A turbine operating adiabatically is fed with steam at 400 °C and 8.0 MPa at the rate of 1000 kg/h. Process steam saturated at 0.5 MPa is withdrawn from an intermediate location in the turbine at a rate of 300 kg/h, and the remaining steam leaves the turbine saturated at 0.1 MPa. The kinetic energies and differences in potential energies of all streams are negligible. What is the power output of the turbine?

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

                [tex]F_{net} = Buoyancy force(F_{b}) - weight force(w)[/tex]

Also, buoyancy force [tex](F_{b})[/tex] = (volume submerged in water × density of water) + (volume in oil × density of oil)

          [tex](F_{b})[/tex] = [tex](0.592 V \times \rho) + (1 - 0.592)V \times 1000 g[/tex]          

                      = [tex](0.592 V \times \rho + 408 V)[/tex] g

As,   W = V × density of graphite × g

It is given that density of graphite is [tex]2.16 g/cm^{3}[/tex] or 2160 [tex]kg/m^{3}[/tex].

So, W = 2160 V g

[tex]F_{net}[/tex] = (0.592 V \rho + 408 V) g - 2160 V g = 0

            [tex]0.592 \rho[/tex] = 1752

     [tex]\rho[/tex] = 2959.46 [tex]kg/m^{3}[/tex] or 2.959 [tex]g/cm^{3}[/tex] is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 [tex]cm^{3}[/tex] oil = [tex]35.3 \times 2.959[/tex] 104.7 g

Hence, in second weighing total mass will be calculated as follows.

                       (124.8 + 104.7) g

                       = 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

Without the density of the lubricating oil, the second weight of the flask after adding the oil cannot be precisely calculated. The solution would typically involve using the formula mass = density × volume. The focus here is on understanding the principles of buoyancy and density in physics.

The question involves calculating the new weight of a flask after adding 35.3 cm3 of a heavy lubricating oil to an initially weighed empty flask. The missing piece of information needed to calculate this directly is the density of the heavy lubricating oil.

However, the question hints at a connection to Archimedes' principle and force balance, which suggests a focus on the concept of buoyancy and density for solving problems involving fluids. The detailed calculations would typically require the density of the oil, which directly correlates with the mass addition from the volume introduced (mass = density × volume).

Without the density of the oil, we can't calculate the second weighting precisely. However, understanding that the density of a substance is key to solving such problems is crucial in physics, especially when dealing with buoyancy and the principles outlined by Archimedes.

Answer : The correct answer is, [tex]4.878\times 10^{-4}[/tex]

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as [tex]5.0\times 10^3[/tex]

889.9 is written as [tex]8.899\times 10^{-2}[/tex]

In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex]  and [tex]8.899\times 10^{-2}[/tex]  are written in the scientific notation.

[tex]8.89\times 10^{-2}[/tex]  this is written in the scientific notation and the standard notation of this number will be, 0.00889.

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

As we are given the 0.000487750 in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 0.000487750=4.878\times 10^{-4}[/tex]

As, the decimal point is shifting to right side, thus the power of 10 is negative.

Hence, the correct answer is, [tex]4.878\times 10^{-4}[/tex]

Answer:

The system is spontaneous in the reverse direction

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S  

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature  

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,  

i.e. ,  

if  

ΔG < 0 , the reaction is Spontaneous

ΔG > 0 , the reaction is non Spontaneous

ΔG = 0 , the reaction is at equilibrium

from the question ,

∆H = 81.95 kJ/mol

( since , 1 KJ = 1000 J )

∆H = 81950 J/mol

∆S =  27.0 J/(mol-K)

The ∆G is calculated from the above formula -

∆G = ∆H -T∆S  

∆G = (81950 J/mol) - [(298 K) x ( 27.0 J/(mol·K))]

∆G = (81950 J/mol) - (8046 J/mol)

∆G = 73904 J/mol

∆G = 73.904 kJ/mol

Since ΔG > 0 , the system is non spontaneous in the forward direction and hence it will be spontaneous in the reverse direction .

