Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert these responses to a pH reading, the electrode must be calibrated against standard solutions of known H+ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4 · H2O; FW 138) and disodium hydrogen phosphate (Na2HPO4; FW 142) needed to prepare 1 L of a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M

Answer:

0.4429 m

Explanation:

Given that mass % of the ethanol in water = 2%

This means that 2 g of ethanol present in 100 g of ethanol solution.

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of [tex]CH_3CH_2OH[/tex] = 2 g

Molar mass of [tex]CH_3CH_2OH[/tex] = 46.07 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

[tex]Moles= \frac{2\ g}{46.07\ g/mol}[/tex]

[tex]Moles\ of\ CH_3CH_2OH= 0.0434\ moles[/tex]

Mass of water = 100 - 2  g = 98 g = 0.098 kg ( 1 g = 0.001 kg )

So, molality is:

[tex]m=\frac {0.0434\ moles}{0.098\ kg}[/tex]

Molality = 0.4429 m

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

        [tex]C_{p}_{liquid}[/tex] = 4.19 [tex]kJ/kg ^{o}C[/tex]

        [tex]C_{p}_{vaporization}[/tex] = 1.9 [tex]kJ/kg ^{o}C[/tex]

Heat of vaporization ([tex]\DeltaH^{o}_{vap}[/tex]) at 1 atm and [tex]100^{o}C[/tex] is 2259 kJ/kg

        [tex]H^{o}_{liquid}[/tex] = 0

Therefore, calculate the enthalpy of water vapor at 1 atm and [tex]100^{o}C[/tex] as follows.

            [tex]H^{o}_{vap}[/tex] = [tex]H^{o}_{liquid}[/tex] + [tex]\DeltaH^{o}_{vap}[/tex]        

                                   = 0 + 2259 kJ/kg

                                   = 2259 kJ/kg

As the desired temperature is given [tex]179.9^{o}C[/tex] and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.

             [tex]H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})[/tex]

                             = [tex]0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)[/tex]

                              = 334.781 kJ/kg

Hence, enthalpy of water vapor at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.

               [tex]H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})[/tex]

             [tex]H^{D}_{vap}[/tex] = [tex]2259 kJ/kg + 1.9 \times (179.9 - 100)[/tex]            

                              = 2410.81 kJ/kg

Therefore, calculate the latent heat of vaporization at 10 bar and [tex]179.9^{o}C[/tex] as follows.

       [tex]\Delta H^{D}_{vap}[/tex] = [tex]H^{D}_{vap} - H^{D}_{liq}[/tex]              

                         = 2410.81 kJ/kg - 334.781 kJ/kg

                         = 2076.029 kJ/kg

or,                      = 2076 kJ/kg

Thus, we can conclude that at 10 bar and [tex]179.9^{o}C[/tex] latent heat of vaporization is 2076 kJ/kg.

Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

Taking logarithm on both the sides, we get:

[tex]\ln k=-\frac{Ea}{RT}+\ln A[/tex]             ............(1)

where,

k = rate constant  = [tex]2.95\times 10^{-3}L/mol.s[/tex]

Ea = activation energy  = ?

T = temperature = 435 K

R = gas constant  = 8.314 J/K.mole

A = pre-exponential factor  = [tex]3.00\times 10^{+10}L/mol.s[/tex]

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

[tex]\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)[/tex]

[tex]Ea=108318.365J/mol=108.318kJ/mol[/tex]

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

Answer: [tex]1027nm[/tex]

Explanation:

Using Weins displacement law:

[tex]\lambda_{max}=\frac{b}{T}[/tex]

where [tex]\lambda_{max}[/tex] = wavelength

b = constant =[tex]2898\micro mK[/tex]

T = Temperature in Kelvin = 2822 K

Putting the values we get:

[tex]\lambda _{max}=\frac{2898\micro mK}{2822K}=1.027\micro m[/tex]

[tex]1\micro m=10^3nm[/tex]

[tex]1.027\micro m=\frac{10^3}{1}\times 1.027=1.027\times 10^3nm[/tex]

Thus wavelength of the spectral maximum in the emission of light by this star is [tex]1027nm[/tex]

Answer:

The answers is the option a: 46.062

Explanation:

First, you must determine the limiting reagent, that is the reagent that is consumed first during the reaction.

So, in first place, you must determine the molecular weight of the molecules CH4, O2, CO2 and H20. You know that

So, to determine the mass of a molecule, you must multiply the individual masses of each atom by the amount present in the molecule. This would be:

In the same way, you can determinate the mass of all reagents and products involved in the reaction.

