Consider the formation of nitrogen dioxide from nitric
oxideand oxygen.
2NO(g) + O2(g) -- 2NO2(g)
If 9L of NO are reacted with excess 02 at STP, what is
thevolume in liters of the NO2 produced?

Answer:

The energy of the photon generated in the transition is 3.14*10⁻¹⁹ J

Explanation:

There are two equation that we need to use in order to solve this problem:

The first one is Planck's equation, which describes the relationship between energy and frequency:

E = h*v    eq. 1)

Where E is energy, h is Planck's constant (6.626 * 10⁻³⁴ J*s) and v is the radiation frequency.

In order to know the frequency, we use the second equation, which is the wave equation:

c = λ*v    eq. 2)

Where c is the speed of light in vacuum (aprx 3 * 10⁸ m/s), and λ is the wavelength. If we solve that equation for v we're left with

v=c/λ    eq. 3)

We replace v in eq. 1):

E= c*h/λ

Lastly we put the data we know and solve the equation, keeping in mind the correct use of units (converting 652 nm into m):

[tex]E=\frac{3*10^{8}ms^{-1} *6.626*10^{-34}Js}{633*10^{-9}m }=3.14*10^{-19}J[/tex]

Final answer:

The energy of the photon generated in the transition using the given formula. Calculating gives approximately 3.04 x 10^-19 joules.

Explanation:

The energy of the photon generated in the transition can be calculated using the formula:

E = hc / λ

Where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light. Substituting the given values:

E = (6.63 x 10^-34 J s x 3 x 10^8 m/s) / (652 x 10^-9 m)

Calculating this gives the energy of the photon as approximately 3.04 x 10^-19 joules.

Answer:

100.223°C is the boiling point of an aqueous solution.

Explanation:

Osmotic pressure of the solution = π = 10.50 atm

Temperature of the solution =T= 25 °C = 298 .15 K

Concentration of the solution = c

van'y Hoff factor = i = 1 (non electrolyte)

[tex]\pi =icRT[/tex]

[tex]c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}[/tex]

c = 0.429 mol/L = 0.429 mol/kg = m

(density of solution is the same as  pure water)

m = molality of the solution

Elevation in boiling point = [tex]\Delta T_b[/tex]

[tex]\Delta T_b=iK_b\times m[/tex]

[tex]\Delta T_b=T_b-T[/tex]

T = Boiling point of the pure solvent

[tex]T_b[/tex] = boiling point of the solution

[tex]K_b[/tex] = Molal elevation constant

We have :

[tex]K_b=0.52^oC/m[/tex] (given)

m = 0.429 mol/kg

T = 100° C (water)

[tex]\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg[/tex]

[tex]\Delta T_b=0.223^oC[/tex]

[tex]\Delta T_b=T_b-T[/tex]

[tex]T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC[/tex]

100.223°C is the boiling point of an aqueous solution.

Answer:

8.83 moles of Hydrogen

Explanation:

We are told that 1.92 moles of H2 occupies 22.2 L at a certain pressure and temperature, and we are asked to calculate the quantity of moles of H2 are necessary to fill a container of 102.1 L. So, if the pressure and temperature are constant, then the equality expressed before 1.92 H2 moles = 22.2 L maintains. Therefore, we can calculate what is being asked as follows:

22.2 L ----- 1.92 H2 moles

102.1 L ---- x = (102.1 L × 1.92 H2 moles)/22.2 L = 8.83 H2 moles

This means that 8.83 H2 moles are necessary to fill a 102.1 L container at a certain pressure and temperature.

