A double pipe heat exchanger is to be designed to heat 1 kg/s of a cold fluid from 30°C to 60°C using 2 kg/s of a hot fluid at 100°C. The two streams have equal specific heat capacities and overall heat transfer coefficient. Calculate the ratio of heat transfer areas of counter current to co-current. a) 1.142 b) 0.875 c) 0.927 d) 1.077

Answer:

The statement is true

Explanation:

Any physical quantity has a unique dimensional representation and when we change the dimensions of any physical quantity it looses it's original form. while as unit conversion is applicable to physical quantities of same nature as speed can be converted from kilometers per hour to miles per hour or meters per second, changing the units of any physical quantity into different unit's but of same dimensional formula only changes a magnitude of the physical quantity as in the above case the speed will be speed no matter whatever be the units.

since [tex]\frac{m^3}{sec}[/tex] are not the units of velocity but of volumetric rate of flow  we cannot convert this quantity into velocity by only unit conversion

Answer:

[CH₃COO⁻]  [H⁺] pH

0,1 M  0,0025 M  6,30

0,1 M  0,005 M  6,02

0,1 M  0,01 M  5,70

0,1 M  0,05 M  4,74

0,01 M  0,0025 M  5,22

0,01 M  0,005 M  4,75

0,01 M  0,01 M  3,38

0,01 M  0,05 M  1,40

Explanation:

The equilibrium of sodium acetate is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵

Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.

For  [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:

[CH₃COO⁻] = 0,1 M - x

[H⁺] = 0,0025 M - x

[CH₃COOH] = x

The expression for this equilibrium is:

Ka = [tex]\frac{[CH3COO^-] [H^+] }{[CH3COOH]}[/tex]

Replacing:

1,8x10⁻⁵ = [tex]\frac{[0,1-x] [0,0025-x] }{[x]}[/tex]

Thus:

0 = x²-0,102518x +2,5x10⁻⁴

Solving:

x = 0,100 ⇒ No physical sense

x = 0,0024995

Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷

pH = - log [H⁺] = 6,30

Following the same procedure changing both  [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:

[CH₃COO⁻]  [H⁺] pH

0,1 M  0,0025 M  6,30

0,1 M  0,005 M  6,02

0,1 M  0,01 M  5,70

0,1 M  0,05 M  4,74

0,01 M  0,0025 M  5,22

0,01 M  0,005 M  4,75

0,01 M  0,01 M  3,38

0,01 M  0,05 M  1,40

I hope it helps!

The pH values for the [tex]0.1 \ M[/tex] sodium acetate solution when [tex]0.0025 \ M[/tex], [tex]0.005\ M, 0.01\ M,[/tex] and [tex]0.05\ M[/tex] [tex]HCl[/tex] are added are approximately [tex]6.35, 6.03, 5.70,[/tex] and [tex]4.75[/tex] respectively. For the 0.01 \ M sodium acetate solution, the pH values are approximately [tex]5.23, 4.75, 2,[/tex] and [tex]1.30[/tex] respectively.

The Henderson-Hasselbalch equation is given by:

[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \][/tex]

For acetic acid ([tex]CH_3COOH[/tex]) and its conjugate base acetate ([tex]CH_3COO^-[/tex]), the [tex]pKa[/tex] is approximately [tex]4.75[/tex] at [tex]25^\circ C[/tex]

Let's calculate the pH for each case:

1. For the [tex]0.1 \ M[/tex] sodium acetate solution:

a. When [tex]0.0025 \ M \ HCl[/tex] is added:

The concentration of acetate ions remains [tex]0.1 \ M[/tex], while the concentration of acetic acid formed by the reaction of [tex]HCl[/tex] with acetate ions is [tex]0.0025 \ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.1}{0.0025} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (40) \][/tex]

[tex]\[ \text{pH} = 4.75 + 1.60 \][/tex]

[tex]\[ \text{pH} = 6.35 \][/tex]

b. When [tex]0.005\ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.095 \ M[/tex], and the concentration of acetic acid is [tex]0.005\ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.095}{0.005} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (19) \][/tex]

[tex]\[ \text{pH} = 4.75 + 1.28 \][/tex]

[tex]\[ \text{pH} = 6.03 \][/tex]

c. When [tex]0.01 \ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.09 \ M[/tex], and the concentration of acetic acid is[tex]0.01\ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.09}{0.01} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (9) \][/tex]

[tex]\[ \text{pH} = 4.75 + 0.95 \][/tex]

[tex]\[ \text{pH} = 5.70 \][/tex]

d. When [tex]0.05 \ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.05 \ M[/tex], and the concentration of acetic acid is [tex]0.05 \ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.05}{0.05} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (1) \][/tex]

[tex]\[ \text{pH} = 4.75 \][/tex]

2. For the [tex]0.01 \ M[/tex] sodium acetate solution:

a. When [tex]0.0025\ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.0075 \ M[/tex], and the concentration of acetic acid is [tex]0.0025 \ M[/tex].