Now you can apply stoichiometry to determine the limiting reagent  using these numbers and observing how many molecules react.

On one side, it is known that, by stoichiometry, 1 mol of CH4 and 2 moles of O2 react. This means that 16 kg/mol of CH4 and 64 kg/kmol ( 2moles* 32 kg/kmol) of O2 react

And it is known that there are 71 kg/hr of CH4 reacting with 67 kg/hr of 02

So, using the stoichiometric information (16 kg/mol of CH4 and 64 kg/kmol of O2), 71 kg/hr of CH4 and The Rule of Three, you can determine the limiting reagent:

16 kg/kmol CH4 ⇒ 64 kg/kmol O2 (stoichiometry)

71 kg/kmol CH4 ⇒ x

So [tex]x=\frac{71*64}{16}[/tex]

x=284 kg/kmol

This means that to react 71 kg/kmol of CH4, 284 kg/kmol of O2 are needed. But you only have 67 kg/mol that can react. That is why O2 is the limiting reagent, because it is consumed first.

Now, you can calculate the rate of CO2 that is generated, using the data of the amount of limiting reagent and stoichiometry. This is:

64 kg/kmol O2 ⇒ 44 kg/kmol CO2 (stoichiometry)

67 kg/kmol O2 ⇒ x

So [tex]x=\frac{67*44}{64}[/tex]

x=46.0625 kg/kmol

This means that 46.0625 kg/kmol of CO2 are generated.

The mass of Benadryl that should be given to a dog weighing 26.6 kg is indeed 65 milligrams.

When calculating the appropriate dosage of Benadryl for a dog weighing 29.5 kg, it's important to consider that the recommended dosage is given per pound. Since the dog's weight is provided in kilograms, it's necessary to convert it to pounds for accurate dosage determination. Converting 29.5 kg to pounds results in approximately 64.7 pounds.

Convert the weight from kilograms to pounds:

29.5 kg × 1 lb / 0.454 kg ≈ 64.7 lbs

Calculate the mass of Benadryl in milligrams using the dog's weight in pounds: (1 mg/lb) × 64.7 lbs ≈ 64.7 mg

Rounding this to 65 mg ensures practicality. This process accounts for the difference in units (kilograms to pounds) and utilizes the given dosage information to arrive at the correct amount of Benadryl needed to treat the dog's itchy skin.

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To determine the appropriate Benadryl dosage for a 26.6 kg dog, you first convert the dog's weight to pounds (26.6 kg = 58.6 lbs). You then multiply the weight in pounds by the recommended dosage, which leads to a recommended dosage of 58.6 mg.

In order to calculate the appropriate dosage of Benadryl for dogs, you first transform the dog's weight from kilograms to pounds, as the given dosage is in milligrams per pound. Given that 1 kilogram is roughly equal to 2.20462 pounds,  you can find the weight of a 26.6 kg dog in pounds as follows:

26.6 kg * 2.20462 lbs/kg = 58.6 lbs

Next, you multiply the weight of the dog in pounds by the recommended dosage:

58.6 lbs * 1 mg/lb = 58.6 mg

So, a dog that weighs 26.6 kg should be administered a 58.6mg dose of Benadryl.

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Answer:

1) Sv = 0.4584 m³/Kg...assuming steam as an ideal gas

% deviation from the values in the steam tables

⇒ % dev = 45 %

2) mass air = 272.617 Kg; assuming air to be an ideal gas

Explanation:

ideal gas:

PV = RTn

molar volume:

⇒ V/n = RT/P

∴ P = 90 psi * ( 0.06895 bar/psi ) = 6.2055 bar

∴ T = 650 F = 343.33 °C = 616.33 K

∴ R = 0.08314 bar.L/mol.K

⇒ V/n = (( 0.08314 )*(616.33 K )) / 6.2055 bar

⇒ V/n = 8.2574 L/mol * ( m³/1000L ) = 8.2574 E-3 m³/mol

specific volume ( Sv ):

∴ Mw = 18.01528 g/mol

⇒ Sv = 8.2574 E-3 m³/mol * ( mol / 18.01528 g ) * ( 1000 g/Kg )

⇒ Sv = 0.4584 m³/Kg

steam table:

∴ P = 6.2055 bar ≅ 6 bar → Sv = 0.3157 m³/Kg

⇒ % deviation = (( 0.4584 - 0.3157 ) / 0.3157) * 100

⇒ % dev = 45.2 %; significant value, assuming  steam to be a ideal gas

2) mass air, assuming ideal gas:

∴ V = 20ft * 35ft * 10ft = 7000ft³ * ( 28.3168 L/ft³ ) = 198217.6 L

∴ P = 17 psi * ( 0.06895 bar/psi ) = 1.172 bar

∴ T = 75 °F = 23.89 °C = 296.89 K

∴ R = 0.08314 bar.L/K.mol

⇒ n air = PV/RT = (( 1.172 )*( 198217.6 )) / (( 0.08314 )*( 296.89 ))

⇒ n air = 9411.616 mol air

∴ Mw air = 28.966 g/mol

⇒ mass air = 9411.616 mol * ( 28.966 g/mol ) = 272616.892 g = 272.617 Kg

Answer:

When you start to make this operations, you will find out that the correct answer is, NaCl 5.56 mM glucose, 16.7 mM.

Explanation:

First of all you should need to find, how many mols are in the first solutions you add: In glucose you have 0.100m, so as you know they are in 1000ml,  how many, in 25 ml? this is 2,5 *10^-3 moles. In NaCl, you should do the same, 1000 ml has 0.5 mols, so how many are, in 15ml?. The answer is 7.5 *10^-3. Now, that you have your mols you have to take account the water which is in 450 ml. So, let's go again, in 450ml you have 2,5 *10^-3 moles of glucose and 7.5 *10^-3 moles of NaCl, how many moles of them, are in 1000 ml. You will get that concentrations are 0,0167 M in NaCl and 5,56 *10^-3 M. Let's see that this numbers are in M, so if u want to get mM, just *1000.

The correct concentrations in the student's mixed solution are 5.10 mM glucose and 15.3 mM NaCl, calculated by using the dilution formula and considering the total volume of the final solution.

To calculate the concentrations of glucose and NaCl in the final solution after mixing 25.0 mL of 0.100 M glucose, 15.0 mL of 0.500 M NaCl, and 450.0 mL water, one would use the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the initial solutions, and M2 and V2 are the molarity and volume of the final solution respectively.

For glucose: moles of glucose = M1V1 = 0.100 mol/L × 0.025 L = 0.0025 mol.

For NaCl: moles of NaCl = M1V1 = 0.500 mol/L × 0.015 L = 0.0075 mol.

Total volume of the final solution = 25.0 mL + 15.0 mL + 450.0 mL = 490.0 mL = 0.490 L (to convert mL to L, divide by 1000).

Concentration of glucose in the final solution = moles of glucose / total volume = 0.0025 mol / 0.490 L = 5.10 mM (since 1 mM = 0.001 M).

Concentration of NaCl in the final solution = moles of NaCl / total volume = 0.0075 mol / 0.490 L = 15.3 mM.

Therefore, the correct concentrations in the solution are 5.10 mM glucose and 15.3 mM NaCl.

Answer:

The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.

Explanation:

Every weak acid or base is at equilibrium with its conjugate base or acid respectively when it is dissolved in water.

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

This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,

[tex]pH = pKa + Log \frac{[A^{-}]}{[HA]}[/tex]

shows the dependency between the pH of the solution, the pKa and the concentration of the species. If the pH decreases the concentration of protons will increase and the ratio between A- and AH will decrease. Instead, if the pH increases the concentration of protons will decreases and the ratio between A- and AH will increase.

Final answer:

The ratio [A-]/[HA] changes with changing pH according to the Henderson-Hasselbalch equation.

Explanation:

The Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), shows how the ratio [A-]/[HA] changes with changing pH. In this equation, [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the weak acid. As the pH increases, the concentration of [A-] increases relative to [HA], resulting in a higher value for the ratio [A-]/[HA]. Conversely, as the pH decreases, the concentration of [A-] decreases relative to [HA], leading to a lower value for the ratio [A-]/[HA].

Answer: Option (c) is the correct answer.

Explanation:

Boiling point is defined as the point at which vapor pressure of a liquid becomes equal to the atmospheric pressure.

Boiling point of water is [tex]100^{o}C[/tex].

Whereas when we heat one mole of a liquid at its boiling point without any change in temperature then the heat required to bring out change from liquid to vapor state is known as heat of vaporization.

Thus, we can conclude that the heat of vaporization of water is the amount of heat/energy required to convert 1 gram of liquid water at [tex]100^{o}C[/tex] to 1 gram of steam at [tex]100^{o}C[/tex].