Answer:

a. T and P remain the same (T=298 K and P=1 bar)

b. 11.23J/K

Explanation:

a. Since the mixing process of an idea gas doesn't present a change in the enthalpy, we could state that no change in neither temperature and pressure are given.

b. It is not necessary to know enthalpy data, the following formula is enough to compute the entropy change:

Δ[tex]S_{mix}=-n_{N_2}R ln(x_{N_2})-n_{O_2}R ln(x_{O_2})[/tex]

Thus, the molar fractions are equal to 0.5, and the result yields:

Δ[tex]S_{mix}=-(1mol)[(8.314J/(mol*K)]ln(0.5)-(1mol)[(8.314J/(mol*K)]ln(0.5)[/tex]

Δ[tex]S_{mix}=11.23J/K[/tex]

Best regards

Answer:

FALSE

Explanation:

Grunge refers to the genre of rock music and the fashion inspired by it. It originated in the mid-1980s in Seattle, Washington State.

Grunge was described as the fusion of punk rock and heavy metal.

This genre of music became popular in the early mid-1990s and included lyrics based on the theme of emotional and social alienation, betrayal, abuse, trauma etc.    

The statement that grunge is a rock style from Detroit is false; grunge originated in Seattle, Washington. This music genre is characterized by a stripped-down aesthetic, distorted guitars, and socially conscious lyrics, gaining popularity in the late 1980s and early 1990s.

Grunge is indeed a style of rock music, but it is not from Detroit; that statement is false. Grunge, often referred to as the Seattle sound, originated in Seattle, Washington and became widely popular in the late 1980s and early 1990s. It represented a departure from the extravagant stage productions and fashion of contemporary rock bands, favoring a more stripped-down sound with an emphasis on distorted electric guitars and direct, often socially conscious lyrics.

The grunge movement produced several iconic bands, including Nirvana, Pearl Jam, and Soundgarden, which significantly shaped the alternative rock landscape. Seattle's geographic isolation played a pivotal role in the development of grunge music. Local bands evolved their unique sound and aesthetic away from the influence of mainstream music industry expectations, which at the time were centered around Los Angeles. This isolation contributed to grunge's authenticity and appeal when it eventually burst onto the national and international stages.

Contrary to the glamorous rock scene in Los Angeles, grunge musicians typically wore simple, unremarkable clothing, often sourced from second hand stores, and their lyrics addressed themes like political issues, mental health, and substance abuse, resonating with the disaffected Generation X.

Answer:

[tex]T=-272.9^{o}C[/tex]

Explanation:

We have the ideal gasses equation [tex]PV=nRT[/tex] and the expression for the specific volume [tex]v=\frac{V}{m}[/tex], that is the inverse of the density, and for definition the number of moles is equal to the mass over the molar mass, that is [tex]n=\frac{m}{M}[/tex]

And we can relate the three equations as follows:

[tex]PV=nRT[/tex]

Replacing the expression for n, we have:

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

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

Replacing the expression for v, we have:

[tex]Pv=\frac{RT}{M}[/tex]

Now resolving for T, we have:

[tex]T=\frac{PvM}{R}[/tex]

Now, we should convert all the quantities to the same units:

-Convert 500kPa to atm

[tex]500kPa*\frac{0.00986923}{1kPa}=4.93atm[/tex]

-Convert 0.2[tex]\frac{m^{3}}{kg}[/tex] to [tex]\frac{L}{kg}[/tex]

[tex]0.2\frac{m^{3} }{kg}*\frac{1L}{1m^{3}}=0.2\frac{L}{kg}[/tex]

- Convert the molar mass M of the water from [tex]\frac{g}{mol}[/tex] to [tex]\frac{kg}{mol}[/tex]

[tex]18\frac{g}{mol}=\frac{1kg}{1000g}=0.018\frac{kg}{mol}[/tex]

Finally we can replace the values:

[tex]T=\frac{(4.93atm)(0.2\frac{L}{kg})(0.018\frac{kg}{mol})}{0.082\frac{atm.L}{mol.K}}[/tex]

[tex]T=0.216K[/tex]

[tex]T=0.216K-273.15\\T=-272.9^{o}C[/tex]

Answer:

The molar mass of the unknown solute is 106,9 g/m

Explanation:

Cryoscopic descent formula to solve this

ΔT = Kf . m

Be careful because units in Kfp are K/m, so let's get the ΔT degrees °C in K

3,91°C  = 3,91 K

It's a difference, in the end it does not matter

For example you can have 5° C as the final temperature and as initial, 1,09 °C -- ΔT is 5 - 1.09 = 3.91

What happens in Kelvin?