Final answer:

To calculate the Gibbs free energy change (ΔG) at 298 K, use the equation ΔG = ΔH - TΔS. With the provided values, the calculation shows ΔG to be 73.904 kJ/mol, indicating that the reaction is not spontaneous at 298 K and is spontaneous in the reverse direction.

Explanation:

To calculate the Gibbs free energy change (ΔG) for the reaction at 298 K, we use the equation ΔG = ΔH - TΔS. Plugging in the values, we get ΔG = 81.95 kJ/mol - (298 K)(27.0 J/(mol·K)). To get the units consistent, we convert 27.0 J to kJ by dividing by 1000, giving us 0.027 kJ/(mol·K). The calculation now is ΔG = 81.95 kJ/mol - (298 K)(0.027 kJ/(mol·K)), which equals ΔG = 81.95 kJ/mol - 8.046 kJ/mol, resulting in a ΔG of 73.904 kJ/mol.

Since the value of ΔG is positive, the reaction is not spontaneous at 298 K and would be spontaneous in the reverse direction under standard conditions.

Explanation:

Molarity is defined as the number of moles of solute present in liter of solution.

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

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

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

It is given that [tex]M_{1}[/tex] is 0.14 M, [tex]V_{1}[/tex] is 0.083 L, and [tex]V_{2}[/tex] is 1.0 L.

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

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

                 [tex]0.14 M \times 0.083 L = M_{2} \times 1.0 L[/tex]

                      [tex]M_{2} = \frac{0.01162 M.L}{1 L}[/tex]

                                  = 0.01162 M

Thus, we can conclude that the molarity of a solution is 0.01162 M.

Answer:

84.6 mL

Explanation:

To get the total volume of soultion we must add the given volumes together.

4.21 mL + 80.4 mL = 84.6 mL

Answer: The percentage yield of aspirin is 38.02 %.

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 salicylic acid [tex](C_7H_6O_3)[/tex] = 10.09 g

Molar mass of salicylic acid [tex](C_7H_6O_3)[/tex] = 138.12 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of salicylic acid}=\frac{10.09g}{138.12g/mol}=0.0730mol[/tex]

The chemical equation for the formation of aspirin follows:

[tex]C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH[/tex]

As, acetic anhydride is present in excess. So, it is considered as an excess reagent.

Thus, salicylic acid is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of salicylic acid produces 1 mole of aspirin.

So, 0.0730 moles of salicylic acid will produce = [tex]\frac{1}{1}\times 0.0730=0.0730mol[/tex] of aspirin

Now, calculating the mass of aspirin from equation 1, we get:

Molar mass of aspirin = 180.16 g/mol

Moles of aspirin = 0.073 moles

Putting values in equation 1, we get:

[tex]0.073mol=\frac{\text{Mass of aspirin}}{180.16g/mol}\\\\\text{Mass of aspirin}=13.15g[/tex]

To calculate the percentage yield of aspirin, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of aspirin = 5.0 g

Theoretical yield of aspirin = 13.15 g

Putting values in above equation, we get:

[tex]\%\text{ yield of aspirin}=\frac{5.0g}{13.15g}\times 100\\\\\% \text{yield of aspirin}=38.02\%[/tex]

Hence, the percent yield of aspirin is 38.01 %.

Explanation:

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,  

d = density ,

From the question , The density of ethylene glycol = 1.11 g / cm³

The unit  1 cm³ = 1 mL

Hence , density = 1.11 g / mL

( a )

The volume of ethylene glycol = 376 mL

The mass of ethylene glycol can be calculated by using the above formula -

d = m / V

m = d * V

m = 1.11 g / mL * 376 mL

m = 417.36 g

( b )

The mass of ethylene glycol = 3.10 Kg

( since , 1 Kg = 1000 g )

The mass of ethylene glycol = 3.10 Kg = 3.10 * 1000 = 3100 g

The Volume of ethylene glycol can be calculated by using the above formula -

d = m / V

V = m / d

V = 3100 g / 1.11 g / mL

V =  2792.79 mL = 2.79279 L

( since , 1 ml = 1/1000 L )

Answer:

K₅ is smaller because the reaction is slower due to steric and electronic effects.

