How much power (energy per unit time) can be provided by a 75 m high waterfall with a flow rate of 10,000 L/s? Give answer in kW rate given here is volume per unit time; 10,000 L/s mean that every second 10,000 L of water go through the water fall

Answer:

20.7

Explanation:

20.7298014 rounded off to 3 sig fig =20.7

Answer:

The dissociation equations for NaBr gives Na+ and Br-

The dissociation equations for ZnCl2 gives Zn2+ and 2 Cl-  

Explanation:

The following pictures shows that the dissociation of one particle of NaBr produces one particle of Na+ (sodium cation) and one particle of Br- (bromine anion).

The dissociation of one particle of ZnCl2 produces one particle of Zn+2 (Zinc cation) and two particles of Cl- (chlorine anion).

Answer : The formula of limiting reagent is, [tex]Mg_3N_2[/tex].

The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.

The mass of excess reactant is, 36 grams.

Solution : Given,

Mass of [tex]Mg_3N_2[/tex] = 101 g

Mass of [tex]H_2O[/tex] = 144 g

Molar mass of [tex]Mg_3N_2[/tex] = 101 g/mole

Molar mass of [tex]H_2O[/tex] = 18 g/mole

Molar mass of [tex]Mg(OH)_2[/tex] = 58 g/mole

First we have to calculate the moles of [tex]Mg_3N_2[/tex] and [tex]H_2O[/tex].

[tex]\text{ Moles of }Mg_3N_2=\frac{\text{ Mass of }Mg_3N_2}{\text{ Molar mass of }Mg_3N_2}=\frac{101g}{101g/mole}=1moles[/tex]

[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{144g}{18g/mole}=8moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]Mg_3N_2+6H_2O\rightarrow 3Mg(OH)_2+2NH_3[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]Mg_3N_2[/tex] react with 6 mole of [tex]H_2O[/tex]

So, given 1 moles of [tex]Mg_3N_2[/tex] react with 6 moles of [tex]H_2O[/tex]

From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Mg_3N_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]Mg(OH)_2[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]Mg_3N_2[/tex] react to give 3 mole of [tex]Mg(OH)_2[/tex]

So, given 1 mole of [tex]Mg_3N_2[/tex] react to give 3 moles of [tex]Mg(OH)_2[/tex]

Now we have to calculate the mass of [tex]Mg(OH)_2[/tex]

[tex]\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2\times \text{ Molar mass of }Mg(OH)_2[/tex]

[tex]\text{ Mass of }Mg(OH)_2=(3moles)\times (58g/mole)=174g[/tex]

The mass of [tex]Mg(OH)_2[/tex] is, 174 grams.

Now we have to calculate the moles of excess reactant [tex](H_2O)[/tex].

Moles of excess reactant = 8 - 6 = 2 moles

Now we have to calculate the mass of excess reactant.

[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]

[tex]\text{ Mass of }H_2O=(2moles)\times (18g/mole)=36g[/tex]

The mass of excess reactant is, 36 grams.

Answer:

11.39

Explanation:

Given that:

[tex]pK_{b}=4.82[/tex]

[tex]K_{b}=10^{-4.82}=1.5136\times 10^{-5}[/tex]

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

[tex]Moles= \frac{1.805\ g}{82.0343\ g/mol}[/tex]

[tex]Moles= 0.022\ moles[/tex]

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.022}{0.055}[/tex]

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

[tex]K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}[/tex]

[tex]1.5136\times 10^{-5}=\frac {x^2}{0.4-x}[/tex]

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

Answer: enthalpy of reaction.

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like [tex]^0C[/tex] and [tex]K[/tex]

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

Free energy is the amount of energy that can be converted into useful work.

Enthalpy of the reaction is the difference between the energy of products and the energy of reactants. it is either the heat released or absorbed during the reaction. It is either positive or negative.