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.0075}{0.0025} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (3) \][/tex]

[tex]\[ \text{pH} = 4.75 + 0.48 \][/tex]

[tex]\[ \text{pH} = 5.23 \][/tex]

b. When [tex]0.005 \ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.005 \ M[/tex], and the concentration of acetic acid is[tex]0.005 \ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.005}{0.005} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (1) \][/tex]

[tex]\[ \text{pH} = 4.75 \][/tex]

c. When [tex]0.01 \ M\ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0 M[/tex] (fully reacted), and the concentration of acetic acid is [tex]0.01 \ M[/tex]. The solution is no longer buffered, and the pH is that of a [tex]0.01 \ M[/tex] acetic acid solution.

[tex]\[ \text{pH} = -\log (0.01) \][/tex]

[tex]\[ \text{pH} = 2 \][/tex]

d. When [tex]0.05 \ M \ HCl[/tex] is added:

The solution is overwhelmed by the strong acid, and the pH is determined by the excess [tex]HCl[/tex]. The pH is approximately that of a [tex]0.05 M HCl[/tex] solution.

[tex]\[ \text{pH} = -\log (0.05) \][/tex]

[tex]\[ \text{pH} = 1.30 \][/tex]

Answer: The de-Broglie's wavelength of a hydrogen molecule is [tex]3.26\AA[/tex]

Explanation:

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

[tex]E=\frac{3}{2}kT[/tex]

where,

E = kinetic energy of the particles  = ?

k = Boltzmann constant  = [tex]1.38\times 10^{-23}J/K[/tex]

T = temperature of the particle = 30 K

Putting values in above equation, we get:

[tex]E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J[/tex]

Conversion factor used:  1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or [tex]2\times 10^{-3}kg[/tex]  

According to mole concept:

[tex]6.022\times 10^{23}[/tex] number of molecules occupy 1 mole of a gas.

As, [tex]6.022\times 10^{23}[/tex] number of hydrogen molecules has a mass of [tex]2\times 10^{-3}kg[/tex]

So, 1 molecule of hydrogen will have a mass of = [tex]\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg[/tex]

[tex]\lambda=\frac{h}{\sqrt{2mE_k}}[/tex]

where,

[tex]\lambda[/tex] = De-Broglie's wavelength = ?

h = Planck's constant = [tex]6.624\times 10^{-34}Js[/tex]

m = mass of 1 hydrogen molecule = [tex]3.32\times 10^{-27}kg[/tex]

[tex]E_k[/tex] = kinetic energy of the particle = [tex]6.21\times 10^{-22}J[/tex]

Putting values in above equation, we get:

[tex]\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}[/tex]

[tex]\lambda=3.26\times 10^{-10}m=3.26\AA[/tex]    (Conversion factor: [tex]1\AA=10^{-10}m[/tex] )

Hence, the de-Broglie's wavelength of a hydrogen molecule is [tex]3.26\AA[/tex]

Answer:

A 71 kg person will get a BAC of 0.05 when drinking 0.3442 L of beer

Explanation:

71 kg * 8/100 = 5.68 kg of blood.

5.68 kg * [tex]\frac{1m^{3}}{1025kg}[/tex] = 0.0055415 m³

We convert m³ into mL, keeping the unit that we want to convert to in the numerator; and the unit that we want to convert in the denominator:

[tex]0.0055415m^{3}*\frac{1000L}{1m^{3}} *\frac{1000mL}{1L}=5541.5mL[/tex]

[tex]\frac{5541.5mL}{100mL}*0.05g[/tex] = 2.77 g of alcohol are needed in the bloodstream in order to have a BAC of 0.05

2.77 g [tex]*\frac{100}{17} =[/tex] 16.29 g of alcohol need to be ingested

[tex]789\frac{kg}{m^{3}}*\frac{1000g}{1kg} *\frac{1m^{3}}{1000L}  =789g/L[/tex]

[tex]16.29g*\frac{1L}{789g}=[/tex] 0.02065 L of alcohol are needed.

[tex]0.02065L_{alcohol}*\frac{100L_{beer}}{6L_{alcohol}} =[/tex]0.3442 L of beer are needed.

To have a Blood Alcohol Concentration (BAC) of 0.05, a 71 kg person would need to consume approximately 271.57 liters of beer with a 6% alcohol content, assuming that 17% of the alcohol ends up in the bloodstream.

First, let's calculate the total mass of blood in the body. The total weight of the body is 71 kg and blood comprises about 8 percent of body weight. Therefore, the mass of blood = 0.08 * 71 = 5.68 kg. Because the blood density is 1025 kg/m³, this implies that the volume of the blood is 5.68/1025 = 0.00554 m³, or 5.54 liters. When the BAC is 0.05, this means that there are 0.05 grams of alcohol in every 100 ml of blood, or 0.05 grams of alcohol per 0.1 liter of blood. Therefore, to have a BAC of 0.05 in 5.54 liters, there needs to be 0.05 * 5.54 * 10 = 2.77 grams of alcohol in the blood. We know that 17 percent of the alcohol that a person drinks goes into their bloodstream. Therefore, the total mass of beer consumed to get this amount of alcohol is = 2.77 / 0.17 = 16.294 grams of pure alcohol. Beer has an alcohol equivalent to 6 % of its volume, so we find that the total volume of beer consumed = 16.294 / 0.06 = 271.57 liters. Because alcohol density is 789 kg/m³, this would be approximately 0.271 m^3 or 271.57 liters of beer.