Explanation:

K is the rate constant of the reaction. The higher the value of log K, the higher the value of K and the faster is the reaction.

The reaction can be represented by the following reaction:

6 NH₃ + [Cu(OH₂)₆]²⁺ → [Cu(NH₃)₆]²⁺ + 6 H₂O

This means that the reaction is exchanging the H₂O ligands by NH₃ ligands.

The more NH₃ ligands we add to the complex, the more difficult (slower) is the substitution. This happend because the addition of NH₃ ligands promotes a steric hindrance and electronic repulsion, which makes it harder for the next NH₃ to approach the complex and substitute the H₂O ligand.

This is the reason why K₅ is negative. The rate of this substitution is extremelhy low.

Answer:

Explanation:

Hello,

At first, consider the balanced reaction for the combustion of the isooctane:

[tex]C_8H_{18}+\frac{25}{2} O_2-->8CO_2+9H_2O[/tex]

Now, the stoichiometric relationship between the hydrocarbon and the oxygen leads to:

[tex]molO_2=230molC_8H_{18}(\frac{\frac{25}{2}mol O_2 }{1mol C_8H_{18}} )=2875molO_2[/tex]

Best regards.

Answer:

Reaction products are shown below

Explanation:

In the Cannizzaro reaction of benzaldehyde, the internal reduction product is an alcohol.

The internal reduction product in the Cannizzaro reaction of benzaldehyde is an alcohol. During the reaction, benzaldehyde undergoes disproportionation in the presence of a strong base like KOH, resulting in the formation of an oxidized carboxylic acid and a reduced alcohol product.

In the Cannizzaro reaction of benzaldehyde, the internal reduction product is an alcohol.

The internal reduction product in the Cannizzaro reaction of benzaldehyde is an alcohol. During the reaction, benzaldehyde undergoes disproportionation in the presence of a strong base like KOH, resulting in the formation of an oxidized carboxylic acid and a reduced alcohol product.

Answer:

Explanation:

You can drive the production rate of an equilibrium reaction by handling several factors: temperature, reactant concentrations, and product concentrations are the main of those factors.

The thermodynamic variable that tells whether a chemical reaction is spontaneous is the free energy, ΔG:

ΔG is related with ΔG° per the expression:

Where Q represents the ratio between the molar concentration of the products (each raised to its stoichiometric coefficient) and the molar concentrations of the reactants (each raised to its stoichiometric coefficient).

Since, you want to increase your production, means you want to favor the forward reaction and that means that you want to make ΔG more negative.

So, you want to make the term RT lnQ more netative.

Logarithm function of a rational expression gets more negative when the numerator decreases or the denominator increases.

So, you want either reduce the amount of products or increase the amount of reactants, which is given by the options B and C:

Enzymes are a kind of catalyst. In an equilibrium reaction, a catalys speeds up both the forward and reverse reaction rates equally, so the equilibrium concentrations will not change. So, adding an enzyme (choice A) would help if you, continually remove products (B) or increase the concentration of reactants (C).

Adding products to get the reaction primed (D) will not help because that just would drive to the consumption of part of the products to obtain some reactants until reaching the equilibrium.

To produce more of an organic molecule from a reaction with a positive ΔG°, the best methods are to continually remove the reaction products and to increase the concentration of reactants. These approaches apply Le Chatelier's principle to push the reaction toward the formation of the desired product.

