want to convert with its respective units:

[tex]Diameter=5.5nm[/tex]

Then you put the conversion factor, always you should put the same unit that you want to convert in the denominator:

Diameter = [tex]5.5nm*\frac{1*10^{-9}m}{1nm}[/tex]

And finally, you should multiply and/or divide the quantities:

Diameter = [tex]5.5*10^{-9}m[/tex]

Answer:

We need add acid to prepare 0.15 acetate buffer with pH 3.7.

Explanation:

As we know that buffer solution is the combination of weak acid with strong base or strong acid with weak base.

We know that CH3COOH is weak acid .Acetate buffer is the combination of weak acid CH3COOH and conjucate base CH3COO- (from salt).

So we have to add acid to achieve the proper final pH value of mixture.We need add acid to prepare 0.15 acetate buffer with pH 3.7.

Answer:

a. 0.14042

b. 0.221

c. equal to a.

5) 0.579 L

3) 8.1e6 ms

Explanation:

Hello,

In the attached photo you'll find the numerical procedure to get the results.

Best regards.

Answer: The mass of air present in the room is 37.068 kg

Explanation :  Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

[tex]V=lbh[/tex]

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

[tex]V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L[/tex]

Conversion used : [tex]1ft^3=28.3168L[/tex]

Now we have to calculate the mass of air in the room.

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]1.19g/L=\frac{Mass}{31148.53L}[/tex]

[tex]Mass=37066.7507g=37.068kg[/tex]

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg

Answer: There are 37 kg of air in the room.

Explanation:

To calculate the volume of cuboid (room), we use the equation:

[tex]V=lbh[/tex]

where,

V = volume of cuboid

l = length of room = 11 ft

b = breadth of room =  10 ft

h = height of room= 10 ft

Putting values in above equation, we get:

[tex]V=10\times 11\times 10=1100ft^3=1100\times 28.3L=31130L[/tex]  (Conversion factor: [tex]1ft^3=28.3L[/tex]

To calculate mass of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

We are given:

Density of air = 1.19 g/L

Volume of air = volume of room =  31130 L

Putting values in above equation, we get:

[tex]1.19g/L=\frac{\text{Mass of air}}{31130L}\\\\\text{Mass of air}=37000g=37.0kg[/tex]    (1kg=1000g)

Hence, the mass of air is 37 kg.

Answer: The mass of air is 0.00260 lbs.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of the gas = 150 psia = 10.2 atm   (Conversion factor:  1 psia = 0.068 atm)

V = Volume of gas = [tex]10in^3=0.164L[/tex]    (Conversion factor:  [tex]1in^3=0.0164L[/tex] )

m = mass of air = ?

M = Average molar mass of air = 28.97 g/mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]440^oF=499.817K[/tex]  (Conversion factor: [tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]  )

Putting values in above equation, we get:

[tex]10.2atm\times 0.164L=\frac{m}{28.97g/mol}\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 499.817K\\\\m=1.18g[/tex]

Converting this mass into lbs, we use the conversion factor:

1 lbs = 454 g

So, [tex]1.18g\times \frac{1lbs}{454g}=0.00260lbs[/tex]

Hence, the mass of air is 0.00260 lbs.

Answer:

0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.

It will take 17,447.41 days long to gain 1 lb of fat by this person.

Explanation:

Calorie intake of an adult in a day =[tex]2.2\times 10^3 calorie[/tex]

Calorie burnt by an adult in a day = [tex]2.0\times 10^3 calorie[/tex]

Excess  nutritional energy in day=

[tex](2.2\times 10^3 calorie)-(2.0\times 10^3 calorie)[/tex]

[tex]=2.0\times 10^2 calorie[/tex]

1 kcal = 4.184 kJ

So , 1000 cal = 4.184 kJ

[tex]1 cal = 4.184\times 10^{-3} kJ[/tex]

[tex]2.0\times 10^2 calorie=2.0\times 10^2\times 4.184\times 10^{-3} kJ[/tex]

[tex]=0.8368 kJ[/tex]

0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.