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Answer:

310.69K

Explanation:

Given parameters:

Initial temperature T₁ = 292K

Initial pressure P₁ = 1.25atm

Final pressure P₂ = 1.33atm

Unknown:

Final temperature T₂ = ?

Solution

To find the unkown, we need to apply the combined gas law. From the combined gas law, it can be deduced that at constant volume, the pressure of a give mass or mole of gas varies directly with the absolute temperature.

Since the same aerosol can is heated, the volume is constant.

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

Now, we have to make T₂ the subject of the formula:

      T₂ =  [tex]\frac{P_{2}  x T_{1}  }{P_{1} }[/tex]

       T₂ =  [tex]\frac{1.33  x 292  }{1.25 }[/tex] =  310.69K

The resulting temperature in the can is approximately 311 K.

To solve this problem, we can use Gay-Lussac's Law, which states that the pressure of a given mass of gas is directly proportional to its absolute temperature (in Kelvin) when the volume is kept constant. Mathematically, this can be expressed as:

[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]

Rearranging the equation to solve for [tex]\( T_2 \),[/tex] we get:

[tex]\[ T_2 = T_1 \times \frac{P_2}{P_1} \][/tex]

Substituting the given values:

[tex]\[ T_2 = 292 \text{ K} \times \frac{1.33 \text{ atm}}{1.25 \text{ atm}} \] \[ T_2 = 292 \text{ K} \times 1.064 \] \[ T_2 \approx 310.608 \text{ K} \][/tex]

Rounding to the nearest whole number, since temperature in Kelvin is typically expressed without decimals:

[tex]\[ T_2 \approx 311 \text{ K} \][/tex]

Explanation:

It is given that peak (maximum) emission of star is 300 nm.

And, it is known that according to Wein's displacement relation between wavelength and temperature is as follows.

                    [tex]\lambda_{max} \times T[/tex] = 2898 micrometer.K

Hence, putting the given values into the above formula as follows.

               [tex]\lambda_{max} \times T[/tex] = 2898 micrometer.K

               [tex]300 \times 10^{-9} m \times T[/tex] = 2898 \times 10^{-6} meter.K

                 T = [tex]\frac{2898000}{300}[/tex] K

                     = 9660 K

Thus, we can conclude that the estimated surface temperature of star is 9660 K.

Using Wien's Law, the estimated surface temperature of a star with a peak emission at 300 nm is approximately 9660 Kelvin, considering a 2% precision.

To estimate the surface temperature of a star based on its peak emission wavelength, we use Wien's Law, which is a principle in physics that describes the relationship between the peak wavelength of emission from a blackbody and its temperature. Wien's Law can be written as:

λmax * T = b

where λmax is the peak wavelength in meters, T is the temperature in Kelvin, and b is Wien's displacement constant, approximately 2.897 × 10-3 m·K.

To find the temperature, we rearrange the formula to solve for T:

T = b / λmax

Converting the peak wavelength from nanometers to meters (300 nm = 3 × 10-7 m), we can calculate the temperature of the star as follows:

T = 2.897 × 10-3 m·K / 3 × 10-7 m
= 9660 Kelvin

Allowing for a 2% precision, the estimated temperature range of the star is approximately 9660K ± 2%.

Answer:

Osmotic pressure is: 2,01 atm

Explanation:

Osmotic pressure is the minimum pressure that you needs to be apply to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. The formula of osmotic pressure  is:

π = M×R×T

Where:

M is molar concentration of dissolved species (units of mol/L). 0,078M

R is ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹, ).

T is the temperature on the Kelvin scale. 41°C +273,15 = 314,15 K

Replacing you have:

π = 0,078M×0.08206 L atm mol⁻¹ K⁻¹×314,15 K

π = 2,01 atm

I hope it helps!

Explanation:

As it is given that W = [tex]m \times g[/tex] = 28.9 lbf

Mass is given as 30.0 lbm

Now, calculate [tex]g_{c}[/tex] as follows.

                [tex]g_{c} = \frac{weight given}{mass}[/tex]

                           = [tex]\frac{28.9 lbf}{30.0 lbm}[/tex]

                           = 0.9633 lbf/lbm

Since, it is known that 1 lbf = lbm \times [tex]32.2 ft/s^{2}[/tex]. Therefore, local gravity (g') will be calculated as follows.

                        [tex]0.9633 \times 32.2 ft/s^{2}[/tex]

                           = [tex]31.02 ft/s^{2}[/tex]    

Thus, we can conclude that the value of local gravity is [tex]31.02 ft/s^{2}[/tex].

Answer: The moles of hydrochloric acid is [tex]1.0346\times 10^{-4}\mu mol[/tex]

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Or,

[tex]\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}[/tex]

We are given:

Molarity of solution = 0.5173 M

Volume of solution = [tex]200\mu L[/tex]

Putting values in above equation, we get:

[tex]0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol[/tex]

Hence, the moles of hydrochloric acid is [tex]1.0346\times 10^{-4}\mu mol[/tex]

Spontaneous reactions result in an increase in entropy, aligning with the second law of thermodynamics.