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically,         Molarity = [tex]\frac{\text{no. of moles}}{\text{Volume in liter}}[/tex]

Also, when number of moles are equal in a solution then the formula will be as follows.

                     [tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]

It is given that [tex]M_{1}[/tex] is 8.00 M, [tex]V_{1}[/tex] is 7.00 mL, and [tex]M_{2}[/tex] is 0.80 M.

Hence, calculate the value of [tex]V_{2}[/tex] using above formula as follows.

                    [tex]M_{1} \times V_{1} = M_{2} \times V_{2}[/tex]

                 [tex]8.00 M \times 7.00 mL = 0.80 M \times V_{2}[/tex]

                      [tex]V_{2} = \frac{56 M. mL}{0.80 M}[/tex]

                                  = 70 ml

Thus, we can conclude that the volume after dilution is 70 ml.

Answer:

Specific gravity of mercury is 13.56 and it is an unit-less quantity.

Explanation:

Mass of the mercury = m = 607.0 lb = 275330.344 g

1 lb = 453.592 g

Volume of the mercury  = v = [tex]0.717 ft^3=20,303.18 mL[/tex]

[tex]1 ft^3 = 28316.847 mL[/tex]

Density of the mercury = d=[tex]\frac{m}{v}=\frac{275330.344  g}{20,303.18 mL}[/tex]

d = 13.56 g/mL

Specific gravity of substance = Density of substance ÷ Density of water

[tex]S.G=\frac{d}{1 g/mL}[/tex]

Specific gravity of mercury :

[tex]S.G=\frac{13.56 g/mL}{1 g/mL}=13.56[/tex] (unit-less quantity)

Answer: The chemical formula for aluminium nitrite is [tex]Al(NO_2)_3[/tex]

Explanation:

The given compound is formed by the combination of aluminium and nitrite ions. This is an ionic compound.

Aluminium is the 13th element of periodic table having electronic configuration of [tex][Ne]3s^23p^1[/tex].

To form [tex]Al^{3+}[/tex] ion, this element will loose 3 electrons.

Nitrite ion is a polyatomic ion having chemical formula of [tex]NO_2^{-}[/tex]

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for aluminium nitrite is [tex]Al(NO_2)_3[/tex]

The formula for aluminum nitrite is Al(NO3)3.

The formula for aluminum nitrite is Al(NO2)3.

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Answer:

Pipe racks are construction in chemical and other industries plants, that support the pipe line, electric cables and instrument cable.

Explanation:

The pipe racks also used to support mechanical equipment as valve and vessels. You can transfer material between equipment and sorage or utility areas. Pipe racks aren´t only non-building constructions that have similiraties to the Steel buildings but also have additional loads style. The requirements found in the building codes apply and dictate some of the design requirements.

Some industry references exist to help the designer apply the intent of the code and follow expected engineering practices.

Pipe racks have design criteria: In most of the United States, the governing building code is the International Building Code.

Includes:

*Dead Loads

*Live Loads

*Thermal Loads

*Earthquake Loads

*Wind Loads

*Rain Loads

*Snow Loads

*Ice Loads

*Load Combinations

Also have Design Considerations:

*Layout

*Seismic

*Seismic System Selection

*Period Calculations

*Analisys Procedure

*Selection

*Equivalent Lateral Force Method Analysis

*Modal Response Spectra Analysis

*Drift

*Seismic Detailing Requirements

*Wind

*Pressures and Forces

*Coatings

*Fire Protection

*Torsion on Support Beams  

3rd Law of Thermodynamics

The correct answer is the third law of thermodynamics.

"The entropy of a perfect crystal is zero when the temperature of the crystal is equal to absolute zero (0 K).”

The temperature scale they refer to is the zero in kelvin degrees, this is what we call absolute zero.

If the entropy is zero, all physic processes stop, and the entropy of the system is minimum and constant.

Final answer:

The concept described refers to the Third Law of Thermodynamics, which states that the entropy of a perfect crystalline substance is zero at absolute zero temperature. This law is often used to compute standard entropy values and predict entropy changes during phase transitions and chemical reactions. Option b.

Explanation:

The statement describing a pure crystalline substance at absolute zero temperature having no movement refers to the Third Law of Thermodynamics. This law states that the entropy of a perfectly ordered, crystalline substance at absolute zero temperature (0 K) is zero. Entropy, often denoted by S, is a measure of the disorder or randomness in a system, and it increases with temperature as molecular motion increases.