[tex]14.6\times 10^3[/tex] kilo joules of excess nutritional energy = 1 lb fat

The 1 kilo joule of excess nutritional energy  = [tex]\frac{1}{14.6\times 10^3}[/tex] lb fat

Excessive nutritional energy of an adult per day = 0.8368 kiloJoules

Amount of fat gained by an adult per day =

= [tex]0.8368 kiloJoules\times \frac{1}{14.6\times 10^3}=5.7315\times 10^{-5} lb[/tex] of fat

In 1 day an adult gains = [tex]5.7315\times 10^{-5} lb[/tex] of fat

Time taken to gain 1 lb fat:

[tex]\frac{1}{5.7315\times 10^{-5}} day=17,447.41 days[/tex]

It will take 17,447.41 days long to gain 1 lb of fat by this person.

Final Answer:

An adult who consumes an excess of 0.2 x 10³ Cal per day, equivalent to 0.8368 x 10³ kJ, will gain 1 lb of fat in approximately 17.5 days, as 1 lb of fat is stored for each 14.6 x 10³ kJ of excess nutritional energy consumed.

Explanation:

An adult consumes approximately 2.2 x 10³ Calories (Cal) per day and burns 2.0 x 10³ Cal per day. To find the excess nutritional energy in kilojoules, we need to subtract the energy burned from the energy consumed and then convert the result from Calories to kilojoules:

Excess energy (Cal) = 2.2 x 10³ Cal - 2.0 x 10³ Cal = 0.2 x 10³ Cal

Excess energy (kJ) = 0.2 x 10³ Cal x 4.184 kJ/Cal = 0.8368 x 10³ kJ

To calculate how long it will take for the adult to gain 1 lb of fat, we divide the amount of kilojoules that corresponds to 1 lb of fat by the daily excess kilojoules:

Days to gain 1 lb = (14.6 x 10³ kJ per 1 lb) / (0.8368 x 10³ kJ/day)

Days to gain 1 lb = 17.4512 days

Therefore, it will take approximately 17.5 days for the adult to gain 1 lb of fat, given the daily excess of 0.2 x 10³ Cal.

Answer:

1,2 dichlorobenzene was used as the solvent for the diels alder reaction: because the co elimination part of the reaction needs high temp and a high boiling point solvent such as 1,2 dichlorobenzene

Explanation:

Diels-Alder Reaction is a useful synthetic tool to prepare cyclohexane rings. It is a process, which occurs in a single step that consists of a cyclic redistribution of its electrons. The two reagents are bond together through a cyclic transition state in which the two new C-C bonds are formed at the same time. For this to occur, most of the time, it is necessary a high temperature and high-pressure conditions. Since 1,2 dichlorobenzene has a boiling point of 180ºC is a good solvent for this type of reactions.

Answer:

[tex]n=2.20x10^{3} mmol[/tex]

Explanation:

Hello,

To compute such moles, one must identify that the concentration unit in this problem is molarity, which is defined by:

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

Since we are inquired to compute the moles, we solve for them as follows:

[tex]n=M*V=11.0mol/L*200.0*\frac{1L}{1000mL}\\n=2.2mol[/tex]

Finally, this value in millimoles turns out into:

[tex]n=2.2mol*\frac{1000mmol}{1mol}\\n=2.20x10^{3} mmol[/tex]

Best regards.

Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

[tex]0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}[/tex]

Answer:

[tex]93.43\times 10^{-9}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.

The given number:

0.00000009345 can be written as [tex]93.425\times 10^{-9}[/tex]

Answer upto 4 significant digits = [tex]93.43\times 10^{-9}[/tex]

Answer: The mass of solid NaOH required is 80 g

Explanation:

Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

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

where,

n = acidity for bases = 1 (For NaOH)

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

[tex]\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq[/tex]

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

[tex]\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}[/tex]

Or,

[tex]\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}[/tex]         ......(1)

We are given:

Given mass of NaOH = ?

Equivalent mass of NaOH = 40 g/eq

Volume of solution = 400 mL

Normality of solution = 5 eq/L

Putting values in equation 1, we get:

[tex]5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g[/tex]

Hence, the mass of solid NaOH required is 80 g

Answer:

NCL3

Explanation:

NCL3

- nitrogen trichloride.

The chemical formula for a compound containing one nitrogen atom for every three chlorine atoms is NCl₃.

The chemical formula NCl₃ represents a compound consisting of one nitrogen atom and three chlorine atoms. In this formula, "N" represents nitrogen, and "Cl" represents chlorine. The subscript "3" indicates that there are three chlorine atoms bonded to each nitrogen atom in the compound.

This arrangement reflects the stoichiometry of the compound, specifying the relative ratios of the elements. Chemical formulas provide a concise way to describe the composition of substances, allowing scientists to understand the types and quantities of atoms present in a compound.