The student is asking about how to drive a reaction with a positive ΔG° towards the production of a desired product. There are several strategies to shift the equilibrium of a reaction so that the product formation is favored:

However, adding an enzyme that does not couple to another reaction or adding some products initially to get the reaction primed will not effectively push the reaction towards product formation if the reaction has a positive ΔG°. The best strategies to promote the production of the organic molecule of interest are therefore options B and C - continually remove products and increase the concentration of reactants.

Answer:

1,39 L

Explanation:

Charle's Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas, for a constant amount of gas we can write:

[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]

As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:

°K=273+°C

°K=5/9(°F-32)+273

Therefore:

V1=1.5 L, T1=273+25=298°K

V2=?,      T2=5/9(36-32)+273=275,2°K

[tex]V2=\frac{V1T2}{T1}=\frac{275,2*1,5}{298}=1,39L[/tex]

The new volume of the balloon is 1,39 L.

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

[tex]A\rightarrow B[/tex]

Expt    [A] M        [B] M         Rate [M/s]

1          3.40         4.16           1.82*10^-4

2         4.59         4.16           3.32*10^-4

3.        3.40          5.46          1.82*10^-4

[tex]Rate = k[A]^{x}[B]^{y}[/tex]

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

[tex]\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2[/tex]

To find y:

Divide rate of expt 3 by expt 1

[tex]\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0[/tex]

Therefore: x = 2, y = 0

[tex]Rate = k[A]^{2}[B]^{0}[/tex]

To find k

Use rate for expt 1:

[tex]k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1[/tex]

Answer:

[tex]524.5903\mu m^{3}[/tex]

Explanation:

Given that, consider the model of a cell as a sphere.

The size of the human cell is [tex]10\mu m[/tex].

And its radius will become [tex]r_{c}=\frac{10\mu m}{2} =5\mu m[/tex]

And the size of the membrane is [tex]4 nm[/tex].

Now the radius of the cell with the membrane is,

[tex]r=r_{c}+r_{m}\\r=5\mu m+4nm\\r=5\mu m+0.004\mu m\\r=5.004\mu m[/tex]

Now volume with the membrane will be,

[tex]V=\frac{4}{3}\pi r^{3} \\V=\frac{4}{3}\pi (5.004)^{3}\\V=\frac{4}{3}\pi\times 125.30024\\V=524.5903\mu m^{3}\\[/tex]

This is the required volume of cell with the membrane.

Answer:

1383.34 kJ/mol is  the energy released on combustion of the organic compound.

Explanation:

Mass of an organic compound = 0.6654 g

Molar mass of organic compound = 46.07 g/mol

Moles of an organic compound = [tex]\frac{0.6654 g}{46.07 g/mol}=0.01444 mol[/tex]

Let heat evolved during burning of 0.6654 grams of an organic compound be -Q.

Heat absorbed by calorimeter = Q' = -Q

The total heat capacity of the calorimeter all  its contents = C

C = 3576 J/°C

Change in temperature of the calorimeter =  

ΔT = 30.589°C - 25.000°C = 5.589°C

[tex]Q'=C\times \Delta T[/tex]

[tex]Q'=3576 J/^oC\times 5.589^oC=19,975.536 J=19.975 kJ[/tex]

Q' =  19.975 kJ

Q = -19.975 kJ (negative sign; energy released)

0.01444 moles of an organic compound gives 19.975 kilo Joule.

The 1 mole of an organic compound will give : [tex]\Delta H_{comb}[/tex]

[tex]\Delta H_{comb}=\frac{-19.975 kilo Joule}{0.01444 mol}[/tex]

[tex]=-1383.34 kJ/mol[/tex]

Answer:

1. Nonmetals tend to gain electrons in reactions.

4. Nonmetals are poor conductors of heat and electricity.

5. Nonmetals can be found as solids, liquids, or gases.

Non-metals are mostly gases . They tend to gain electrons in reactions. Non-metals are poor conductors of heat and electricity.