5°C + 273 = 278 K

1,09° C + 273 = 274,09 K

ΔT = 278 K - 274,09 K = 3,91 K

3,91 K = 20,1 K/m * m

3,91 K / 20,1 m/K = m

0,194 = m (molality)

Molality means moles from solute in 1 kg of solvent.

1kg = 1000 g

1000 g ________  0,194 moles

50 g _________ x

x = (50 g * 13,77 moles) / 1000 g = 9,72 *10-3 moles

Moles = mass / molar mass

Molar mass = mass / moles

Molar mass = 1,04 g / 9,72 *10-3 moles

Molar mass = 106,9 g/m

Answer:

The correct answer is: PCl₃ and PCl₅

Explanation:

Law of multiple proportions, also known as the Dalton's Law, states that the when the two different chemical elements combine in order to form two or more than two chemical compounds, then the ratio of mass of the chemical element that combines with the fixed mass of other chemical element is the ratio of small whole numbers.

Example: PCl₃ and PCl₅

Answer:

Ka = [tex]2.32 \times 10^{-4}[/tex]

pH = 2.12

Explanation:

Calculation of Ka:

% Dissociation = 3% = 0.03

Concentration of solution = 0.25 M

HA dissociated as:

       [tex]HA \rightarrow H^+ + A^{-}[/tex]

      C(1 - 0.03)    C×0.03   C×0.03

HA] after dissociation = 0.25×0.97 = 0.2425 M

[tex][H^+]= 0.25\times0.03 = 0.0075 M[/tex]

[tex][A^{-}]= 0.25 \times 0.03 = 0.0075 M[/tex]

[tex]Ka= \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}[/tex]

Calculation of pH of the solution

[tex]pH = -log [H^+][/tex]

[tex]H^+[\tex] = 0.0075 M[/tex]

[tex]pH = -log 0.0075 = 2.12[/tex]

The Ka of the weak acid is calculated using the concentrations of the dissociated and non-dissociated parts of the weak acid and the pH is derived from the concentration of H+ ions.

The subject in concern relates to the dissociation of a weak acid and the calculation of the Acid dissociation constant (Ka) and the pH of the solution.

We are given that 3% of the weak acid, HA, is dissociated. Meaning, 3% of 0.25 M HA dissociates into H+ and A-. This will be 0.03 * 0.25 M = 0.0075 M.

The equilibrium for the reaction is HA <--> H+ + A-. Due to water autoionization, the concentration of water is approximately taken as constant in the denominator of Ka calculation. Hence, Ka for HA can be approximated as [H+][A-] / [HA]. We know that [H+] = [A-] = x, and [HA] = 0.25 - x. We approximate x as 0.0075 since only 3% of HA dissociates. Hence Ka=~(0.0075)^2/(0.25-0.0075).

The pH of the solution is -log[H+] = -log(0.0075).

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Answer:

a) mass flow = 56940 Kg/h

b) mass flow = 202.5 mol/s

Explanation:

∴ δ C6H6 = 876 Kg/m³,,,,,wwwcarlroth.com

⇒ 65m³/h * 876 Kg/m³ = 56940 Kg/h

⇒ 56940 Kg/h * ( 1000 g/Kg ) * ( mol/ 78.11 g) * ( h/3600s )= 202.5 mol/s

The mass flow rate of benzene is calculated using its density of 0.876 kg/m³, resulting in 56.94 kg/h. The molar flow rate is determined using the molar mass of benzene, 78.11 g/mol, which yields 20.34 mol/s.