NCl₃, nitrogen trichloride, is a covalent compound with notable chemical properties, and its formula encapsulates the balanced proportion of nitrogen and chlorine within its molecular structure.

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Answer:

See explanation

Explanation:

Sea water = a mixture of different substances in water. If the water evaporates, some of those substances remain as salts. It's not a compound because it's a mixture of compounds and substances. An example of a compound is water ( because it has hydrogen atoms and an oxygen atom)

Calcium = Calcium is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because we can find it on the left side of the periodic table, in group 2.  It's an alkaline earth metal, what makes that calcium is a reactive metal.

Argon = Argon is found on the periodic table, so it can't be a mixture or compound. It is a non-metallic element. We can find it in group 18 on the periodic table. Argon is a noble gas, so non-metallic.

Water = a compound because its only made of 2 atoms : oxygen and hydrogen.  Reasons why water is a compound and not a mixture are:

 - The ormation of a compound is a chemical change which is followed by absorption of energy or evolution of energy, in case of water, electricity is required.

- Mixtures can be separated by physical separation techniques ,Water can not be separated into it its elements by physical separation techniques. But by the absorption of chemical energy.

Air = Mixture because it can be separated into different atoms, molecules,..  like oxygen, nitrogen etc. by the physical separation techniques (fractional distillation).

Carbon Monoxide = Carbon Monoxide is not found on the periodic table so it cannot be an element. It's made of 2 elements, this means caron monoxide is a compound. It's not a mixture since the elements cannot be separated by physical separating techniques.

Iron = Iron is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because:

-High melting point

-  Some metals form a dull oxide layer, this explains the shiny luster surface

- Electrical conductivity and thermal conductivity

Sodium chloride = NaCl cannot be found on the periodic table, so it isn't an element. It's a compound because it only has 2 atoms (elements). Those elements cannot be seperated by physical separating techniques, but would require electricity. So it's a compound.

Diamond = is a solid form of the element Carbon.  It's an allotrope of carbon. They have the same physical state but in distinct form. Technically diamond is a non-metallic element. Since it's seen as carbon.

Brass = Brass is a mixture of the elements of copper and zinc. Those elements can be separated by physical techniques.

Copper = Copper is found on the periodic table so it cannot be a compound or mixture. It's a metallic element because:

-High melting point

-  Some metals form a dull oxide layer, this explains the shiny luster surface

- Electrical conductivity and thermal conductivity

Dilute sulphuric acid = This is a mixture. Sulphuric acid is a compound but to dilute it's added in water, what is another compound. So it's a mixture of different compounds.

Sulphur = Can be found in the periodic table so it cannot be a mixture or compound. It's part of the metalloids, therefore, it can be concluded that sulfur is a non-metal. It belongs to the oxygen family.

Oil = Oil is a mixture of hydrocarbon compounds which varies in lengths.

Nitrogen = Nitrogen is found on the periodic table, so it can't be a mixture or compound. It can be found as a gas so it is a non-metallic element.

Ammonia = a compound of nitrogen and hydrogen with the formula NH3. Those elements cannot be separated with physical separating techniques.

Explanation:

Orifice meters require to compensate for pressure and temperature when one uses these meters to measure the steam or the gas flow in the pipes with the variable operating pressure as well as temperature conditions.

Normally chemists do not have online density measurement tool and thus to avoid complications, the chemists consider density as constant parameter.

In the steam or the gas flow measurement, density of steam or gas changes as the pressure and the temperature change. This significant  change in the density can affect accuracy of measured flow rate if the change is uncompensated and thus, this has to be fixed in order to avoid errors. Therefore, it is important to compensate for the pressure and the temperature when orifice is used to measure the steam or the gas flow.