At absolute zero, all molecular motion ceases, meaning a perfectly crystalline substance has only a single microstate available to it (W = 1). Since there is just one possible arrangement for the particles, the entropy is zero as per the Boltzmann equation. This is part of the Third Law of Thermodynamics, which can be used to calculate entropy changes for phase transitions and chemical reactions under standard conditions.

Answer:

Carbon generally forms four covalent chemical bonds.

Explanation:

Carbon is a chemical element that belongs to group 14 of the periodic table and has atomic number 6. It is a member of the p-block and is nonmetallic in nature.

The ground-state electronic configuration of carbon atom is 1s²2s²2p². Thus, it has 4 valence electrons and is said to be tetravalent.

Therefore, carbon generally forms four covalent chemical bonds.

Answer:

Vinegar has 4.09% of acetic acid.

Explanation:

The neutralization reaction is:

[tex]CH_{3}COOH + KOH=>CH_{3}COOK + H_{2}O[/tex]

Each mol of KOH reacts with each mole of acetic acid so the quantity of moles of acetic acid is:

[tex]M= 0.46 \frac{mol}{l}*37ml*\frac{1l}{1000ml} =0.01702 mol[/tex]

The mass of acetic acid is:

0.01702 mol×60.02g/mol=1.0215 g Acetic Acid

Finally, the percentage is:

%=1.0215 g Acetic Acid÷25g vinegar(solution)=4.09%

Answer:

A) [tex]2.7\frac{km}{day}[/tex]

B) [tex]1.03*10^{2}\frac{ft}{s}[/tex]

C) [tex]31\frac{m}{s}[/tex]

D) [tex]2.05*10^{3}\frac{yd}{min}[/tex]

Explanation:

A) Convert [tex]70\frac{mi}{hr} to \frac{km}{day}[/tex]

[tex]70\frac{mi}{hr}*\frac{24hr}{1day}*\frac{1.60934km}{1mi}=2.7\frac{km}{day}[/tex]

B) Convert  [tex]70\frac{mi}{hr} to \frac{ft}{s}[/tex]

[tex]70\frac{mi}{hr}*\frac{1hr}{3600s}*\frac{5280ft}{1mi}=1.03*10^{2}\frac{ft}{s}[/tex]

C) Convert [tex]70\frac{mi}{hr} to \frac{m}{s}[/tex]

[tex]70\frac{mi}{hr}*\frac{1hr}{3600s}*\frac{1609.34m}{1mi}=31\frac{m}{s}[/tex]

D) Convert [tex]70\frac{mi}{hr} to \frac{yd}{min}[/tex]

[tex]70\frac{mi}{hr}*\frac{1hr}{60min}*\frac{1760yd}{1mi}=2.05*10^{3}\frac{yd}{min}[/tex]

To convert the speed limit of 70 mi/hr into alternative units, multiply the speed by the appropriate conversion factors.

To convert the speed limit of 70 mi/hr into alternative units:

Answer:  The mass of the gas is 18.3 g/mol.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]

[tex]\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55[/tex]

[tex]\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}[/tex]

[tex]1.55=\sqrt{\frac{44}{M_{X}}[/tex]

Squaring both sides and solving for [tex]M_{X}[/tex]

[tex]M_{X}=18.3g/mol[/tex]

Hence, the molar mas of unknown gas is 18.3 g/mol.

Answer: The molar mass of the unknown gas is 18.3 g/mol

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]

We are given:

[tex]\text{Rate}_{\text{(unknown gas)}}=1.55\times \text{Rate}_{C_3H_8}[/tex]

We know that:

Molar mass of propane = 44 g/mol

Taking the ratio of the rate of effusion of the gases, we get:

[tex]\frac{\text{Rate}_{\text{(unknwon gas)}}}{\text{Rate}_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{\text{(unknown gas)}}}}[/tex]

Putting values in above equation, we get:

[tex]\frac{1.55\times \text{Rate}_{C_3H_8}}{\text{Rate}_{C_3H_8}}=\sqrt{\frac{44}{M_{\text{(unknown gas)}}}}[/tex]

[tex]1.55=\sqrt{\frac{44}{M_{\text{(unknown gas)}}}}\\\\1.55^2=\frac{44}{M_{\text{unknwon gas}}}\\\\M_{\text{unknwon gas}}=\frac{44}{2.4025}\\\\M_{\text{unknwon gas}}=18.3g/mol[/tex]

Hence, the molar mass of the unknown gas is 18.3 g/mol

Answer:

(0,653±0,002) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =

0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,653±0,002) M of HNO₃

I hope it helps!