Answer: The molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]

Explanation:

We are given:

Empirical formula of the compound = [tex]CH_2O[/tex]

Empirical mass of the compound = [tex][(1\times 12)+(2\times 1)+(1\times 16)]=30g/mol[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

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

We are given:

Mass of molecular formula = 180.12 g/mol

Mass of empirical formula = 30 g/mol

Putting values in above equation, we get:

[tex]n=\frac{180.12g/mol}{30g/mol}=6[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 6)}H_{(2\times 6)}O_{(1\times 6)}=C_6H_{12}O_6[/tex]

Hence, the molecular formula of glucose is [tex]C_6H_{12}O_6[/tex]

Answer: Option (d) is the correct answer.

Explanation:

According to Bronsted-Lowry an acid is defined as the specie which is able to donate hydrogen ions when dissolved in water.

For example, [tex]HCl \rightarrow H^{+} + Cl^{-}[/tex]

On the other hand, bases are the species which are able to donate hydroxide ions when dissolved in water.

For example, [tex]NaOH \rightarrow Na^{+} + OH^{-}[/tex]

Thus, we can conclude that a definition of an acid is that it donates H ions when dissolved in water.

Answer:

(a) Pressure at point 200 m downstream: 113.4 kPa

(b) Pressure at point 200 m downstream: 105.4 kPa

Explanation:

The point of reference, that will be named Point A, is the pressure gage.

The point 200 m downstream will be named Point B.

With a slope of 1:50, the height difference between A and B is (1/50)*200=4m.

(a) In the case there is no friction losses, we can write

[tex]p_a+\frac{\rho V^2}{2} +\rho*g*h_a=p_b+\frac{\rho V^2}{2} +\rho*g*h_b[/tex]

Because ther is no change in the diameter of the pipe, V=constant.

Rearranging

[tex]p_a+\rho*g*h_a=p_b+\rho*g*h_b\\\\\Delta p = \rho * g * \Delta h\\\\\Delta p = 0.8505 * 1000\frac{kg}{m^3} *9.81\frac{m}{s^2}*4 m\\\\ \Delta p = 33.373.62 kg/(m*s^2)=33.373.62 Pa=33.4 kPa[/tex]

Then we have an increase in of 33.4 kPa, and pressure in Point B (downstream) is:

[tex]p_b=p_a+\Delta p = 80 kPa + 33.4 kPa = 113.4 kPa[/tex]

(b) If we consider 10% of the total initial head as friction loss, we have

[tex]p_a+\rho*g*h_a=p_b+hf +\rho*g*h_b\\\\hf=0.1*p_a\\\\\Delta P = \rho*g*\Delta h-0.1*p_a\\\\\Delta P = 0.8505*1000 kg/m^3*9.81m/s2*4m-0.1*80kPa\\\\\Delta P = 33.4kPa-8kPa=25.4kPa[/tex]

In this case the rise in pressure is 25.4 kPa, due to the friction losses.

The pressure at point B is

[tex]p_b=p_a+\Delta p = 80 kPa + 25.4 kPa = 105.4 kPa[/tex]

Answer: Isotopes are the species which have same number of protons but differs in the number of neutrons. Or in other words isotopes have same atomic number but different mass number. Isotopes are generally unstable and decays to an stable isotope. When the isotope is unstable it releases energy in the form of radiations. It emits alpha particle, gamma rays, and beta particles.

Therefore, the correct option is (A),(B) and (C)

Answer:

23.4 mph

Explanation:

The conversion factors are multiplied so that the units cancel out:

(100 meters / 9.58 sec) x (1 mile / 1609 meters) x (60 sec / min) x (60 min /1 hr) = 23.4 mph

Answer:

6.113 atm

Explanation:

Data provided in the question:

Molarity of the solution = 0.25 M

Temperature, T = 25° C = 25 + 273 = 298 K

Now,

Osmotic pressure (π) is given as:

π = MRT

where,

M is the molarity of the solution

R is the ideal gas constant = 0.082057 L atm mol⁻¹K⁻¹

on substituting the respective values, we get

π = 0.25 × 0.082057 × 298

or

π = 6.113 atm

Answer:

(a) more than

Explanation:

Conduction:

Transfer of heat due to direct contact between two mediums at different temperatures, without having any of the bodies traveling. Therefore, conduction heat transfer occurs by the transfer of momentum (molecular) from always the same group of molecules in one medium to another group of molecules in another medium.