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =[tex]25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)[/tex]

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= [tex]25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}[/tex]

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x [tex]\frac{40.67 KJ}{mol H20}[/tex] x [tex]\frac{1 mol H20}{18 g}[/tex]

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

Answer:

P=atm

[tex]b=\frac{L}{mol}[/tex]

Explanation:

The problem give you the Van Der Waals equation:

[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]

First we are going to solve for P:

[tex](P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}[/tex]

[tex]P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}[/tex]

Then you should know all the units of each term of the equation, that is:

[tex]P=atm[/tex]

[tex]n=mol[/tex]

[tex]R=\frac{L.atm}{mol.K}[/tex]

[tex]a=atm\frac{L^{2}}{mol^{2}}[/tex]

[tex]b=\frac{L}{mol}[/tex]

[tex]T=K[/tex]

[tex]V=L[/tex]

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

[tex]P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}[/tex]

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

[tex]P=\frac{L.atm}{L-L}-atm[/tex]

Then operate the fraction subtraction:

P=[tex]P=\frac{L.atm-L.atm}{L}[/tex]

[tex]P=\frac{L.atm}{L}[/tex]

And finally you can find the answer:

P=atm

Now solving for b:

[tex](P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT[/tex]

[tex](V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]

[tex]nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}[/tex]

[tex]b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}[/tex]

Replacing units:

[tex]b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}[/tex]

Multiplying and dividing units,(please see the second photo below), we have:

[tex]b=\frac{L-\frac{L.atm}{atm}}{mol}[/tex]

[tex]b=\frac{L-L}{mol}[/tex]

[tex]b=\frac{L}{mol}[/tex]

Answer: The correct answer is Option B.

Explanation:

The chemical equation for the reaction of lead (II) nitrate and ammonium chloride follows:

[tex]Pb(NO_3)_2(aq.)+2NH_4Cl(aq.)\rightarrow PbCl_2(s)+2NH_4NO_3(aq.)[/tex]

Ionic form of the above equation follows:

[tex]Pb^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)[/tex]

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

[tex]Pb^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)[/tex]

Solubility rules:

Hence, the correct answer is Option B.

Answer:

Significant hydrogen bonding is possible in [tex]NH_{2}Cl[/tex]

Explanation:

The only substance from above in which significant hydrogen bonding possible is chloramine ( NH2Cl )

Organic compounds are substances which contain carbon and hydrogen. Some few groups of organic compounds include:

So therefore, the only substance from above in which significant hydrogen bonding possible is chloramine ( NH2Cl )

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Answer:

Volume of HCl required = 28.4 mL

Explanation:

[tex]Zn(s) + 2HCl (aq) \rightarrow ZnCl_2(aq) + H_2(g)[/tex]

Mass of zinc ore = 4.65 g

% of zinc in zinc ore = 50 %

So, mass of zinc in zinc ore = 0.50 × 4.65 = 2.325 g

No. of moles of Zn = [tex]\frac{2.325}{65.40} = 0.0355\ mol[/tex]

So, as per the reaction coefficient,

1 mol of zinc reacts with 2 mol of HCl

0.0355 mol of zinc reacts with

                                    = 0.0355 × 2 = 0.071 mol of HCl

Molarity of HCl = 2.50 M

Volume of HCl = [tex]\frac{Moles}{Concentration}[/tex]

[tex]Volume\ of\ HCl = \frac{0.071}{2.50} = 0.0284\ L[/tex]

1 L = 1000 mL

0.0284 L = 1000 × 0.0284 = 28.4 mL

Answer:

The vapor pressure at 298 K for this solution is 50,49 Torr.

Explanation:

We have to apply Raoult's law to solve this which its formula is:

P pure sv° - P sl = P pure sv° . X (molar fraction)

ΔP = P pure sv° . X (molar fraction)

58,9 Torr - P sl = 58,9 Torr . X

Molar mass urea CO(NH2)2: 60,06 g/m

Molar mass ethanol: 46,07 g/m

Moles for urea: 42,4 g /  60,06 g/m = 0,705 moles

Moles for ethanol: 195 g / 46,06 g/m = 4,233 moles

X (molar fraction): moles from solute / moles from solute + moles from solvent

X (molar fraction): 0,705 / 0,705 + 4,233 = 0,142

58,9 Torr - P sl = 58.9 Torr . 0,142

58,9 Torr - 8,40 Torr = P sl

50,49 Torr = P sl

Answer:

0.7681

Explanation:

Given that the mass percentage of methanol = 85.5 %

which means that 85.5 g of methanol is present in 100 g of the solution.