Answer:

Force of attraction = 35.96 [tex]\times 10^{27}[/tex]N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = [tex]10^{-9}[/tex] m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

[tex]F = \dfrac{k \times q_1 q_2}{ \r^2}[/tex] ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = [tex]8.99 \times 10^9 \ Nm^{2}C^{-2}[/tex]

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

[tex]F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }[/tex]

Force of attraction = 35.96 [tex]\times 10^{27}[/tex]N

Answer: The values of 'x' are 0.074 and -0.132

Explanation:

The equation given to us is:

[tex]\frac{x^2}{0.160-x}=0.058[/tex]

Rearranging the above equation, we get a quadratic equation:

[tex]x^2+0.058x-0.009744=0[/tex]

To solve this equation, we use quadratic formula, which is:

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

where,

a = coefficient of [tex]x^2[/tex] = 1

b = coefficient of x = 0.058

c = constant = 0.009744

Putting values in above equation, we get:

[tex]x=\frac{-0.058\pm \sqrt{(0.058)^2-4(1)(0.009744)}}{2\times 1}\\\\x=0.074,-0.132[/tex]

Hence, the values of 'x' are 0.074 and -0.132

Answer:

The velocity of flow in 10in pipe is 4.16 ft/s.

Explanation:

Given that

Specific gravity = 0.8

Viscosity =0.00042[tex]lbf/ft^2[/tex]

For pipe 1

[tex]V_1=6 ft/s,d_1=12\ in[/tex]

For pipe 1

[tex]V_2,d_1=10\ in[/tex]

If we assume that flow in the both pipe is laminar

For laminar flow through circular pipe

[tex]\dfrac{\Delta P}{L}=\dfrac{32V\mu }{d^2}[/tex]

So same pressure drop we can say that

[tex]\dfrac{V_1 }{d^2_1}=\dfrac{V_2}{d^2_2}[/tex]

[tex]\dfrac{6}{12^2}=\dfrac{V_2}{10^2}[/tex]

[tex]V_2=4.16 ft/s[/tex]

So the velocity of flow in 10in pipe is 4.16 ft/s.

The vapor pressure of acetone at 25.0°C under 1 atm of pressure is approximately 2.05 atm.

The molar entropy of vaporization of acetone under 1 atm of pressure can be calculated using the Clausius-Clapeyron equation. The equation is given as:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

Where P1 and P2 are the initial and final vapor pressures, ΔHvap is the molar enthalpy of vaporization, R is the ideal gas constant, T1 and T2 are the initial and final temperatures respectively. By rearranging the equation, we can solve for the final vapor pressure:

P2 = P1 * exp(ΔHvap/R * (1/T1 - 1/T2))

Substituting the given values, we have:

P1 = 1 atm

T1 = 56°C = 329 K

T2 = 25°C = 298 K

ΔHvap = 31.3 kJ/mol = 31,300 J/mol

R = 8.314 J/(mol·K)

Plugging these values into the equation:

P2 = 1 atm * exp(31300 J/mol / (8.314 J/(mol·K)) * (1/329 K - 1/298 K))

Simplifying the equation gives:

P2 ≈ 2.05 atm

Therefore, the vapor pressure of acetone at 25.0°C is approximately 2.05 atm.

The molar concentration of O2 in the surface water of a mountain lake at 20 °C and an atmospheric pressure of 665 torr is approximately 1.21×10-3 M.

To calculate the molar concentration of O2 in the surface water of a mountain lake using Henry's law, we first need to understand how pressure affects the solubility of gases and vice versa.

As per Henry's law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.  The partial pressure of O2 in air at sea level is 0.21 atm. This means that when the atmospheric pressure is 1 atm, the molar concentration of O2 is 1.38×10−3 M. At higher altitudes, the atmospheric pressure reduces. The given atmospheric pressure at the mountain lake is 665 torr, which is approximately 0.875 atm.

Using these values in Henry's law, the molar concentration of O2 can be calculated as:

C = P * x

where C is molar concentration, P is atmospheric pressure, and x is given solubility at 1 atm. Substituting the values:

C = (0.875 atm) * (1.38×10−3 M) = 1.21x10-3 M approximately

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Using Henry's law, the molar concentration of O₂ in water at a 20°C mountain lake with an atmospheric pressure of 665 torr is approximately 2.5×10⁻⁴ M. This is calculated by converting the pressure to atm and then using the given solubility at standard conditions. The relationship is direct proportionality between solubility and partial pressure.