Example: Heat transfer INSIDE a solid.

Convention:

Transfer of heat or mass due to at least one traveling medium, where the transfer of momentum is not bounded anymore to the same groups of molecules. Molecules moving to transfer their momentum and keep flowing to the next group, also allowing other molecules behind to do the same. Example: heat transfer by the wind.

Hence, the Mass transfer rate in convection is more than mass transfer in conduction

Final answer:

The mass transfer rate in convection is generally more than (option a) that in conduction due to the macroscopic movement of mass that facilitates faster heat transfer.

Explanation:

The question asks to compare the mass transfer rate in convection to the mass transfer rate in conduction. The correct comparison is that the mass transfer rate in convection is generally more than the mass transfer rate in conduction. This is because conduction is the process by which heat is directly transmitted through a substance when there is a difference of temperature, without movement of the material. In contrast, convection involves the movement of mass (typically a fluid such as air or water) which contributes to a faster transfer of heat. An empirical equation for the rate of heat transfer by forced convection can be represented by Q/t = hA(T₂ – T₁), showing that this rate is dependent on the temperature difference (T₂ – T₁), contact surface area (A), and the convective heat transfer coefficient (h).

Answer:

a.

silver iodide BC

manganese(II) hydroxide A

b.

silver iodide Ksp = S²

manganese(II) hydroxide Ksp = 4S³

Explanation:

Ksp (Solubility products) are the equilibrium constants for poorly soluble compounds. As every equilibrium constant, it is formed by the product of the products raised to their stoichiometric coefficients divided by the product of reactants raised to their stoichiometric coefficients. We only include in the constant gases and aqueous species. So, to solve this task, we need to write each reaction and its Ksp.

1. Silver Iodide

AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)

When AgI is put in water, the concentration of Ag⁺ and I⁻ that actually dissolve is known as solubility (S). So the concentration of both Ag⁺ and I⁻ would be S.

AgI(s) ⇄ Ag⁺(aq) + I⁻(aq)

                 S             S

We can replace this in the Ksp expression:

Ksp = [Ag⁺].[I⁻] = S.S = S²

We can follow the same steps to find out the relationship between Ksp and S for each compound.

2. Mn(OH)₂(s) ⇄ Mn²⁺(aq) + 2OH⁻(aq)

                                S             2S

In this case, the concentration of OH⁻ is 2S because 2 moles are produced along with 1 mole of Mn²⁺.

Ksp = [Mn²⁺].[OH⁻]² = S.(2S)² = 4S³

A. Fe(OH)₂(s) ⇄ Fe²⁺(aq) + 2OH⁻(aq)

                              S              2S

Ksp = [Fe²⁺].[OH⁻]² = S.(2S)² = 4S³

B. CaSO₃(s) ⇄ Ca²⁺(aq) + SO₃²⁻(aq)

                             S              S

Ksp = [Ca²⁺].[SO₃²⁻] = S.S = S²

C. NiCO₃(s) ⇄  Ni²⁺(aq) + CO₃²⁻(aq)

                            S              S

Ksp = [Ni²⁺].[CO₃²⁻] = S.S = S²              

D. Ba₃(PO₄)₂(s) ⇄ 3 Ba²⁺(aq) + 2 PO₄³⁻(aq)

                                   3S              2S

Ksp = [Ba²⁺]³.[PO₄³⁻]²= (3S)³.(2S)²= 108S⁵

a. The salts that can be compared using the same Ksp expressions are:

silver iodide BC

manganese(II) hydroxide A

b.

silver iodide Ksp = S²

manganese(II) hydroxide Ksp = 4S³

None of the listed salts can be compared to silver iodide using Ksp values. Manganese(II) hydroxide can be compared to Fe(OH)2. The Ksp expressions in terms of solubility, s, are s^2 for silver iodide and 4s^3 for manganese(II) hydroxide.

We can directly compare salts by considering their solubility product constants, or Ksp values

. 1. Silver iodide: No salts on the right share the same ions with silver iodide, so none can be directly compared.
2. Manganese(II) hydroxide: It contains Mn2+ and OH- ions. Thus, it can be compared to the Fe(OH)2 salt which contains Fe2+ and OH- ions, making the matching answer A.