Thus, mass of water = 100 g - 85.5 g = 14.5 g

The formula for the calculation of moles is shown below:

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

Mass of methanol = 85.5 g

Molar mass of methanol = 32.04 g/mol

Moles of methanol = 85.5 g / 32.04 g/mol = 2.6685 moles

Mass of water = 14.5 g

Molar mass of methanol = 18 g/mol

Moles of methanol = 14.5 g / 18 g/mol = 0.8056 moles

So, according to definition of mole fraction:

[tex]Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}[/tex]

Applying values as:

[tex]Mole\ fraction\ of\ methanol=\frac {2.6685}{2.6685+0.8056}[/tex]

Mole fraction of methanol in the solution = 0.7681

Answer:

% of mass of electrons in C = 0.0272

Explanation:

Atomic no. of carbon = 6

So, no. of electron = 6

Mass of an electron = [tex]\frac{1}{1836} amu[/tex]

Mass of 6 electrons = [tex]6 \times \frac{1}{1836} = 0.003268 amu[/tex]

Mass of C = 12.0107 u

% of mass of electrons in C = [tex]\frac{Mass\ of\ all\ electrons}{Mass\ of\ C}\times 100[/tex]

% of mass of electrons in C = [tex]\frac{0.003268}{12.0107}\times 100=0.0272 \%[/tex]

Answer: The mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of water = 4.05 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{4.05g}{18g/mol}=0.225mol[/tex]

For the given chemical reaction:

[tex]2H_2O\rightarrow 2H_2+O_2[/tex]

By Stoichiometry of the reaction:

2 moles of water is producing 2 moles of hydrogen gas

So, 0.225 moles of water will produce = [tex]\frac{2}{2}\times 0.225=0.225mol[/tex] of hydrogen gas.

Now, calculating the mass of hydrogen gas by using equation 1, we get:

Moles of hydrogen gas = 0.225 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

[tex]0.225mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=0.45g[/tex]

By Stoichiometry of the reaction:

2 moles of water is producing 1 mole of nitrogen gas

So, 0.225 moles of water will produce = [tex]\frac{1}{2}\times 0.225=0.1125mol[/tex] of nitrogen gas.

Now, calculating the mass of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.1125 mol

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]0.1125mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=3.15g[/tex]

Hence, the mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Answer:

a) n-butane has a higher boiling point

b) 1-butanol has a higher boiling

Explanation:

Given the molecules, propane (C3H8) and n-butane (C4H10), n-butane has a higher boiling point mainly due to greater molar mass and longer chain (more interactions between each molecule).

Given the molecules, diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH), 1-butanol has a higher boiling point mainly due to hydrogen bonding forces. The OH group is more electronegative than ether group.

The larger and more complex n-butane has a higher boiling point due to London dispersion forces, while 1-butanol has a higher boiling point due to hydrogen bonding.

Given the molecules, propane (C3H8) and n-butane (C4H10), n-butane has a higher boiling point mainly due to its larger size and increased London dispersion forces, a type of intermolecular force.

London dispersion forces increase with increased molecular size and shape because larger and more complex molecules tend to have more electrons, leading to larger temporary fluctuations in charge distribution.

Given the molecules diethyl ether (CH3CH2OCH2CH3) and 1-butanol (CH3CH2CH2CH2OH), 1-butanol has the higher boiling point mainly due to the presence of hydrogen bonding, another type of intermolecular force. Hydrogen bonding is generally stronger than other types of intermolecular forces, leading to higher boiling points.

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Answer:

The least amount of energy emitted in this case is 0.6 eV.

The corresponding quantum number n would be n=4.

The wavelenght asociated to the emitted photon would be 2.06 [tex]\mu[/tex]m, corresponding to the Infrared spectrum.

Explanation:

For calculating the energy of an electron emitted/absorbed in an electronic transition of the hydrogen atom, the next equation from the Bohr model can be used:

[tex]E=E_{0} Z^2 [\frac{1}{n_1^2}-\frac{1}{n_2^2}][/tex]

, where E is the photon energy, [tex]E_0[/tex] is the energy of the first energy level (-13.6 eV), Z is the atomic number, [tex]n_1[/tex] is the quantum number n of the starting level and [tex]n_2[/tex] the quantum number n of the finishing level. In this case, [tex]n_2=3[/tex], and [tex]n_1=4[/tex], because this excited level is the next in energy to n=3.

Considering that [tex]1 eV= 1.60217662x10^{-19} J[/tex], and using the Planck equation [tex]E=h\nu=\frac{hc}{\lambda}[/tex], you can calculate the wavelenght or the frequency associated to that photon. Values in the order of [tex]\mu[/tex]m in wavelenght belong to the Infrared spectrum, wich can not being seen by humans.