The true statement about the relationship between entropy and spontaneity is: Spontaneous reactions tend to lead to higher entropy. This concept is encapsulated by the second law of thermodynamics, which states that all spontaneous changes cause an increase in the entropy of the universe. The increase in entropy relates to how energy and matter become more spread out and disordered over time in a spontaneous process, such as a chemical reaction or heat flow between two objects.

Answer:

The correct answer is that letter.  (μ)

Explanation:

Hello!

According to the international system of units, we use prefixes to indicate the number of places we ran the point.

The letter  μ indicates [tex]10^{-6}[/tex]

The correct answer is that letter.  (μ)

It is not in the options but it is correct.

Answer:

Explanation:

Hello,

Polonium has the atomic number 83, thus, it has 83 protons and electrons; in such a way, the amount of neutrons is given by the subtraction:

[tex]neutrons=atomic.mass - atomic.number\\neutrons=208-83\\neutrons=125[/tex]

Now, the addition between the neutrons and electrons is the same atomic mass considering the previous formula, thus:

[tex]electrons+neutrons=83+125=208[/tex]

So this procedure could be avoided by just looking at the given information "208Po" which contains the result.

Best regards.

To calculate the sum of neutrons and electrons in the 208Po ion, presume it is a neutral atom due to the lack of charge information, which would result in 124 neutrons and 84 electrons. The sum of these particles is then 208.

The question asks for the sum of the number of neutrons and electrons in the ion 208Po. To find the number of neutrons, we subtract the atomic number (number of protons) from the mass number. Polonium (Po) has an atomic number of 84, which is also the number of protons in a neutral atom of Polonium. The mass number of this isotope is 208, so the number of neutrons is 208 - 84 = 124. As for the number of electrons, since it is an ion and not neutral, we need to know the charge to determine this. However, as the charge is not provided in the question, we will assume that 208Po represents a neutral atom. Therefore, the number of electrons would also be 84. The sum of neutrons and electrons in the ion 208Po assuming it is uncharged is 124 + 84 = 208.

Answer:

The molar composition of the equilibrium mixture is

NO: 0.338 = 33.8%

Br2: 0.169 = 16.9%

NOBr: 0.493 = 49.3%

Explanation:

The reaction of nitrosyl bromide formation can be written as

[tex]NO+0.5\,Br_2\rightarrow NOBr[/tex]

To form 1 mol of NOBr, we need 1 mol of NO and 0.5 mol of Br2.

A sample of 0.0524 mol NO with 0.0262 mol Br2 gives an equilibrium mixture containing 0.0311 mol NOBr.

Then, to form 0.0311 mol NOBr, were needed 0.0311 mol NO and 0.01555 mol of Br2.

The amount of NO that stay in the same form is (0.0524-0.0311)=0.0213 mol NO.

The amount of Br2 that stay in the same form is (0.0262-0.01555)=0.01065 mol Br2.

The total amount of mol is (0.0213 mol NO + 0.01065 mol Br2 + 0.0311 mol NOBr) = 0.06305.

The molar composition is

NO: 0.0213/0.06305 = 0.338 = 33.8%

Br2: 0.01065/0.06305 = 0.169 = 16.9%

NOBr: 0.0311/0.06305 = 0.493 = 49.3%

Answer:

The increasing order of Bronsted acids is:

(b) < (c) < (e) < (d) < (a)

Explanation:

A Bronsted acid is a substance who donated protons.

[tex]AH + H_{2}O[/tex] ⇄ [tex]A^{-} + H_{3}O^{+}[/tex]

The acidity of a compound is determined by the acidity constant and every factor that stabilizes the product of the acid reaction (the anion) will contribute to increase the acidity of the compound.

First, it is necesary to identify the donor group:

(a) is -SO3H

(b) is -OH

(c), (d) and (e) is -COOH

For these groups the order of acidity is

-SO3H > -COOH > -OH

The strongest is the group -SO3H because it comes from a strong inorganic acid (H2SO4). Second is the group -COOH, carboxylic acids are the strongest organic acids because of the resonance structure of the anion at acid equilibrium, this equilibrium shifts the reaction to products favoring deprotonation. Third is the group of -OH because alcohols are weak acids.

Further, between (c), (d) and (e) it is necessary to analyze the molecule substituents. The -Cl substituent is an electronegative group that stabilizes the anion of the equilibrium because of its inductive effect. Thus, the molecule with the -Cl near to de acidic group will be more acidic. Therefore the order of acidity between these three compounds will be: (d) > (e) > (c)

Summarised all, the final answer is (b) < (c) < (e) < (d) < (a)

In increasing order of acidity, the compounds are b. CH3CH OH, a. CH3CH2SO3H, c. CH3CH2COOH, d. CH3CHCICOOH, e. CICH CH2COOH.

The acidity of a compound is determined by the stability of its conjugate base. The more stable the conjugate base, the stronger the acid. In this case, we can rank the compounds in increasing acidity as follows:

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Answer:

One urea molecule can make hydrogen bonding with 8 water molecules

Explanation:

Urea can form up to 4 hydrogen bonds with water molecules.