Non-metals are elements in periodic table which are located in the right side. They are extremely different from metals and most them are gases and some in between non-metals are metalloids showing properties intermediate metals and non-metals.

Metals are electropositive elements with peculiar features such as ductility, shiny, good conductivity, malleability etc. Metals lose electrons when they bind with other elements.

Non-metals are almost electronegative and they will gain electrons in reactions to attain stability. They are poor conductors or are insulators and have no physical features such as shining, ductility etc. Thus, options 1 and 4 are correct.

To find more on non-metals, refer here:

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Explanation:

It is known that relation between wavelength and frequency is as follows.

                  [tex]\lambda = \frac{c}{\nu}[/tex]

where,        [tex]\lambda[/tex] = wavelength

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

                  [/tex]\nu[/tex] = frequency

It is given that frequency is [tex]7.2 \times 10^{11} s^{-1}[/tex]. Hence, putting this value into the above formula and calculate the wavelength as follows.

                    [tex]\lambda = \frac{c}{\nu}[/tex]

                    [tex]\lambda = \frac{3 \times 10^{8} m/s}{7.2 \times 10^{11} s^{-1}}[/tex]

                             = [tex]0.416 \times 10^{-3} m[/tex]

or,                         = [tex]4.16 \times 10^{-4} m[/tex]

Thus, we can conclude that wavelength of given radiation is [tex]4.16 \times 10^{-4} m[/tex].

Answer:

[tex]t_2 = 14.4 min[/tex]

Explanation:

Given data:

Number of frame is 12+ 8 = 20

Total filtrate is 360 lb

we know

For 1 plate area [tex]= \frac{65}{12} ft^2[/tex]

Then for 20 are PLATE [tex]= 20\tmes \frac{65}{12} ft^2 = 108.33 ft^2[/tex]

filtrate equation is given as

[tex]t\frac{\mu_s\times(\alpha v) V^2}{A^2(-\Delta P)}[/tex]

In the above equation only t and A is variable , all other terms are constant

Therefore we have

[tex]\frac{t_2}{t_1}= \frac{A_1}{A_2} = [\frac{65}{105.33}]^2 = 0.36[/tex]

[tex]t_2 = 0.36\times 40 = 14.4 min[/tex]

[tex]t_2 = 14.4 min[/tex]

Final answer:

To calculate the time it would take to produce 360lb of filtrate after adding 8 more frames to the filter, we can use the concept of filtration rate and the given information. Using the formula (360 lb * 40 mins) / (65 ft^2 + 8 * A ft^2), we find the time is approximately 14.40 mins.

Explanation:

To calculate the time it would take to produce 360lb of filtrate after adding 8 more frames to the filter, we can use the concept of filtration rate. The filtration rate is the amount of filtrate produced per unit of time per unit of filtration area. In this case, the filtration rate is given as 360 lb/40 mins/65 ft2.

After adding 8 more frames to the filter, the total filtration area becomes 65 ft2 + 8 * area of each frame. Let's assume the area of each frame is A ft2. The new filtration rate can be calculated by dividing the total filtration area by the time it would take to produce 360lb of filtrate, which is T mins. So, (360 lb/T mins) = (65 ft2 + 8 * A ft2)/(40 mins).

To find the value of T, we can rearrange the equation and solve for T: T = (360 lb * 40 mins) / (65 ft2 + 8 * A ft2). Substituting the initial values, we get T = (360 lb * 40 mins) / (65 ft2 + 8 * 0.5 ft2). After calculating, we find that T is approximately 14.40 mins.

Answer:

b) Ion-dipole

Explanation:

Intermolecular forces are the forces of attraction or repulsion between molecules, they are significantly weaker than intramolecular forces like covalent or ionic bonds.

Considering this information we can conclude that Ion-Dipole interaction is the strongest force among intermolecular forces.

I hope this information is useful to you!

Answer:

At the start of the process, the volume not occupied by the water is 2 m3

Explanation:

At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).

Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).