Mass Flow and Molar Flow Calculations

To calculate the mass flow fed in kg/h for benzene, the density of benzene is required. Benzene has a density of approximately 0.876 kg/m³ at room temperature. Using this density, the mass flow can be calculated as follows:

Mass flow (kg/h) = Volumetric flow (m³/h) × Density (kg/m³)

Mass flow = 65 m³/h ×  0.876 kg/m³ = 56.94 kg/h

To calculate the molar flow in mol/s, we use the molar mass of benzene which is approximately 78.11 g/mol:

Molar flow (mol/s) = Mass flow (kg/h) × (1000 g/kg) / (Molar mass (g/mol) × (3600 s/h))

Molar flow = (56.94 kg/h × 1000 g/kg) / (78.11 g/mol ×  3600 s/h) = 20.34 mol/s

Note: The molecular formula of benzene is C₆H₆, not CH as mentioned in the information provided. Therefore, the combustion analysis would yield different amounts of CO₂ and H₂O than suggested by the empirical formula CH. It's important to use the correct molecular formula for accurate calculations.

Explanation:

The given data is as follows.

        Mass of gold = 645 lb,       Volume = 0.5348 ft3

         Density of water = 62.4 lbs/ft3

It is known that specific gravity is defined as density of substance divided by the density of standard fluid.

Mathematically,       Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water}}[/tex]

               Specific gravity = [tex]\frac{\text{density of gold}}{62.4 (lbs.ft^{-3})}[/tex]

Now, calculate the density of gold then from density we will calculate specific gravity as follows

Since,             Density = [tex]\frac{mass}{volume}[/tex]

                        Density = [tex]\frac{645 lbs}{0.5348 ft^{3}}[/tex]

                                    = 1206.06 [tex]lbs/ft^{3}[/tex]

As,        Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water (standard fluid)}}[/tex]

                                      = [tex]\frac{1206.06 (lbs/ft^{3})}{62.4 (lbs/ft^{3})}[/tex]

                                      = 19.32

Therefore, the value of specific gravity is 19.32.

Specific gravity has no units as it is density divided by density. Hence, all the units get canceled out.

Explanation:

STP means standard temperature and pressure where values are as follows.

                      T = 273.15 K,         P = 1 atm

According to ideal gas equation, PV = nRT. Since, it is given that mass is 48.3 g and we have to find the volume as follows.

                  n = [tex]\frac{mass}{\text{molar mass}}[/tex]

So,                       PV = nRT

               V = [tex]\frac{mass}{\text{molar mass}} \times \frac{RT}{P}[/tex]                  

                   = [tex]\frac{48.3 g}{28 g/mol} \times \frac{0.0821 L atm/mol K \times 273.15 K}{1 atm}[/tex]             (molar mass of CO = 28 g/mol)

                   = 38.68 L                

Thus, we can conclude that volume of  48.3 g of carbon monoxide at STP is 38.68 L.

To calculate the mass of iodine solid that participates in a chemical reaction, multiply the number of moles of iodine by its molar mass.

To calculate the mass of iodine solid that participates in a chemical reaction, we need to use the concept of moles and the molar mass of iodine. Given that 0.0200 moles of iodine solid participate in the reaction, we can convert moles to grams by multiplying the number of moles by the molar mass of iodine. The molar mass of iodine (I2) is 253.8 grams per mole. Multiplying 0.0200 moles by 253.8 grams/mole gives us a mass of 5.08 grams of iodine solid that participates in the chemical reaction.

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Answer: The mass of water produced in the reaction is 47.25 grams.

Explanation:

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

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

Given mass of ammonia = 29.7 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ammonia}=\frac{29.7g}{17g/mol}=1.75mol[/tex]

The given chemical reaction follows:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

By stoichiometry of the reaction:

4 moles of ammonia produces 6 moles of water.

So, 1.75 moles of ammonia will produce = [tex]\frac{6}{4}\times 1.75=2.625mol[/tex] of water.

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

Molar mass of water = 18 g/mol

Moles of water = 2.625 moles

Putting values in equation 1, we get:

[tex]2.625mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=47.25g[/tex]

Hence, the mass of water produced in the reaction is 47.25 grams.

The complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water by converting the given mass of NH3 to moles, using the balanced chemical equation to find the ratio of NH3 to H2O, and then converting the moles of H2O to grams.

To find out how many grams of water are produced in the complete reaction of 29.7 grams of ammonia (NH3), we need to use stoichiometry, which involves several steps:

Therefore, the complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water (H2O).

Answer: The molarity of sulfuric acid solution is 2.3 M

Explanation:

The relationship between specific gravity and density of a substance is given as:

[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]

Specific gravity of sulfuric acid solution = 1.13

Density of water = 1.00 g/mL

Putting values in above equation we get:

[tex]1.13=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.13\times 1.00g/mL)=1.13g/mL[/tex]

We are given:

20% (m/m) sulfuric acid solution. This means that 20 g of sulfuric acid is present in 100 g of solution

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

Density of solution = 1.13 g/mL

Mass of Solution = 100 g

Putting values in above equation, we get:

[tex]1.13g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.13g/mL}=88.5mL[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Mass of solute (sulfuric acid) = 20 g

Molar mass of sulfuric acid = 98 g/mol

Volume of solution = 88.5 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{20g\times 1000}{98g/mol\times 88.5mL}\\\\\text{Molarity of solution}=2.3M[/tex]

Hence, the molarity of sulfuric acid solution is 2.3 M

The answer is unknown

Answer:

Ethyl alcohol and isopropyl alcohol are the most commonly used agents.

Explanation:

Isopropyl alcohol is used in a 10% concentration.

- not true, to be useful as a chemical control agent its is the most potent  with a (water) solution containing 70% isopropyl alcohol.

Ethyl alcohol and isopropyl alcohol are the most commonly used agents.

-true

Dipping small instruments in ethyl alcohol for a few seconds will result in disinfection.

- ethyl alcohol is denaturing proteins, inhibiting metabolic processes so it has bactericidal and fungicidal properties. However the proces is not so fast so the instruments are not disinfected in few seconds.

Alcohols will eliminate bacterial spores.

-alcohols are not sporicidal, however they do inhibit the processes of sporulation and germination but they do not eliminate the spores.

Answer:

5.45 g is 0.0221 moles of magnesium sulfate heptahydrate.

Explanation:

The molecular formula of magnesium sulfate heptahydrate is Mg SO₄·7H₂O.

The molar mass of this compound is calculated adding the molar mass of each element:

Mg: 24. 3 g

S: 32. 1 g

O: 16 g

H: 1 g

Then the mass of a mole of Mg SO₄·7H₂O is:

24.3 g + 32.1 g +4(16 g) + 7(2(1) + 16) = 246.4 g.

if 246.4 g is 1 mole of Mg SO₄·7H₂O, then 5.45 g will be:

5.45 g *(1 mol / 246.4 g) = 0.0221 mol.

Answer: The volume of oxygen gas at STP is 446 mL

Explanation:

STP conditions are:

Pressure of the gas = 1 atm

Temperature of the gas = 273 K

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas

[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas

We are given:

[tex]P_1=1.50atm\\V_1=346mL\\T_1=45^oC=(45+273)K=318K\\P_2=1atm\\V_2=?\\T_2=273K[/tex]

Putting values in above equation, we get:

[tex]\frac{1.50atm\times 346mL}{318K}=\frac{1atm\times V_2}{273K}\\\\V_2=446mL[/tex]

Hence, the volume of oxygen gas at STP is 446 mL

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 346 mL,     [tex]T_{1}[/tex] = [tex]45.0^{o}C[/tex] = (45 + 273) K = 318 K,

      [tex]P_{1}[/tex] = 1.50 atm,   [tex]V_{2}[/tex] = ?,   [tex]T_{2}[/tex] = 273 K,

      [tex]P_{2}[/tex] = 1 atm

And, according to ideal gas equation,

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

Now, putting the given values into the above formula and we will calculate the final volume as follows.