The maximum theoretical number of water molecules with which one urea molecule can hydrogen bond is 4.

This is because urea has 4 hydrogen bond donor sites (NH groups) and 2 hydrogen bond acceptor sites (carbonyl oxygen).

Each hydrogen bond requires one hydrogen bond donor and one hydrogen bond acceptor, so one urea molecule can form up to 4 hydrogen bonds with water molecules.

Answer:

a) 40,75 atm

b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

                                     PV = nRT

where:

Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.

The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the intermolecular forces and the space occupied by the gas

[tex]\frac{Pv}{RT} = 1 + \frac{B}{v}[/tex]

where:

a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

[tex]\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}[/tex]

and v = 3 L/5 moles = 0,6 L/mol

[tex]P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm[/tex]

Explanation:

The given data is as follows.

          [tex]\lambda[/tex] = 211 nm,          [tex]\sum = 6.17 \times 10^{3} mol/L/cm[/tex]

           l = 1 cm,             7% < Transmittance < 85%

Suppose the aqueous solution follows Lambert-Beer's law. Therefore,

                   Absorbance = [tex]-log \frac{\text{Percentage transmittance}}{100}[/tex]

Hence, for 7% transmittance the value of absorbance will be as follows.

                  Absorbance = [tex]-log \frac{7}{100}[/tex]

                           [tex]A_{1}[/tex] = 1.155

For 85% transmittance the value of absorbance will be as follows.

                 Absorbance = [tex]-log \frac{85}{100}[/tex]

                           [tex]A_{2}[/tex] = 0.07058

According to Lambert-Beer's law.

                  A = [tex]\sum \times l \times C[/tex]

where,       A = absorbance

                 [tex]\sum[/tex] = molar extinction coefficient

                 C = concentration

Therefore, concentration for 7% absorbance is as follows.

                    [tex]A_{1} = \sum \times l \times C_{1}[/tex]

                  [tex]C_{1}[/tex] = [tex]\frac{1.155}{6.17 \times 10^{3} \times 1}[/tex]

                                  = [tex]0.187 \times 10^{-3} mol/L[/tex]

                                  = 0.187 mmol/L

Concentration for 85% absorbance is as follows.

                   [tex]A_{2} = \sum \times l \times C_{2}[/tex]

                  [tex]C_{2}[/tex] = [tex]\frac{0.07058}{6.17 \times 10^{3} \times 1}[/tex]

                                  = [tex]0.01144 \times 10^{-3} mol/L[/tex]

                                  = 0.01144 mmol/L

Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.

The linear range of phenol concentrations for the given conditions is from [tex]\( 1.14 \times 10^{-5} \) M to \( 1.87 \times 10^{-4} \) M[/tex].

To calculate the linear range of phenol concentrations, we need to use the Beer-Lambert law, which relates the absorbance (A) of a solution to its concentration (c), the molar absorptivity (µ), and the path length (l) of the cell:

[tex]\[ A = \varepsilon \cdot c \cdot l \][/tex]

The transmittance (T) is related to absorbance by the equation:

[tex]\[ T = 10^{-A} \][/tex]

Given that the molar absorptivity (µ) is [tex]\( 6.17 \times 10^3 \)[/tex] L/mol/cm and the path length (l) is 1 cm, we can rearrange the Beer-Lambert law to solve for concentration (c):

[tex]\[ c = \frac{A}{\varepsilon \cdot l} \][/tex]

First, we need to find the absorbance values corresponding to 85% and 7% transmittance:

For 85% transmittance:

[tex]\[ A = -\log(T) = -\log(0.85) \][/tex]

[tex]\[ A \approx 0.0706 \][/tex]

For 7% transmittance:

[tex]\[ A = -\log(T) = -\log(0.07) \][/tex]

[tex]\[ A \approx 1.1549 \][/tex]

Now we can calculate the concentration range:

For the lower limit of transmittance (7%):

[tex]\[ c_{min} = \frac{A_{min}}{\varepsilon \cdot l} = \frac{1.1549}{6.17 \times 10^3 \cdot 1} \] \[ c_{min} \approx 1.87 \times 10^{-4} \text{ M} \][/tex]

For the upper limit of transmittance (85%):

[tex]\[ c_{max} = \frac{A_{max}}{\varepsilon \cdot l} = \frac{0.0706}{6.17 \times 10^3 \cdot 1} \] \[ c_{max} \approx 1.14 \times 10^{-5} \text{ M} \][/tex]

Therefore, the linear range of phenol concentrations for the given conditions is from [tex]\( 1.14 \times 10^{-5} \) M to \( 1.87 \times 10^{-4} \) M[/tex].

Answer:

In a volumetric flask of marking 500.0 mL add 138.75 grams of calcium chloride and add small amount of water to dissolve solute completely. After the solute gets completely soluble add more water up till the mark of 500 ml.

