Not yet answered Marked out of 2.00 What is the concentration of NH4+ in 60.0 mL of a 0.50 M solution of (NH4)3PO4? (To write your answer using scientific notation use 1.0E-1 instead of 1.0 x 10-?) Answer: Answer

Answer:

0.9

Explanation:

The pka represents the force by which the molecules need to dissociate for the acids ,

Hence , lower the pka stronger will be the acid and that therefore will  dissociate completely and vice versa , for a weak acid higher the pka .

And in case of a base , it will be completely reversed , lower pKa , weaker base ,

and higher pKa , stronger base .

From the data of the question ,

0.9 is the lowest value of the pKa , hence , weakest base .

Explanation:

(a)  A number which is dimensionless and provides us an estimate of the degree of conversion which can be achieved in CSTR, that is, continuous stirred tank reactor is known as Damkohler number.

This number is denoted as Da.

Mathematically,     Da = [tex]\frac{\text{reaction rate}}{\text{convection rate}}[/tex]

             Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]

Now, for first order system, Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]

                        = [tex]\frac{k \times CA_{o} \times V}{v \times CA_{o}}[/tex] = Tk

where,      rA = rate of reaction

                V = volume of reactor

                [tex]Fa_{o}[/tex] = molar flow rate of component A

                 k = rate constant

               [tex]CA_{o}[/tex] = initial concentration of A

                  v = volumetric flow rate of A

                  T = residence time

(b)   Since, from a given Damkohler number we can figure out the possible conversion of CSTR, that is, continuous stirred tank reactor.

So, if we have a low Damkohler number then the system will give us a less conversion formula. As the conversion is as follows.

                       X = [tex]\frac{Da}{Da + 1}[/tex]

Hence, we can conclude that [tex]Da \leq 0.1[/tex] will give less than 10% conversion as calculated by using above formula.

Final answer:

The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. This means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.

Explanation:

The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. It is defined as the ratio of the characteristic time for a chemical reaction to occur to the characteristic time for a system to transport material through itself.

A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. In practical terms, this means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.

For example, if we have a packed bed reactor (a reactor where reactants flow through a bed of solid catalyst particles), a low Damkohler number implies that the reaction is so fast that the reactants are fully converted into products even before they can flow through the bed, resulting in high conversion levels.

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:

2.25 g

Explanation:

The mass of the solid X must be the total mass (beaker + solid X) less than the mass of the beaker. Then:

mass of the solid X = 34.40 - 32.15

mass of the solid X = 2.25 g

The difference of 0.25 g must occur for several problems: an incorrect weight in the balance, the configuration of the balance, the solid can be hydrophilic and absorbs water, and others.

Answer:

24.76 kJ/g

844.8 kJ/mol

Explanation:

The heat produced by the burning of the solid magnesium is equal to the heat absorbed by the calorimeter, which causes its temperature to increase.

First, we calculate the heat absorbed by the calorimeter, where C is the heat capacity and Δt is the temperature change.

Q = CΔt = (3024 J·°C⁻¹)(1.126 °C) = 3405 J

This is the same amount of heat that was produced by burning the magnesium

Now we can calculate the heat produced per gram of magnesium:

(3405 J)(1 kJ/1000 J) / (0.1375 g) = 24.76 kJ/g

We can convert grams to moles using the atomic weight of Mg (24.305 g/mol).

(24.76 kJ/g)(24.305 g/mol) = 844.8 kJ/mol

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

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

where,

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

n = number of moles = 1.0 mol

[tex]T_f[/tex] = final temperature = 300 K

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

[tex]C_{p,m}[/tex] = heat capacity of chloroform = [tex]91.47+7.5\times 10^{-2}(T/K)[/tex]

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

[tex]\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}[/tex]

[tex]\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}[/tex]

[tex]\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)[/tex]

[tex]\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)[/tex]

[tex]\Delta S=8.626+2.025[/tex]

[tex]\Delta S=10.651J/K.mol[/tex]

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

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:

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 : 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

Answer:

a) 63,0%

b) 54,4 Pounds HNO₃ per cubic foot of solution

c) 13,8 M

Explanation:

a) Weight percent is the ratio solute:solution times 100:

[tex]\frac{1,704 kg HNO_3}{2,704 kg Solution}[/tex] = 63,0%

b) Pounds HNO₃ per cubic foot of solution at 20 °c

Pounds HNO₃:

1,704 kg [tex]\frac{2,20462 pounds}{1 kg}[/tex] = 3,7567 pounds

Cubic foot:

2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex]x[tex]\frac{1 cubic foot }{28,3168 L}[/tex] = 0,069 ft³

Thus:[tex]\frac{3,7567 pounds}{0,069ft^3} =[/tex] = 54,4 Pounds HNO₃ per cubic foot of solution

c) Moles of HNO₃:

1704 g HNO₃ [tex]\frac{1 mol HNO_3}{63,01 g }[/tex] = 27,04 moles

Liters of solution:

2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex] = 1,96 L of solution

Molarity:

[tex]\frac{27,04 mol}{1,96 L}[/tex] = 13,8 M

Final answer:

The composition of the solution can be expressed as 63.02 wt% HNO3, 54.28 lb HNO3/ft3, and a molarity of 14.2 M at 20 °C.

Explanation:

To express the composition of a water solution with 1.704 kg of HNO3 per kg of H2O and a specific gravity of 1.382 at 20 °C, we can calculate the following:

Weight Percent HNO3

Weight percent (wt%) is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100. For 1.704 kg HNO3 in 1 kg of water, the total mass of the solution is 1.704 kg + 1 kg = 2.704 kg. The weight percent HNO3 is then ((1.704 kg) / (2.704 kg)) × 100 = 63.02 wt%.