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

           [tex]\frac{1.50 atm \times 346 mL}{318 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]

            [tex]V_{2}[/tex] = 445.56 mL

Thus, we can conclude that the volume of this [tex]O_{2}[/tex] gas sample at STP is 445.56 mL.

Answer:

The pump work is 3451 kJ/s

Explanation:

Pump work (W) is calculated as

[tex] W = (h_f - h_i) \times \dot{m}[/tex]

where

[tex] h_f [/tex] is the enthalpy of water at its final state

[tex] h_i [/tex] is the enthalpy of water at its initial state

[tex]\dot{m} = 10 kg/s [/tex] is water mass flow

For liquids, properties are evaluated as saturated liquid. From the figure attached, it can be seen that

[tex] h_f = 762.81 kJ/kg [/tex]

[tex] h_i = 417.46 kJ/kg [/tex]

Replacing

[tex] W = (762.81 kJ/kg - 417.46 kJ/kg) \times 10 kg/s[/tex]

[tex] W = 3451 kJ/s [/tex]

Answer:

[α] = -77.5° / [tex]\frac{\textup{dm-g}}{\textup{mL}}[/tex]

Explanation:

Given;

Mass of optically pure substance in the solution = 10 g

Volume of water = 500 mL

Length of the polarimeter, l = 20 cm = 20 × 0.1 dm = 2 dm

measured rotation = - 3.10°

Now,

The specific rotation ( [α] ) is given as:

[α] = [tex]\frac{\alpha}{c\times l}[/tex]

here,

α is the measured rotation = -3.10°

c is the concentration

or

c = [tex]\frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}[/tex]

or

c =  [tex]\frac{10}{500}[/tex]

or

c = 0.02 g/mL

on substituting the values, we get

[α] = [tex]\frac{-3.10^o}{0.02\times2}[/tex]

or

[α] = -77.5° / [tex]\frac{\textup{dm-g}}{\textup{mL}}[/tex]

Answer : The molecular weight of a substance is 157.3 g/mol

Explanation :

As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.

Mass of solute = 7 g

Mass of solution = 100 g

Mass of solvent = 100 - 7 = 93 g

Formula used :  

[tex]\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}[/tex]

where,

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

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

[tex]T_f[/tex] = temperature of solution = [tex]-0.89^oC[/tex]

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

m = molality

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

[tex](0-(-0.89))^oC=1.86^oC/m\times \frac{7g\times 1000}{\text{Molar mass of substance(solute)}\times 93g}[/tex]

[tex]\text{Molar mass of substance(solute)}=157.3g/mol[/tex]

Therefore, the molecular weight of a substance is 157.3 g/mol

Explanation:

The given data is as follows.

                 [tex]mass_{1}[/tex] = 5 g,        [tex]Volume_{1}[/tex] = 1 L

                 [tex]mass_{2}[/tex] = 1 g,         [tex]Volume_{1}[/tex] = ?

No. of moles of helium present in 5 g helium gas are as follows.

                   No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                         = [tex]\frac{5 g}{4 g/mol}[/tex]  

                                         = 1.25 mol

No. of moles of helium present in 1 g helium gas are as follows.

                 No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                         = [tex]\frac{1 g}{4 g/mol}[/tex]  

                                         = 0.25 mol

According to the ideal gas equation, PV = nRT. And, since temperature and pressure are held constant. Therefore,

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

                      [tex]\frac{1 L}{1.25 mol} = \frac{V_{2}}{0.25 mol}[/tex]

                             [tex]V_{2}[/tex] = 0.2 L

Thus, we can conclude that the new volume of the balloon is 0.2 L.

The final volume of the balloon after adding 5.00 g of helium to the already present 1.00 g of helium, under constant temperature and pressure, is 6.00 L.

The subject of your question pertains to the ideal gas law which is represented by the formula PV = nRT, where P is pressure, V is volume, n is the number of moles, R refers to the gas constant, and T is temperature. Since the question states that there is no change in temperature or pressure, you can use Avogadro's law which states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.