Explanation:

Concentration of calcium chloride = 2.5 M

Volume of the solution = 500.0 ml = 0.5000 L

Moles of calcium chloride = n

[tex]c=\frac{n}{V(L)}[/tex]

n = moles of solute

c = concentration of solution

V = volume of the solution in L

[tex]2.5 M=\frac{n}{0.5000 L}[/tex]

[tex]n=2.5 M\times 0.5000 L = 1.2500 mol[/tex]

[tex]n=\frac{\text{Mass of calcium chloride}}{\text{Molar mass of calcium chloride}}[/tex]

[tex]1.2500 mol=\frac{\text{Mass of calcium chloride}}{111 g/mol}[/tex]

Mass of calcium chloride = 111 g/mol × 1.2500 mol = 138.75 g

In a volumetric flask of marking 500.0 mL add 138.75 grams of calcium chloride and add small amount of water to dissolve solute completely. After the the solute gets completely soluble add more water up till the mark of 500 ml.

Answer : The concentration of [tex]NH_4^+[/tex] ion is [tex]0.15E^1M[/tex]

Explanation :

First we have to calculate the moles of [tex](NH_4)_3PO_4[/tex].

[tex]\text{Moles of }(NH_4)_3PO_4=\text{Concentration of }(NH_4)_3PO_4\times \text{Volume of solution}=0.50M\times 0.06L=0.03mole[/tex]

The balanced chemical reaction will be:

[tex](NH_4)_3PO_4\rightleftharpoons 3NH_4^++PO_4^{3-}[/tex]

From the reaction we conclude that,

1 mole of [tex](NH_4)_3PO_4[/tex] dissociate to give 3 moles of [tex]NH_4^+[/tex] ion and 1 mole of [tex]PO_4^{3-}[/tex] ion

So,

0.03 mole of [tex](NH_4)_3PO_4[/tex] dissociate to give [tex]3\times 0.03=0.09[/tex] moles of [tex]NH_4^+[/tex] ion and 0.03 mole of [tex]PO_4^{3-}[/tex] ion

Now we have to calculate the concentration of [tex]NH_4^+[/tex] ion.

[tex]\text{Concentration of }NH_4^+=\frac{\text{Moles of }NH_4^+}{\text{Total volume}}[/tex]

[tex]\text{Concentration of }NH_4^+=\frac{0.09mole}{0.06L}=1.5M=0.15E^1M[/tex]

Therefore, the concentration of [tex]NH_4^+[/tex] ion is [tex]0.15E^1M[/tex]

Answer:

2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.

Explanation:

We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.

Now we devise the following reasoning:

If          18 g of phthalic acid are dissolved in 100 mL of water at 99 °C

Then   0.5 g of phthalic acid are dissolved in X mL of water at 99 °C

X = (0.5 × 100) / 18 = 2.77 mL of water

Final answer:

To dissolve 500 mg of phthalic acid, a student could use a minimum of approximately 2.78 mL of boiling water, considering the solubility of 18 g per 100 mL at 99 ℃.

Explanation:

The question asks for the smallest volume of boiling water (at 99 ℃) needed to dissolve 500 mg of phthalic acid, with the solubility of phthalic acid provided as 0.54 g at 14 ℃ and 18 g at 99 ℃. To find this volume, we start by converting the mass of phthalic acid to be dissolved (500 mg or 0.5 g) to grams because the solubility is given in grams. Knowing the solubility of phthalic acid at 99 ℃ is 18 g per 100 mL, we calculate the volume required for 0.5 g as follows:

Solubility equation: (Volume of water) x (Solubility of compound in g/100mL) = Mass of compound to be dissolved

Thus, (Volume of water) x (18 g/100 mL) = 0.5 g

Rearranging for Volume of water gives: Volume of water = (0.5 g) / (18 g/100 mL) = 2.78 mL

Therefore, the student could use a minimum of approximately 2.78 mL of boiling water to dissolve 500 mg of phthalic acid.

Answer: A. -396 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

[tex]2SO_2(g)+O_2 (g)\rightarrow 2SO_3(g)[/tex]  [tex]\Delta H^0=198kJ[/tex]

Reversing the reaction, changes the sign of [tex]\Delta H[/tex]

[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2 (g)[/tex]  [tex]\Delta H^0=-198kJ[/tex]

On multiplying the reaction by 2, enthalpy gets multiplied by 2:

[tex]4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)[/tex]   [tex]\Delta H=2\times -198kJ=-396kJ[/tex]

Thus the enthalpy change for the reaction [tex]4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)[/tex]  is -396 kJ.

The change in enthalpy when 4.00 mol of sulfur trioxide decomposes into sulfur dioxide and oxygen gas is 792 kJ.

The question is asking for the change in enthalpy when 4.00 mol of sulfur trioxide decomposes into sulfur dioxide and oxygen gas. Looking at the provided data, the change in enthalpy (ΔHº) for the decomposition of one mole of sulfur trioxide is 198 kJ. Enthalpy is an extensive property, meaning it depends on the amount of substance involved. Therefore, if the reaction involves 4.00 mol of sulfur trioxide, the total change in enthalpy will be 4 times 198 kJ, which is 792 kJ. Hence, the correct answer is E. 792 kJ.

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Explanation:

The specie produced when a strong acid is dissociated in water along with the proton is the conjugate base of that acid.

Consider a strong acid such as HCl which donates the proton so readily as it is a strong base and there is no tendency for conjugate base, which is [tex]Cl^-[/tex] to re accept that proton.  