The volume in time will be

[tex]V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3][/tex]

Answer:

The correct option is: D)10 mL

Explanation:

A measuring cylinder or a graduated cylinder, is a narrow cylinder-shaped, chemical resistant, transparent equipment which is used to measure the volume of a given liquid.

Since the 10 mL graduated cylinder has 0.1 mL grading divisions. Therefore, a 10 mL graduated cylinder will precisely measure 8 mL liquid.

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

First, we will convert 3.00 g of CaCl₂ to moles using its molar mass (110.98 g/mol).

[tex]3.00 g \times \frac{1mol}{110.98g} = 0.0270 mol[/tex]

The heat of solution (ΔHsoln) of CaCl₂ is −82.8 kJ/mol. The heat released by the solution of 0.0270 moles is:

[tex]0.0270 mol \times \frac{-82.8kJ}{mol} = -2.24 kJ[/tex]

According to the law of conservation of energy, the sum of the heat released by the solution (Qs) and the heat absorbed by the calorimeter (Qc) is zero.

[tex]Qs + Qc = 0\\\\Qc = -Qs = 2.24 kJ[/tex]

Assuming the density of water is 1 g/mL, we have 100 mL (100 g) of water and 3.00 g of CaCl₂. The mass of the solution (m) is:

[tex]m = 100g + 3.00 g = 103 g[/tex]

Finally, we can calculate the final temperature of the system using the following expression.

[tex]Qc = c \times m \times (T_2 - T_1)[/tex]

where,

c: specific heat of the solution (same as water 4.18 J/g.°C)

T₁ and T₂: initial and final temperature

[tex]T_2 = \frac{Qc}{c \times m} + T_1 = \frac{2.24 \times 10^{3}J }{(\frac{4.18J}{g.\° C} ) \times 103 g} + 23.0 \° C = 28.2 \° C[/tex]

When 3.00 g of CaCl₂ is added to a calorimeter containing 100 mL of water at 23.0°C, the final temperature is 28.2 °C.

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To determine the amount of heat involved in the dissolution of CaCl₂ and the final temperature of the solution, we can use the formula q = m * c * ΔT. First, we need to calculate the heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water. Assuming the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat.

To determine the amount of heat involved in the dissolution of CaCl₂ and the final temperature of the solution, we can use the formula q = m * c * ΔT, where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat of the solution, and ΔT is the change in temperature.

First, we need to calculate the heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water. The mass of the solution is 55.0 g (50.0 g + 5.00 g), and the change in temperature is 39.2°C - 23.0°C = 16.2°C. Assuming the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat using the formula: q = 55.0 g * 4.18 J/g°C * 16.2°C = 3660.36 J.

Since 1 kJ = 1000 J, we can convert the heat to kilojoules: 3660.36 J ÷ 1000 = 3.66 kJ. Therefore, the amount of heat absorbed or released by water when 5.00 g of CaCl₂ is dissolved in 50.0 g of water is 3.66 kJ.

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Answer:

15766

Explanation:

The average molecular weight of the polypropylene = 663419 g/mol.

The mass of the repeating unit , monomer which is propylene = 42.08 g/mol

The degree of polymerization is:

[tex]DP_n=\frac {Total\ molecular\ weight\ of\ the\ polymer}{Molecular\ weight\ of\ the\ monomer}[/tex]

[tex]DP_n=\frac {663419\ g/mol}{42.08\ g/mol}[/tex]

Degree of polymerization = 15766

The new volume of the balloon that rose from a temperature of 25°C and 1.00 atm to a temperature of 20°C and 0.30 atm is calculated using the ideal gas law and is found to be 8.25 L.

In this problem, we can use the ideal gas law which states that the pressure times the volume of a gas is equal to the number of moles of gas times the gas constant times the temperature. Since the number of moles of gas and the gas constant remain constant, we can modify this equation to P1V1/T1 = P2V2/T2, which expresses the relationship between the initial and final state of the gas relative to its pressure, volume, and temperature.