For the solubility expressions for each salt, when Multiplied we put the number first and then combine all exponents:

For silver iodide (AgI), the equilibrium is AgI ↔ Ag+ + I-, so the Ksp expression is Ksp = s x s = s^2.

For manganese(II) hydroxide (Mn(OH)2), the equilibrium is Mn(OH)2 ↔ Mn2+ + 2OH-, so the Ksp expression is Ksp = 4s^3.

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Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

[tex]S.G=\frac{D}{d_w}[/tex]

[tex]d_w[/tex] = density of water = 1 g/mL

[tex]D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL[/tex]

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

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

[tex]=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L[/tex]

1 mL = 0.001 L

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

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

[tex]Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L[/tex]

Answer : The  speed in miles per hour is 22 mile/hr.

Explanation :

The conversion used from meters to miles is:

[tex]1m=100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}[/tex]

The conversion used from second to hour is:

[tex]1s=\frac{1}{60}min\times \frac{1hr}{60min}[/tex]

The conversion used from meter per second to mile per hour is:

[tex]1\frac{m}{s}=\frac{100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}}{\frac{1}{60}min\times \frac{1hr}{60min}}[/tex]

[tex]1m/s=2.2mile/hr[/tex]

As we are given the speed 10.00 meter per second. Now we have to determine the speed in miles per hour.

As, [tex]1m/s=2.2mile/hr[/tex]

So, [tex]10.00m/s=\frac{10.00m/s}{1m/s}\times 2.2mile/hr=22mile/hr[/tex]

Therefore, the speed in miles per hour is 22 mile/hr.

Answer:

[ H2O ]eq = 0.298 mol/L

Explanation:

∴ V = 0.501 L

∴ T = 1043 K

at equilibrium:

∴ n CH4 = 0.255 mol

⇒ [ CH4 ]eq = 0.255 mol /  0.501 L = 0.509 mol/L

∴ n CO = 0.169 mol

⇒ [ CO ]eq = 0.169 / 0.501 = 0.337 mol/L

∴ n H2 = 0.257

⇒ [ H2 ] eq = 0.257 / 0.501 =  0.513 mol/L

∴ Kc = [ H2 ]³ * [ CO ] /  [ CH4 ] * [ H2O ] = 0.30

⇒ [ H2O ] = [ H2 ]³ * [ CO ] / [ CH4 ] * 0.30

replacing the value of the concentration in Kc:

⇒ [ H2O ] = ( 0.513 )³ * ( 0.337 ) / ( 0.509 ) * 0.30

⇒ [ H2O ] = 0.298 mol/L

The equilibrium concentration of water vapor in the conversion of methane to other fuels at 1043 K is determined to be 0.6196 M by using the equilibrium constant expression and the known concentrations of the other substances.

The student's question involves determining the water concentration at equilibrium for the conversion of methane to other fuels involving a reaction of methane (CH4) and water vapor (H2O). First, we need to write a balanced chemical equation for the reaction:

CH4(g) + 2H2O(g) ⇌ CO2(g) + 4H2(g)

To find the water concentration, we use the equilibrium constant expression (Kc):

Kc = [tex][CO]^1[H2]^4 / [CH4]^1[H2O]^2[/tex]

Given that Kc = 0.30 at 1043 K, and the equilibrium amounts of the substances are CH4 = 0.255 mol, CO = 0.169 mol, and H2 = 0.257 mol in a 0.501 L flask, we can calculate their concentrations and solve for [H2O]:

[CH4] = 0.255 mol / 0.501 L = 0.509 M

[CO] = 0.169 mol / 0.501 L = 0.337 M

[H2] = 0.257 mol / 0.501 L = 0.513 M

Plugging these into the Kc expression and solving for [H2O], we get:

0.30 = (0.337)[([tex]0.513)^4] / [(0.509)([H2O]^2)][/tex]

After calculations, the water concentration at equilibrium, [H2O], is found to be approximately 0.6196 M.