In your problem, you are adding more helium to the balloon. Therefore, the volume of the balloon will increase proportionally. Since the initial volume of 1.00 L corresponds to 1.00 g of Helium, 5.00 g of Helium will correspond to 5.00 L. This is because the amount of Helium has increased by a factor of 5, and thus, the volume will also increase by a factor of 5.

Therefore, the final volume of the balloon will be the sum of the initial volume and the increase, which equals 1.00 L + 5.00 L = 6.00 L.

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Answer:

Amount of Carbon dioxide equals 4.49625 grams.

Explanation:

From the basic stichometric reaction between carbon and oxygen we know that 1 mole of carbon combines with 1 mole of oxygen to form 1 mole of carbon dioxide.

Thus we can say that 12 grams of carbon combines with 32 grams of oxygen to form 44 grams of carbon dioxide.

In the given question assuming that there is no limited supply of carbon we can find the find the amount of carbon dioxide formed from 3.27 grams of Oxygen using ratio and proportion method.

As we can see that 32 grams of oxygen form 44 grams of carbon dioxide thus we can say 1 gram of oxygen yields [tex]\frac{44}{32}grams[/tex] Carbon Dioxide

Thus the carbon dioxide formed by 3.27 grams of Oxygen equals

[tex]3.27\times \frac{44}{32}=4.49625grams[/tex]

Answer:

[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]

Explanation:

given data:

temperature  = 552 degree celcius  = 825 Kelvin

energy for vacancy formation is given as 0.63 eV/atom

fraction of atom can be obtained from following formula

[tex]N_C = N e^{\frac{-Q_V}{kT}}[/tex]

Where, K IS BOLTZMAN CONSTANT [tex]= 8.62\times 10^{-5}eV/K[/tex]

[tex]\frac{N_C}{N} =e^{\frac{-0.63}{8.62\times 10^{-5} \times 825}[/tex]

[tex]\frac{N_C}{N} = e^{-8.8858}[/tex]

[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]

The fraction of atom sites that are vacant for silver at 552°C is approximately [tex]\( 7.347 \times 10^{-5} \).[/tex]

The fraction of atom sites that are vacant for silver at a given temperature can be calculated using the formula derived from the statistics of a canonical ensemble:

[tex]\[ f = \exp\left(-\frac{\Delta G_{\text{form}}}{k_B T}\right) \][/tex]

Now, we can plug in the values:

[tex]\[ f = \exp\left(-\frac{0.63 \text{ eV/atom}}{(8.617 \times 10^{-5} \text{ eV/K}) \times (825.15 \text{ K})}\right) \] \[ f = \exp\left(-\frac{0.63}{8.617 \times 10^{-5} \times 825.15}\right) \] \[ f = \exp\left(-\frac{0.63}{0.0658}\right) \] \[ f = \exp\left(-9.574\right) \] \[ f \approx \exp\left(-9.574\right) \] \[ f \approx 7.347 \times 10^{-5} \][/tex]

The given data suggests that the decay of Chlorine oxide (ClO) is a first-order reaction. The rate constant can be calculated using the first-order rate law equation, which in this case gives a value of approximately 70000 s⁻¹.

The reaction order is determined by the relationship between the rate of reaction and the concentration of the reactants. By observing the given data, it appears that the decay of Chlorine oxide (ClO) is halving approximately. This suggests that it could be a first-order reaction, where the rate of the reaction is directly proportional to the concentration of one reactant.

To calculate the rate constant of the reaction, we can use the first-order rate law equation: k = -1/[t]*ln([A]t/[A]0), where 'k' is the rate constant, '[t]' is the elapsed time, '[A]t' is the concentration at time 't' and '[A]0' is the initial concentration.