We will use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid. The formula is:

Concentration = kH × P

Given:

The solubility of O₂ in water at 20°C with 1 atm O₂ pressure (P₀) is 1.38×10⁻³ M (C₀).

Partial pressure of O₂ in air at sea level (P₀) is 0.21 atm.

Atmospheric pressure at the mountain lake is 665 torr.

First, convert 665 torr to atm:

665 torr × (1 atm / 760 torr) = 0.875 atm

Next, calculate the partial pressure of O₂ in air at the mountain lake:

Partial Pressure of O₂ = 0.21 atm × 0.875 atm = 0.184 atm

Using Henry's law, we can find the new concentration (C):

C = kH × P = 1.38×10⁻³ M/atm × 0.184 atm = 2.5×10⁻⁴ M

Therefore, the molar concentration of O₂ in the mountain lake's surface water at 20°C is approximately 2.5×10⁻⁴ M.

In order to ingest 0.150g of mercury from water contaminated with mercury at twice the legal limit (0.004 mg/L), one would need to consume 37500 liters of this contaminated water.

Your question is looking to find out how much water contaminated with mercury at twice the legal limit would need to be consumed to ingest 0.150g of mercury. First, we need to convert the g of mercury you want to find out to the same unit as the water contamination level, which is mg/L. So, 0.150 g = 150 mg.

Then, we will divide this amount by the contamination level, which is 0.004 mg/L. Therefore, 150 divided by 0.004 = 37500 L. So, one would have to consume 37500 liters of water at that level of contamination to ingest 0.150 g of mercury.

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One would have to consume 37,500 litres of water to ingest 0.150 grams of mercury at a concentration of 0.004 mg/L.

To solve this problem, we need to calculate the volume of water that contains 0.150 grams of mercury when the concentration of mercury is twice the legal limit, which is 0.004 mg/L.

First, we convert the mass of mercury from grams to milligrams, since the concentration is given in milligrams per litre.

1 gram = 1000 milligrams

So, 0.150 grams of mercury is equivalent to:

[tex]\[ 0.150 \text{ g} \times 1000 \text{ mg/g} = 150 \text{ mg} \][/tex]

Now, we know that the water is contaminated at a concentration of 0.004 mg/L, which means there are 0.004 milligrams of mercury in every litre of water.

To find out how many litres of water contain 150 mg of mercury, we set up the following proportion:

[tex]\[ \frac{0.004 \text{ mg}}{1 \text{ L}} = \frac{150 \text{ mg}}{x \text{ L}} \][/tex]

Solving for ( x ) (the volume of water in litres), we get:

[tex]\[ x = \frac{150 \text{ mg}}{0.004 \text{ mg/L}} \]\\[/tex]

[tex]\[ x = 37500 \text{ L} \] \\[/tex]

[tex]\[ x = \frac{150 \text{ mg}}{0.004 \text{ mg/L}} \] \\[/tex]

[tex]\[ x = 37500 \text{ L} \][/tex]

Answer:

As can be seen in the explanation, there will be a greater increase in the enthalpy of the working fluid in the adiabatic process than in the isothermal process, since in the adiabatic process the change of internal energy is added; In the isothermal process there is no change in internal energy that is added to the value of enthalpy.

Explanation:

∴ Q = 0  ⇒ ΔU = W......first law

⇒ ΔU = Cv*ΔT

⇒ W = P*dV  ∴ P = nRT/V

⇒ ΔH = ΔU + P*ΔV = Cv*ΔT + nRT/V*ΔV

∴ ΔT = 0 ⇒ ΔU = 0

⇒ Q = - W = nRTLn(V2/V1)

⇒ ΔH = PΔV = nRT/V*ΔV

Final answer:

For a greater increase in the enthalpy of the working fluid in a reversible flow compressor, the isothermal process is more favorable than the adiabatic process. This is because the isothermal process allows for heat exchange with the surroundings, directly increasing the enthalpy, unlike the adiabatic process where there's no heat exchange.

Explanation:

To determine which condition between adiabatic and isothermal will lead to a greater increase in the enthalpy of the working fluid, it is essential to understand the concepts of adiabatic and isothermal processes in the context of a reversible flow compressor. In an adiabatic process, the system is thermally insulated, leading to no heat exchange with the surroundings, which in turn means that any work done on the system results directly in changes to the system's internal energy and hence, its temperature. On the other hand, during an isothermal process, the system's temperature is kept constant by allowing heat exchange with the surroundings, effectively absorbing or releasing heat as work is done on or by the system.