Explanation:

(a)   According to the mole concept, 1 mole of an atom contains [tex]6.022 \times 10^{23}[/tex] atoms.

Hence, number of atoms present in [tex]4.48 \times 10^{-23}[/tex] g will be as follows.

                 [tex]4.48 \times 10^{-23} \times 6.022 \times 10^{23}[/tex]

                 = 26.97 g

or,              = 27 g

This means that 1 mole of aluminium atoms contain 27 g.

(b)     Mass of air plane is 5000 kg of [tex]5000 \times 1000 g[/tex] (as 1 kg = 1000 g).

As mass of 1 mole aluminium is calculated as 27 g and mass of air plane is given as 5000000 g.

Therefore, calculate the number of moles of aluminium atoms as follws.

               No. of moles of Al atoms = [tex]\frac{\text{mass of small air plane}}{\text{mass of 1 mol Aluminium}}[/tex]

                                             = [tex]\frac{5000000g}{27 g}[/tex]

                                              = 185185.18

So, the answer in three significant digits will be 185000 moles of aluminum atoms have a mass equal to the mass of a small airplane.

The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.

The correct answer is described below

According to the question, the answer of the first question is:-

The number of atoms will be:-

[tex]4.48*10^{-23}*6.022*10^{23}\\\\=26.97[/tex]

The correct answer is 27

The answer to the second question is as follows:-

The mass of the air is [tex]5000*1000g[/tex]. hence, As the mass of 1-mole aluminum is calculated as 27 g and mass of airplane is given as 5000000 g.

No moles will be

:- [tex]\frac{500000}{27} \\\\=185185.18[/tex]

Hence, the correct answer is mentioned above.

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Answer:

The concentration of chloride ions in the final solution is 3 M.

Explanation:

The number of moles present in a solution can be calculated as follows:

number of moles = concentration in molarity * volume

In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻

For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:

number of moles of Cl⁻ = 2 * number of moles of  CaCl₂

number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻

The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.

Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:

Concentration = number of moles of Cl⁻ / volume

Concentration = 150.2 mol / 50.1 l = 3.0 M

To find the molar concentration of chloride ions in the solution, calculate the moles of chloride ions in each solution and add them together, then divide by the total volume of the final solution.

To determine the molar concentration of chloride ions in the solution, we need to calculate the number of moles of chloride ions and divide it by the total volume of the solution.

First, calculate the moles of chloride ions in KCl solution:

Moles of chloride ions = molarity of KCl solution * volume of KCl solution

Next, calculate the moles of chloride ions in CaCl2 solution:

Moles of chloride ions = (molarity of CaCl2 solution * volume of CaCl2 solution) * 2 (since CaCl2 has 2 chloride ions per formula unit)

After finding the moles of chloride ions in each solution, add them together and divide by the total volume of the final solution to get the molar concentration of chloride ions.

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Final answer:

The importance of an energy balance in a chemical process facility lies in ensuring efficient energy use, maintaining safety, optimizing chemical processes for increased yield, and reducing environmental impact.

Explanation:

An energy balance is crucial in a chemical process facility because it ensures the efficient use of energy, minimizes waste, and helps maintain the safety of the operation. Since energy is neither created nor destroyed (law of conservation of energy), it's important to know where energy is being consumed and produced within chemical processes. This understanding allows engineers to optimize the process, increase the yield of the desired product, and reduce environmental impact by minimizing wasted energy and reducing unwanted by-products.

An energy balance aids in maintaining the operation within the desired range of conditions, preventing unsafe levels of energy which could lead to accidents. Moreover, certain reactions require specific amounts of energy to produce raw materials or to synthesize products, and tracking energy use is essential for economic and environmental reasons.

Answer:

The correct answer is option a.

Explanation:

Diffusion flux is defined as movement of mas of atoms diffusing through the unit area in per unit time.It measured in ([tex]kg/m^2 s[/tex]).

[tex]J=\frac{1}{A}\frac{dM}{dt}[/tex]

J = diffusion flux

A = Unit area A through which atoms moves.

M = mass of atoms passes in t interval of time.