Using the initial and final concentrations given (at 4.26 × 10−3 s and 6.74 × 10−3 s), the equation for the rate constant becomes: k = -1/(6.74 × 10⁻³ - 4.26 × 10⁻³)*ln((4.01 × 10⁻⁶)/(7.73 × 10⁻⁶)). Calculating gives a rate constant value of approximately 70000 s-1. Remember, these values may vary depending on specific calculation approaches.

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The decomposition of ClO is determined to be first-order with a rate constant of approximately 0.263 s⁻¹. We used the method of initial rates and plotting ln[ClO] versus time to ascertain the reaction order and calculate the rate constant.

To determine the reaction order and rate constant for the decomposition of chlorine oxide (ClO) according to the equation 2ClO(g) → Cl₂(g) + O₂(g), we need to analyze the given concentration data over time.

The reaction is **first-order** with respect to ClO, and the rate constant, **k**, is approximately 0.263 s⁻¹.

Answer:

Biuret test does not work for all kinds of proteins. It only works with proteins and peptides that have 2 or more peptide bonds.

Explanation:

The Biuret test is used to determine proteins and polypeptides that contain 2 or more peptide bonds.

It is made of potassium hydroxide and cupric sulfate, along with sodium and potassium tartrate.

It consists of treating a protein with Cu ++ in alkaline medium, producing a violet coloration by forming a coordination complex between the Cu ++ and the free electron pairs of the nitrogens of the amino groups of the peptide junction. At least two peptide bonds are necessary for the reaction to take place.

Therefore, this test does not serve to determine the amount of amino acids and peptides with a single peptide bond such as: tyrosine (amino acid), tryptophan (amino acid), alanine (amino acid), aspartame (peptide)

Answer: The number of atoms of carbon present in given number of moles are [tex]2.350\times 10^{20}[/tex]

Explanation:

We are given:

Number of moles of carbon = [tex]3.900\times 10^{-4}mol[/tex]

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, [tex]3.900\times 10^{-4}mol[/tex] of carbon will contain = [tex]3.900\times 10^{-4}\times 6.022\times 10^{23}=2.350\times 10^{20}[/tex] number of atoms.

Hence, the number of atoms of carbon present in given number of moles are [tex]2.350\times 10^{20}[/tex]

Answer : The correct option is, (A) -x kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

[tex]A\rightarrow P[/tex] [tex]\Delta H=x\text{ kJ/mole}[/tex]

Now we have to determine the value of [tex]\Delta H[/tex] for the following reaction i.e,

[tex]P\rightarrow A[/tex] [tex]\Delta H'=?[/tex]

According to the Hess’s law, if we reverse the reaction then the sign of [tex]\Delta H[/tex] change.

So, the value [tex]\Delta H'[/tex] for the reaction will be:

[tex]\Delta H'=-(x\text{ kJ/mole})[/tex]

[tex]\Delta H'=-x\text{ kJ/mole}[/tex]

Hence, the value of [tex]\Delta H[/tex] for the reaction is -x kJ/mole.

Answer:

Head loss in turbulent flow is varying as square of velocity.

Explanation:

As we know that head loss in turbulent flow given as

[tex]h_F=\dfrac{FLV^2}{2gD}[/tex]

Where

F is the friction factor.

L is the length of pipe

V is the flow velocity

D is the diameter of pipe.

So from above equation we can say that

[tex]h_F\alpha V^2[/tex]

It means that head loss in turbulent flow is varying as square of velocity.

We know that loss in flow are of two types

1.Major loss :Due to surface property of pipe

2.Minor loss :Due to change in momentum of fluid.

Answer:

The correct answer is head varies directly with square of velocity of flow

Explanation:

The head loss in a pipe as given by Darcy Weisebach equation is

[tex]h_L=\frac{flv^2}{2gD}[/tex]

where

'f' is friction factor whose value depends on the nature of flow (Laminar/turbulent)

'L' is the length of the section in which the head loss is calculated

'v' is the velocity of the flow

'D' is the diameter of the duct

Thus we can see that the head loss varies with square of velocity of the fluid.