By comparing the expressions dU = Tds - PdV and dH = TdS + VdP, and knowing that dH represents the increase in enthalpy (the heat added at constant pressure), it's evident that during an isothermal process, where temperature is constant, the system can absorb or release heat without changing temperature, ideally through reversible compressions or expansions that allow it to maintain equilibrium with a thermal reservoir. This contrasts with an adiabatic process where any work done on the system directly influences its internal energy and, consequently, temperature, without the exchange of heat with its surroundings.

Therefore, for a greater increase in the enthalpy of the working fluid, the isothermal process is favorable. This is because it allows for heat exchange, facilitating a direct increase in enthalpy, as opposed to the adiabatic process, where the increase in internal energy does not directly translate to an increase in enthalpy due to the lack of heat exchange.

Answer:

The rate law for second order unimolecular irreversible reaction is

[tex]\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }[/tex]

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

[tex]v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}[/tex]

rearranging the ecuation

[tex]-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}[/tex]

Integrating between times 0 to t and between the concentrations of [tex][A]_{0}[/tex] to [A].

[tex]\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}[/tex]

Solving the integral

[tex]\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }[/tex]

Using the integrated rate law for second-order reactions, we can determine the concentration of butadiene gas after a certain time period by applying the rate constant and initial concentration into the equation 1/[A] = kt + 1/[A]0 where k is the rate constant, t is the time and [A]0 is the initial concentration.

The integrated rate law for second-order reactions in kinetics helps us understand how the concentration of a reactant changes over time in a unimolecular and irreversible reaction. In the example provided, using the second-order integrated rate law equation, we want to find the concentration of butadiene (C4H6) after 10 minutes given its initial concentration and reaction rate constant.

The second-order integrated rate equation can be written as:

1/[A] = kt + 1/[A]0

With a y-intercept of 1/[A]0 and slope of the rate constant k.

To solve for the concentration after 10 minutes, we can plug in the given values:

Then we calculate [A] after 10.0 min using the equation:

1/[A] = (5.76 × 10-2 L mol-1 min-1 × 10.0 min) + (1/0.200 M)

The above calculation will yield the final concentration [A] of butadiene after 10 minutes.

Answer : The amount of heat changes is, 56.463 KJ

Solution :

The conversions involved in this process are :

[tex](1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(4):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(5):H_2O(g)(100^oC)\rightarrow H_2O(g)(125^oC)[/tex]

Now we have to calculate the enthalpy change.

[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})][/tex]

where,

[tex]\Delta H[/tex] = enthalpy change or heat changes = ?

n = number of moles of water = 1 mole

[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]2.09J/g^oC[/tex]

[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]

[tex]c_{p,g}[/tex] = specific heat of liquid water = [tex]1.84J/g^oC[/tex]

m = mass of water

[tex]\text{Mass of water}=\text{Moles of water}\times \text{Molar mass of water}=1mole\times 18g/mole=18g[/tex]

[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

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

[tex]\Delta H=[18g\times 4.18J/gK\times (0-(-25))^oC]+1mole\times 6010J/mole+[18g\times 2.09J/gK\times (100-0)^oC]+1mole\times 40670J/mole+[18g\times 1.84J/gK\times (125-100)^oC][/tex]

[tex]\Delta H=56463J=56.463KJ[/tex]     (1 KJ = 1000 J)

Therefore, the amount of heat changes is, 56.463 KJ

Answer : The change in entropy is 6 J/K

Explanation :

To calculate the change in entropy we use the formula:

[tex]\Delta S=\int \frac{dQ}{T}[/tex]

and,

[tex]Q=nC_pdT[/tex]

[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p}dT}{T}[/tex]

where,

[tex]\Delta S[/tex] = change in entropy

n = number of moles = 2 moles

[tex]T_f[/tex] = final temperature = 303 K

[tex]T_i[/tex] = initial temperature = 273 K

[tex]C_{p}[/tex] = heat capacity at constant pressure = [tex]0.1\times T(J/K.mol)[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta S=2\int\limits^{303}_{273}{\frac{(0.1\times TdT}{T}[/tex]

[tex]\Delta S=2\times 0.1\int\limits^{303}_{273}dT[/tex]

[tex]\Delta S=2\times 0.1\times [T]^{303}_{273}[/tex]

[tex]\Delta S=2\times 0.1\times (T_f-T_i)[/tex]

[tex]\Delta S=2\times 0.1\times (303-273)[/tex]

[tex]\Delta S=6J/K[/tex]

Therefore, the change in entropy is 6 J/K