A strong base is the one which accepts proton and holds it firmly. Thus, the strong acid has weak conjugate base.

A strong acid must have a weak conjugate base because once the strong acid donates its proton, the remaining base does not readily recapture it. This inverse relationship ensures that the acid can fully dissociate. For instance, HCl, a strong acid, has a weak conjugate base [tex]Cl^-[/tex].

A strong acid is characterized by its ability to donate protons easily. This means that once the acid has donated a proton, the resulting conjugate base is left behind. If this conjugate base were strong, it would readily attract and recapture the proton, preventing the acid from fully dissociating.

For instance, when HCl (a strong acid with a small [tex]pK_a[/tex]  of -7) dissociates, it forms [tex]Cl^-[/tex] ions. The [tex]Cl^-[/tex] ions do not easily regain a proton, making them a weak base. Therefore, we can say that the conjugate base of a strong acid is invariably weak.

Understanding the Relationship with [tex]pK_a[/tex]

Acids with small or negative [tex]pK_a[/tex] values are strong acids. Their corresponding conjugate bases are weak bases because they do not have a strong tendency to grab protons back. This is why HCl, with a very low [tex]pK_a[/tex], has a conjugate base ([tex]Cl^-[/tex] ) that is very weak.

Acid-Base Equilibrium

An important concept in acid-base chemistry is that an acid-base reaction favors the formation of the weaker acid and base. This means in an equilibrium reaction, the side with the stronger acid and stronger base converts into the weaker acid and weaker base.

To summarize:

Answer:

The number of mol corresponding to 2,20 g Cl2 is 0,062.

Explanation:

As we know that the molar mass of chlorine is 35.45 g/mol and in this case we have 2.20 grams of gas, if we divide the mass of the chlorine over the molar mass we will have the moles of our compound.

moles = mass / molar mass

moles = 2,20 g / 35,45 g/mol

moles = 0,062

Answer:

0.153 atm or 116.1 torr

Explanation:

Hello, in the attached photo you will find the numerical procedure to obtain the pressure in the second state.

- Take into account that in the first part, the moles are needed based on the ideal gas equation, because they are subsequently used in the computation of the second state's pressure.

Best regards.

Using Boyle's Law, the new pressure of the gas when it's allowed to expand to a 13.0-L container is calculated to be 116.15 torr or 0.15 atm.

The subject of this question involves the concept of gas laws, specifically Boyle's Law which states that at a constant temperature, the pressure and volume of a gas are inversely related. The initial pressure of the Hydrogen Sulfide gas is 755.0 torr and its initial volume is 2.00L. When the volume increases to 13.0L, the pressure will reduce accordingly.

To calculate the new pressure, the formula of Boyle's Law which is P1V1 = P2V2 can be used, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. Thus, P2 becomes (P1V1/V2) = (755.0 torr * 2.00L) / 13.0L = 116.15 torr.

However, as the question specifies that the answer is needed in atm, the result should be converted from torr to atm. As 1 atm is approximately equal to 760 torr, the final converted pressure is 116.15 torr / 760 torr/atm = 0.15 atm.

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Answer:

0.068 g

Explanation:

The equation that is used to determine the fraction q remaining in water of volume V₁ after n extractions of volume V₂ is:

qⁿ = (V₁/(V₁ + KV₂))ⁿ

Substituting in the values gives the fraction of caffeine left in the water phase:

q³ = (4.0mL/(4.0mL + 4.6(2.0mL))³ = 0.0278

The fraction of caffeine extracted into the methylene chloride phase is:

1 - 0.0278 = 0.972

The amount of caffeine extracted into the methylene chloride is:

(0.070g)(0.972) = 0.068 g

Answer:

18840 lbf

Explanation:

Considering the physical equation:

[tex]F = PA[/tex]

It is possible to determine the force exerted by the piston if the maximum pressure developed ([tex]P_{max}[/tex]) and the area ([tex]A[/tex]) are previously known.

The area of the piston can be known through:

[tex]A = \pi *r^{2}[/tex]

Being [tex]r = \frac{d}{2} =  1~in[/tex]

Then

[tex]A = \pi * r^{2} = 3,14~in^{2}[/tex]

On the other hand, physical units of pressure in PSI are [tex]\frac{lbf}{in^{2}}[/tex], where the force is given in lbf.

Finally, the exerted force is:

[tex]F = 6000~PSI \cdot 3,14~in^{2} = 18840~lbf[/tex]

Answer:

Ethanol most easily forms hydrogen bonds.

Explanation:

The difference among the alcohols in this question is the size of carbonic chain and the position of the -OH group.

Ethanol has 2 carbons and the -OH group is terminal. The other alcohols have more carbons and the -OH group is not terminal. This means that the approximation of molecules will be facilitated for ethanol, and the interaction through hydrogen bons will be easier. However, for the other molecules, there will be steric hindrance, which will make it more difficult for the molecules to make hydrogen bonds.

The figure attached shows the alcohol structures.