Diffusion flux is the mass of gas passing through a unit surface area per unit time, influenced by the concentration gradient, surface area, and particle travel distance.

The diffusion flux can be defined as the amount of gas passing through a given area over a unit of time. Therefore, the correct answer is a) Mass, per unit surface area, per unit time. The rate of diffusion is dictated by several factors such as the concentration gradient, the surface area available for diffusion, and the distance the gas particles must travel. In addition, it's important to note that the time required for diffusion is inversely proportional to the diffusion rate.

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Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and [tex]7.2m^3[/tex] respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

[tex]P_1[/tex] = initial pressure of gas = 10 kPa

[tex]P_2[/tex] = final pressure of gas = 15 kPa

[tex]V_1[/tex] = initial volume of gas = [tex]10m^3[/tex]

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]50^oC=273+50=323K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]75^oC=273+75=348K[/tex]

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

[tex]\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}[/tex]

[tex]V_2=7.2m^3[/tex]

The new volume of carbon dioxide gas is [tex]7.2m^3[/tex]

Now we have to calculate the new density of carbon dioxide gas.

[tex]PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}[/tex]

Formula for new density will be:

[tex]\rho_2=\frac{P_2M}{RT_2}[/tex]

where,

[tex]P_2[/tex] = new pressure of gas = 15 kPa

[tex]T_2[/tex] = new temperature of gas = [tex]75^oC=273+75=348K[/tex]

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

[tex]\rho[/tex] = new density

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

[tex]\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}[/tex]

[tex]\rho_2=0.2281g/L[/tex]

The new density of carbon dioxide gas is 0.2281 g/L

Answer:

ΔHf = 3301 kJ/mol

Explanation:

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

In a reaction ΔHf = ∑ n ΔHproducts - ∑ n ΔHreagents

Where n are moles

For the reaction:

1 C₆H₆ + 7.5 O₂ → 6 CO₂ + 3 H₂O

The ΔHf = -285,8 kJ/mol × 3 - 393,5 kJ/mol×6 - [82,6 kJ/mol×1 -0kJ/mol×7,5]

ΔHf = 3301 kJ/mol

I hope it helps!

Answer:

The correct option is: c. petroleum jelly, d. Polyethylene glycol 4000/600 mixture                

Explanation:

Topical medications are used for the treatment of ailments and include ointments, gels, lotions creams etc. that can applied directly on the surface of the body i.e. skin.

An ointment base medication gets rapidly absorbed into the skin. Some of the examples of ointment bases include water-soluble bases: polyethylene glycol, hydrocarbon bases: petroleum jelly, paraffin wax.

Answer:

yo mama

Explanation:

Answer:

It's an open system, tranfering heat through a rigid, diathermal wall and matter through an imaginary and permeable wall, and it is not at steady state.

Explanation:

I hope you find interesting and useful this information! good luck!

Answer:

The concentration of the new solution will be 0.31 M

Explanation:

The number of moles per liter of the solution is 0.50 mol. Then, in 50 ml there will be:

50 ml · (0.50 mol / 1000 ml) = 0.025 mol.

If we add 30 ml and assuming that the solution is an ideal solution, the final volume will be 80 ml.

Then, 0.025 mol will be present in 80 ml solution. In 1 l there will be:

1000 ml · (0.025 mol / 80 ml) = 0.31 mol.

The concentration of the new solution will be 0.31 M.

Answer:

200 mL = 200 cm³

Explanation:

The relationship between cm³ and mL is 1:1.

1 cm³ = 1 mL

Thus, 200 mL is converted to cm³ as follows:

(200 mL)(1 cm³/1 mL) = 200 cm³

Answer:

8.125 atm

Explanation:

Hello,

Boyle's law states that at constant temperature:

[tex]P_1V_1=P_2V_2[/tex]

In this case:

[tex]P_1=1 atm, V_1=27.3L, V_2=3.36L[/tex]

Solving for [tex]P_2[/tex]:

[tex]P_2=\frac{P_1V_1}{V_2}=\frac{1atm*27.3L}{3.36L}\\P_2=8.125atm[/tex]

Best regards!