First, we need to convert the temperatures from Celsius to Kelvin since the ideal gas law requires temperatures in Kelvin. So our initial temperature, T1, is 25°C + 273.15 K = 298.15 K and our final temperature, T2, is 20°C + 273.15 K = 293.15 K. Our initial and final pressures (P1 and P2) are 1.00 atm and 0.30 atm, respectively. And our initial volume (V1) is 2.50 L.

Plugging these values into the equation, we have: (1.00 atm * 2.50 L) / 298.15 K = (0.30 atm * V2) / 293.15 K.

By cross multiplying and solving for V2, we get: V2 = (1.00 atm * 2.50 L * 293.15 K) / (0.30 atm * 298.15 K) = 8.25 L. So the new volume of the balloon is 8.25 L.

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Answer:

The composition of the original mixture in molepercent is 80% of H₂ and 20% of O₂.

Explanation:

We need to combine the ideal gas law (PV = nRT) and Dalton's law of partial pressure (Pt = Pa +Pb +Pc+...).

The total pressure of the mixture is Pt = P (H₂) + P (O₂)

The number of moles can be found by Pt = nt RT/V, in which nt = n (H₂) +n (O₂).

If Pt is 1 atm, nt is 1.0 mol.

Now we need to consider the chemical reaction below:

H₂ + 0.5O₂ → H₂O

This shows that for each mol of O₂ we need two mol of H₂.

We know that the remaining gas is pure hydrogen and that its pressure is 0.4atm. Since PV = nRT, by the end of the reaction, 0.4 mol of H₂ remains in the system.

This means that in the beginning we have n mol of H₂, and when x mol of H₂ reacts with 0,5x mol of O₂, 0.4 mol of H₂ reamains.

If we have 1 mol in the begining and 0.4 mol in the end, the total amount of gas that reacted (x + 0.5X) is equal to 0.6 mol

x + 0.5X = 0.6 mol ∴ x = 0.6 mol / 1.5 ∴ x = 0.4 mol

0.4 mol of H₂ reacted with 0.2 mol of O₂ and 0.4 mol of H₂ remained as excess.

Therefore, in the beginning we had 0.8 mol of H₂ and 0.2 mol of O₂. Thus the molepercent of the mixture is 80% of H₂ and 20% of O₂.

Answer:

Without doing any calculations it is possible to determine that silver chromate is more soluble than C,D , and silver chromate is less soluble than none .

It is not possible to determine whether silver chromate is more or less soluble than A,B by simply comparing Kspvalues

Explanation:

For this kind of question you need to obtain the balanced equation for the solubility, first, you will write the dissociation equation for every molecule, obtaining the ions in which it dissociates. Then you substitute in the Kps equation

[tex]A_{n}B_{m} -> nA^{m+}+mB^{n-} \\K_{s}=[A^{m}]^{n}[B^{n}]^{m}[/tex]

Considering the corresponding stoichiometry you'll substitute in your dissociation equations as follow.

[tex]Ag_{2}CrO_{4}(s)->2Ag^{+}(aq)+CrO_{4} ^{2-}\\        s  ->2s+s\\s->[2s]^{2}[s]=4s^{3}  \\\\A)  PbF_{2}(s)->Pb^{2+} (aq)+2F^{-}\\s->s+2s\\s->[s][2s]^{2}=4s^{3}\\\\\\B)Ag_{2}SO_{3}(s)->2Ag^{+}(aq)+SO_{3} ^{2-}\\        s  ->2s+s\\s->[2s]^{2}[s]=4s^{3}  \\\\\\C)NiCO_{3}(s)->Ni^{2+}(aq)+CO_{3} ^{2-}\\        s  ->s+s\\s->[s][s]=s^{2}  \\\\\\D)AgCl(s)->Ag^{+}(aq)+Cl^{-}(aq)\\        s  ->s+s\\s->[s][s]=s^{2}  \\[/tex]

IIn this case, the silver chromate has a Kps of [tex]4sx^{3}[/tex], same as the compound in options A and B, comparing these numbers you can't determine which one is bigger. Finally, options C and D have a kps of [tex]s^{2}[/tex], this value is smaller than silver chromate's kps.