Answer:

a) gas price = 2.96 Dollar US/gal;  we could wait until we return to Pembina.

b ) v = 100 Km/h = 62.1371 miles/h = 27.777 m/s

Explanation:

a) gas price = 103.6 cents Cnd / L * ( L / 0.264172 gal ) = 392.17 cents/gal

⇒ gas price = 392.17 cents Cnd/gal * ( dollar Cnd/100cents ) = 3.9217 dollar Cnd/gal

⇒ gas price = 3.9217 dollar Cnd/gal * ( 0.754 dollar US/dollar Cnd)

⇒ gas price = 2.96 dollars US/gal

the difference between the price of gas in Winnipeg and Pembina is 0.32 dollars US;  we could wait until we return to Pembina to fill the tank.

b) velocity = 100 Km/h

⇒ 100 Km/h * ( 0.621371 miles/Km ) = 62.1371 miles/h

⇒ 100 Km/h * ( 1000 m/Km ) * ( h/3600s) = 27.777 m/s

The cost of gasoline in US dollars per gallon in Winnipeg is $5.20 US per gallon, which is higher than the gas price in Pembina ($2.64 US per gallon). Therefore, it is advisable to wait until you get back to Pembina to fill up your gas tank. The speed limit in Winnipeg is 62.14 miles per hour or 27.78 meters per second.

To convert the cost of gasoline from Canadian dollars per liter to US dollars per gallon, we need to consider the exchange rate and volume conversion.

First, we need to convert Canadian cents to Canadian dollars. Since there are 100 cents in a dollar, the cost of gasoline in Canadian dollars is 103.6 cents / 100 = 1.036 Canadian dollars per liter.

To convert Canadian dollars per liter to US dollars per gallon, we need to consider the exchange rate. Since 1 Canadian dollar is equivalent to 0.754 US dollars, the cost of gasoline in US dollars per liter is 1.036 Canadian dollars / 0.754 US dollars = 1.373 US dollars per liter.

Since there are roughly 3.79 liters in 1 gallon, the cost of gasoline in US dollars per gallon is 1.373 US dollars per liter * 3.79 liters = 5.20 US dollars per gallon.

Comparing the cost of gasoline in Winnipeg (103.6 cents Cnd per liter) to Pembina ($2.64 US per gallon), it is cheaper to fill up in Winnipeg as the cost in US dollars per gallon is 5.20 US dollars per gallon, which is less than $2.64 US per gallon in Pembina.

To convert the speed limit from kilometers per hour to miles per hour, we can use the conversion factor 1 kilometer = 0.6214 miles. The speed limit in miles per hour is 100 km/hr * 0.6214 miles/km = 62.14 miles/hr.

To convert the speed limit from kilometers per hour to meters per second, we can use the conversion factor 1 kilometer = 1000 meters and 1 hour = 3600 seconds. The speed limit in meters per second is 100 km/hr * (1000 m/km) / (3600 s/hr) = 27.78 m/s.

The classifications of Substance A: Compound, Substance B: Compound, Substance C: Compound, Substance D: Compound, Substance E: Compound, Substance F: Element or Compound (depends on further information), Substance G: Element or Compound (depends on further information), Substance H: Element (most likely), Substance I: Element, Substance J: Element and Substance K: Pure Element or Compound (depends on further information)

This classification is based on the definitions of elements and compounds in chemistry.

To classify the pure substances as elements or compounds, we can follow these steps based on the information provided:  

Substance A: Contains two elements. Since it is a pure substance that can be separated into its components, it is classified as a compound.  

Substance B and C: These substances react to form a new substance D. Since they undergo a chemical change to give a new substance, both B and C are also compounds.  

Substance D: This is the new substance formed from the reaction of B and C and will also be classified as a compound because it is a product of the chemical reaction.  

Substance E: Decomposes upon heating to produce substances F and G, indicating it is made of more than one element. Therefore, it is classified as a compound.  

Substance F and G: These are the products resulting from the decomposition of substance E. Without additional information on whether they can be broken down into simpler substances, they are initially considered elements or compounds; however, since they are products of a compound's decomposition, if they consist of one type of atom, they could be elements.  

Substance H: Heating it to 1000 °C causes no change, indicating it might be a pure element or possibly a stable compound that does not decompose at this temperature. Without more information, it's uncertain, but it is likely to be an element.  

Substance I: Cannot be broken down into simpler substances by chemical means, which classifies it as a pure element.  

Substance J: Cannot be broken down into simpler substances by physical means, indicating it is a pure element.  

Substance K: Changes from a solid to a liquid upon heating to 500 °C. This phase change suggests that it is a pure substance, but without information, it could be either a pure element (e.g., a metal) or a compound (e.g., a salt).

The molecular size of succinic acid is larger than that of the decomposed product which is evident from the production of a gas and a solid with a different melting point upon heating.

In the process described where succinic acid is heated and gives off a gas, then leaves behind a solid that melts at a different temperature, suggests a chemical reaction has occurred, possibly a decomposition reaction. The succinic acid molecules are likely larger than the molecules of the white solid produced because decomposition breaks larger molecules into smaller ones. This change in melting point and the evolution of gas support that the molecular structure of the succinic acid has been altered, creating a substance with new properties, including a new melting point that typically involves smaller molecules compared to the original substance.