Answer:

The weight percent of NaBr is 25,7%

Explanation:

Molality is a way to express the concentration of a chemical in terms of moles of substances per kg of solution. Weight percent is other way to express concentration in terms of mass of substance per mass of solution per 100

Thus, to obtain weight percent you must convert moles of NaBr to grams with molar mass and these grams to kilograms, thus:

2,50 mol / kg solution × (102,9 g / 1mol) × (1 kg / 1000 g) =

[tex]\frac{0,257 kg NaBr}{kg solution }[/tex] × 100 = 25,7 % (w/w)

I hope it helps!

Answer:

0.182 moles of acetic acid are needed, this means 10.93 g.

0.318 moles of sodium acetate are needed, this means 26.08 g.

Explanation:

The Henderson–Hasselbalch (H-H) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:

pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.

From the problem we know that

[A⁻] + [HA] = 250 mM = 0.250 M     eq. 1

We use the H-H equation, using the data we know, to describe [A⁻] in terms of [HA]:

5.0 = 4.76 + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

[tex]0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}][/tex]        eq.2

Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:

1.74 [HA] + [HA] = 0.250 M

[HA] = 0.091 M

Then we calculate [A⁻]:

[A⁻] + 0.091 M = 0.250 M

[A⁻] = 0.159 M

Using the volume, we can calculate the moles of each substance:

Using the molecular weight, we can calculate the grams of each substance:

Answer:

The amount of released is 1,182 kJ.

Explanation:

When heat is released at constant pressure, this change in energy is known as enthalpy (ΔH°) of the reaction. Enthalpy is an extensive property, so it depends on the amount of reacting material. Let's take a look at the provided equation:

2 NO(g) + O₂ ⇄ 2 NO₂(g)   ΔH° = -114.4 kJ

Since this equation is balanced with 2 moles of NO₂(g), we can say that 114.4 kJ are released every 2 moles of NO₂(g) produced. By convention, when enthalpies are negative, it means that energy is released and the reaction is exothermic. Conversely, positive enthalpies mean energy is absorbed and the reaction is endothermic.

We can calculate the amount of energy released taking into account the previous relationship (-114.4 kJ/2 moles of NO₂(g)), the mass of NO₂(g) produced (951.1g) and its molar mass (46.00g/mol). The calculations would be:

[tex]951.1g.\frac{1molNO_{2} }{46.00g} .\frac{-114.4kJ}{2molesNO_{2} } =-1,182kJ[/tex]

The amount of heat released at constant pressure when 951.1 g of [tex]NO_2[/tex] is produced is approximately -544.5 kJ.

The molar mass of [tex]NO_2[/tex] can be calculated from its molecular weight:

Nitrogen (N) has an atomic mass of approximately 14.01 g/mol,

Oxygen (O) has an atomic mass of approximately 16.00 g/mol.

Thus, the molar mass of [tex]NO_2[/tex] is:

[tex]\[ \text{Molar mass of NO}_2 = 14.01 \text{ g/mol} + 2 \times 16.00 \text{ g/mol} = 46.01 \text{ g/mol} \][/tex]

Now, we can calculate the number of moles of [tex]NO_2[/tex] produced:

[tex]\[ \text{moles of NO}_2 = \frac{\text{mass of NO}_2}{\text{molar mass of NO}_2} = \frac{951.1 \text{ g}}{46.01 \text{ g/mol}} \approx 20.67 \text{ mol} \][/tex]

The thermochemical equation given is:

[tex]\[ 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g); \quad \Delta H = -114.4 \text{ kJ} \][/tex]

This equation tells us that 2 moles of NO2 are produced for every -114.4 kJ of heat released at constant pressure. Therefore, the heat released for the production of 20.67 moles of NO2 is:

[tex]\[ \text{Heat released} = \text{moles of NO}_2 \times \frac{-114.4 \text{ kJ}}{2 \text{ mol}} \][/tex]

[tex]\[ \text{Heat released} = 20.67 \text{ mol} \times \frac{-114.4 \text{ kJ}}{2 \text{ mol}} \approx -544.5 \text{ kJ} \][/tex]