I hope you find this information useful! good luck!

The solubility of silver chromate compared to PbF2, Ag2SO3, NiCO3, AgCl cannot be definitively determined without empirical data or additional information. Factors influencing solubility include temperature, nature of the solute and solvent, pressure, and presence of other substances. Ksp values alone don't provide sufficient comparisons.

Without empirical data or additional information, it isn't possible to definitively determine which of the compounds (PbF2, Ag2SO3, NiCO3, AgCl) silver chromate is more or less soluble than. Solubility often depends on various factors, including temperature, nature of the solute and solvent, pressure, and presence of other substances. The Ksp (solubility product constant) values, which are a measure of the amount of a compound that can dissolve in a solution, are unique to each substance and are empirically determined in a lab. Therefore, we cannot just compare the Ksp values to decide which compounds are more or less soluble than silver chromate.

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Answer:

W = 141.333 BTu/h

Explanation:

W = Q ( 1 - TL/TH )......carnot

∴ Q = 106 BTu/h

∴ TH = 600 °F

∴ TL = 70 °F

⇒ W = 160 BTu/h ( 1 - 70/600 )

⇒ W = 141.333 BTu/h

Explanation:

As per the given data, at a higher temperature, at [tex]24^{o}C[/tex], the solution will occupy a larger volume than at [tex]20^{o}C[/tex].

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At [tex]20^{o}C[/tex], density of water=0.9982071 g/ml  

Therefore, [tex]\frac{concentration}{density}[/tex] will be calculated as follows.

                 = [tex]\frac{C_{1}}{d_{1}}[/tex]

                 = [tex]\frac{1.000 mol/L}{0.9982071 g/ml}[/tex]

                 = 1.0017961 mol/g  

At [tex]24^{o}C[/tex], density of water = 0.9972995 g/ml

Since, [tex]\frac{concentration}{density}[/tex] = [tex]\frac{C_{2}}{d_{2}}[/tex]

                           = [tex]\frac{C_{2}}{0.9972995}[/tex]

Also,             [tex]\frac{C_{1}}{d_{1}}[/tex] = [tex]\frac{C_{2}}{d_{2}}[/tex]

so,                   1.0017961 mol/g = [tex]\frac{C_{2}}{0.9972995}[/tex]

                      [tex]C_{2} = 1.0017961 \times 0.9972995[/tex]

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of [tex]KNO_{3}[/tex] present is calculated as follows.

             [tex]C_{2}[/tex] = [tex]\frac{concentration}{volume}[/tex]

               0.9990907 mol/L = [tex]\frac{concentration}{0.5 L}[/tex]  

               concentration = 0.49954537 mol

Hence, mass (m'') = [tex]0.49954537 mol \times 101.1032 g/mol[/tex] = 50.5056 g       (as molar mass of [tex]KNO_{3}[/tex] = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = [tex]m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}[/tex]

where,          [tex]d_{air}[/tex] = density of air = 0.0012 g/ml

                     [tex]d_{weight}[/tex] = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = [tex]50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}[/tex]

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of [tex]KNO_{3}[/tex] needs to be measured.

Answer:

Distillation

Explanation:

The most most widely used industrial separation operation is Distillation. There are different tipes of distillations can be simple or multiple and also azeotropic or flash. The method is based in the difference of the boiling points between the components that we want to separate from a mixture. It is used in all kind of industries and processes for example in the petroleum industry to separate the different fractions of crude oil, in the chemical industry to separate a component of interest after a chemical reaction and for the separation of gases, as in the production of oxygen and nitrogen from air.