Answer:

1. [tex]E=1.059x10^{-19}J[/tex]

2. [tex]n=4[/tex]

3. The associated wavelength belongs to the infrared spectrum which is invisible for humans.

Explanation:

Hello,

1. At first, according to the equation:

[tex]E=E_oZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]

Whereas [tex]E[/tex] is the emitted energy, [tex]E_o[/tex] the first level energy, [tex]Z[/tex] the atomic number, [tex]n_1[/tex] the first level and [tex]n_2[/tex] the second level. In such a way, the closest the [tex]n[/tex]'s are, the least the amount of emitted energy, therefore, [tex]n_1=3[/tex] (based on the statement) and [tex]n_2=4[/tex], thus, the least amount of energy turns out being:

[tex]E=(2.179x10^{-18}J)(1^2)(\frac{1}{3^2}-\frac{1}{4^2})\\E=1.059x10^{-19}J[/tex]

2. Secondly, and based on the first question, the quantum number for the excited state is mandatorily [tex]n=4[/tex]

3. Finally, to substantiate why we cannot observe the emitted photons we apply the following equation:

[tex]E=\frac{hc}{\lambda} \\\lambda=\frac{hc}{E}=\frac{(6.62607004x10^{-34} m^2 kg / s)(299 792 458m/s)}{1.059x10^{-19}m^2 kg / s^2}  \\\lambda=1.88x10^{-6}m[/tex]

Whereas the obtained wavelength corresponds to the infrared spectrum which is not observable by humans.

Best regards.

Answer:

The final dilution is 1:400

Explanation:

Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.

So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400

This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.

Answer : The rate law is 4.232 mol/L.s

Explanation :

First we have to calculate the value of rate constant.

According to the question, the expression for rate law will be:

[tex]Rate=k[A]^2[B][/tex]

where,

k = rate constant

As we are given :

Rate law = 0.23 mol/L.s

Initial concentration of A = 1.0 mol/L

Initial concentration of B = 1.0 mol/L

Now put all the given values in the above rate law expression, we get:

[tex]0.23mol/L.s=k\times (1.0mol/L)^2\times (1.0mol/L)[/tex]

[tex]k=0.23M^{-2}s^{-1}[/tex]

Now we have to calculate the rate law when initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

[tex]Rate=k[A]^2[B][/tex]

where,

k = rate constant = [tex]0.23M^{-2}s^{-1}[/tex]

Rate law = ?

Initial concentration of A = 2.0 mol/L

Initial concentration of B = 4.6 mol/L

Now put all the given values in the above rate law expression, we get:

[tex]Rate=(0.23M^{-2}s^{-1})\times (2.0mol/L)^2\times (4.6mol/L)[/tex]

[tex]Rate=4.232mol/L.s[/tex]

Therefore, the rate law is 4.232 mol/L.s

The study of chemicals and bonds is called chemistry. When the amount of reactant and product are equal then is said to be an equilibrium state.

The correct answer is 4.232.

According to the question, the expression for rate law will be:

[tex]Rate =k[A]^2[B][/tex]

where,

The data is given as follows:-

Place all the values in the equation and solve it

[tex]0.23= k*1^2*1\\\\k= 0.23[/tex]

Now we have to calculate the rate law when the initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

[tex]Rate =k[A]^2[B][/tex]

where,

[tex]Rate = 0.23*2^2*4.6[/tex]

Hence, the rate will be 4.232.

For more information about the rate of the equation, refer to the link:-

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The energy of the photons emitted by the AM and FM radio stations can be calculated using Planck's equation (E = hf). The calculated energies show that the FM radio station's photons have higher energy than the AM radio station's photons.

To calculate the energy of the photons emitted by radio stations, we use the formula:

E = hf, where E is the energy, h is Planck's constant (6.62607015 × 10⁻³⁴ Js) and f is the frequency.

(For the sake of this calculation, we are going to convert the frequency from kHz and MHz to Hz.)

Comparing these energy values, it's clear that the FM radio station's emitted photons have a higher energy than the AM radio station's photons, due to its higher frequency.

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Answer:

Dispersion system is a system in which certain particles are scattered in a continuous liquid or solid medium. The two phases present in this system are the dispersed particles and dispersion medium. These phases may or may not be present in the same state.

In a dispersion system, the particles that are dispersed are known as the dispersed particles and the medium in which the particles are dispersed is known as the dispersion medium.