Pounds HNO3 per Cubic Foot of Solution at 20 °C

Firstly, we need to convert the specific gravity to density in g/mL: 1.382 (Specific Gravity) × 1.000 g/mL (density of water) = 1.382 g/mL. To find the density in lb/ft3, we multiply by 62.43 lb/ft3 which is the conversion factor from g/mL to lb/ft3. Then, multiply the density of the solution by the weight percent of HNO3 to find pounds HNO3 per cubic foot of solution: (1.382 g/mL) × (62.43 lb/ft3) × (63.02%) = 54.28 lb HNO3/ft3.

Molarity of HNO3 at 20 °C

Using the density of the concentrated HNO3 solution, 1.42 g/mL, and the given formula for molarity which is Molarity = [(%) (d)/ (Mw)] × 10, we substitute the values: Molarity = [(63.02) (1.42 g/mL) / (63.01)] × 10 = 14.2 M.

Answer:

For ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = 206

Explanation:

Lead, chemical symbol Pb, is a chemical element which belongs to the group 14 of the periodic table. The atomic number of lead is 82 and it is a member of the p-block.          

The isotope of lead with the mass number 208, has 126 neutrons.

Since, atomic number = number of protons =  number of electrons for neutral atom

Therefore, for ²⁰⁸Pb: number of electrons= 82

So, for ²⁰⁸Pb²⁺ cation: number of electrons= 82 - 2 = 80

Therefore, for ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = number of electrons + number of neutrons = 80 electrons + 126 neutrons = 206.

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: 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].

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:

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.

Explanation:

The given data is as follows.

          [tex]\Delta_{r} G[/tex] = -16.7 kJ/mol = [tex]-16.7 \times 10^{3}[/tex],       T = 298 K

          R = 8.314 J/mol K,       [tex]K_{eq}[/tex] = ?

Relation between [tex]\Delta_{r} G[/tex] and [tex]K_{eq}[/tex] is as follows.

                [tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]

Hence, putting the values into the above equation as follows.

                [tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]

                        [tex]-16.7 \times 10^{3} J/mol[/tex] = [tex]-8.314 J/mol K \times 298 K ln K_{eq}[/tex]

                     [tex]ln K_{eq}[/tex] = [tex]\frac{-16.7 \times 10^{3} J/mol}{-8.314 J/mol K \times 298 K}[/tex]    

                                         = 6.740

                            [tex]K_{eq}[/tex] = antilog (6.740)

                                            = 846

Thus, we can conclude that [tex]K_{eq}[/tex] for given values is 846.

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.

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:

i need more to solve this

Explanation:

Options for full question:

(A) % oxygen is 32.1%

(B) % hydrogen is 4%

(C) total percent composition of all elements is approximately 100%

(D) % carbon is 15.9%

Answer:

Options B and C

Explanation:

Information Given;

Mass of sample - 50g

Mass of Oxygen - 32.1g

Mass of Hydrogen - 2g

Mass of Carbon - 15.9g

The percentage composition of an element in a compound is the mass percentage of the element present in the compound. It tells the mass percentage of each element present in a compound.

The formular is given as;

Percentage composition = (Mass of element / Mass of compound) * 100

For oxygen;

Percentage Composition = (32.1 / 50) * 100 = 0.642 * 100= 64.2%

For Hydrogen;

Percentage Composition = (2 / 50) * 100 = 0.04 * 100= 4%

For Carbon;

Percentage Composition = (15.9 / 50) * 100 = 0.318 * 100= 31.8%

Total percentage composition of all elements = 4% + 31.8% + 64.2% = 100%

Based on this,

Option A is incorrect

Option B is Correct.

Option C is Correct.

Option D is incorrrect

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

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: Wavenumber of the radiation emitted  is [tex]0.08\times 10^{8}m^{-1}[/tex]

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

[tex]E=\frac{hc}{\lambda}[/tex]

where,

E = energy of the radiation = [tex]1.634\times 10^{-18}J[/tex]

h = Planck's constant  = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] = wavelength of radiation = ?

Putting values in above equation, we get:

[tex]1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m[/tex]

[tex]\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}[/tex]

Thus wavenumber of the radiation emitted  is [tex]0.08\times 10^{8}m^{-1}[/tex]

Answer:

b. 0.100M

Explanation:

The balanced chemical reaction is: NaOH + HA ⇒ H₂O + NaA

The NaOH and HA react in a 1:1 molar ratio, so at the equivalence point, the amount of NaOH added equals the amount of HA that was present in the solution.

The amount of NaOH that was added can be calculated and set equal to the amount of HA that must have been present to react with it.

n = CV = (0.200 mol/L)(12.5 mL) = 2.50 mmol NaOH = 2.50 mmol HA

Thus, there were 2.50 mmol of HA in 25.0 mL. The concentration can be calculated as follow:

C = n/V = (2.50 mmol)/(25.0mL) = 0.100 M

Answer:

i=2 for HCl and i=1 for ethanol

Explanation:

Hello,

Since the hydrochloric acid is composed by sodium and chloride ions which are completely dissociated, two types of ions are present, that's why i becomes 2.

On the other hand, as long as the ethanol doesn't present dissociation in aqueous solution, the Van't Hoff factor becomes 1.

Best regards.

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: [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]

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:

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  

Answer: Molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 0.1015 M

[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex]solution = 200 ml

[tex]M_2[/tex] = molarity of dilute [tex]HNO_3[/tex] solution = ?

[tex]V_2[/tex] = volume of  dilute [tex]HNO_3[/tex]  solution = (2300 +200 )ml = 2500 ml

Putting in the values we get:  

[tex]0.1015M\times 200=M_2\times 2500[/tex]

[tex]M_2=8.12\times 10^{-3}M[/tex]

Thus the molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]

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