If the density of alcohol is 0.79 g/mL, what is the mass in grams of 1.0 L of alcohol?

Explanation:

The given data is as follows.

Diameter of pipe, D = 0.01 m

Velocity V = 0.03 m/s

Fluid density = 36.4 [tex]kg/m^{3}[/tex]

Viscosity = 0.651 Pa-s

Formula to calculate Reynold number is as follows.

Re = [tex]diameter \times velocity \times \frac{density}{viscosity}[/tex]

     = [tex]0.01 \times 0.03 \times \frac{36.4}{0.651}[/tex]

     = 0.01677

Since,  Re < 2100. This means that the flow is laminar.

So, for laminar flow in pipes,

Darcy friction factor f = [tex]\frac{64}{Re}[/tex]

                                   = [tex]\frac{64}{0.01677}[/tex]

                                    = 3816

Thus, we can conclude that the Darcy friction factor value is 3816.

Answer:

6.82% is the mass percent NaOH in the solution.

Explanation:

Mass of the solute that is NaOH = m = 20 g

Mass of the solution = M

Volume of the solution = V = 255 mL

Density of the solution = d = 1.15 g/mL

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

[tex]1.15 g/mL=\frac{M}{255 mL}[/tex]

M = 293.25 g

Mass percentage of solute :(w/w)%

[tex]\frac{m}{M}\times 100[/tex]

[tex](w/w)\%=\frac{20 g}{293.25 g}\times 100=6.82\%[/tex]

6.82% is the mass percent NaOH in the solution.

Answer:

I hope you find these examples useful! good luck!

Answer:

As the average kinetic energy increases, the particles move faster and collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature

HOPE THIS HELPED!!!!!!!!!!!!!! XDDDDD

Answer:

pH = 0.22

Explanation:

The pH of the a solution is related to the concentration of the hydronium ion as follows:

pH = -log([H₃O⁺])

HNO₃ is a strong acid that reacts to completion with water as follows:

HNO₃ + H₂O ⇒ H₃O⁺ + NO₃⁻

The molar ratio between nitric acid and the hydronium ion is 1:1, so a 0.60 M nitric acid solution has a hydronium ion concentration of 0.60 M.

The pH is calculated:

pH = -log([H₃O⁺]) = -log(0.60) = 0.22

Answer:

12,000 units of heparin are received by patient in 24 hours.

Explanation:

10,000 units of heparin in 1000ml of D5W.

Units of heparin in 1 mL = [tex]\frac{10,000}{1000}=10 units[/tex]

Rate at heparin drip set at = 50 ml/hour

Units of heparin in 50 ml of solution = [tex]50\teimes 10 units =500 units[/tex]

500 units of heparin are received by patient in an every hour.So, in 24 hours patient will receive:

[tex]500units\times 24 = 12,000 units[/tex] of heparin

12,000 units of heparin are received by patient in 24 hours.

Answer: Thus rate of disappearance of [tex]H_2=2\times {\text {rate of disappearance of}O_2}[/tex]

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]

Given: Order with respect to [tex]H_2[/tex] = 2

Order with respect to [tex]O_2[/tex] = 1

Thus rate law is:

[tex]Rate=k[H_2]^2[O_2]^1[/tex]

k= rate constant

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

[tex]Rate=-\frac{1d[H_2]}{2dt}=-\frac{1d[O_2]}{dt}[/tex]

[tex]Rate=-\frac{1d[H_2]}{dt}=2\times -\frac{1d[O_2]}{dt}[/tex]

Thus rate of disappearance of [tex]H_2=2\times {\text {rate of disappearance of}O_2}[/tex]

Answer : a) [tex]\frac{[A^-]}{[HA]}=10^{4}[/tex]

b) [tex]\frac{[A^-]}{[HA]}=0.01[/tex]

Explanation : Given,

a) pH = 7.4

[tex]pK_a[/tex] for aspirin = 3.4

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

Concentration of salt [tex](A^-)[/tex] = ?

Concentration of acid [tex](HA)[/tex] = ?

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

Now put all the given values in this expression, we get:

[tex]7.4=3.4+\log \frac{[A^-]}{[HA]}[/tex]

[tex]\frac{[A^-]}{[HA]}=10^{4}[/tex]

b) pH = 1.4

Using Henderson Hesselbach equation :

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

Now put all the given values in this expression, we get:

[tex]1.4=3.4+\log \frac{[A^-]}{[HA]}[/tex]

[tex]\frac{[A^-]}{[HA]}=0.01[/tex]

The pKa is the acid dissociation constant. The ratio of A to HA in the blood is, 10000, and in the stomach is 0.01.

The acid dissociation constant or pKa has been the dissociation of the compound into ions at specified pH.

The relation between pKa and pH can be given as:

[tex]\rm pH=pKa\;+\;log\;\dfrac{acid}{salt}[/tex]

The pKa of aspirin is 3.4.

[tex]\rm 7.4=3.4\;+\;log\;\dfrac{A}{HA} \\\dfrac{A}{HA}=10000[/tex]

[tex]\rm 7.4=1.4\;+\;log\;\dfrac{A}{HA} \\\dfrac{A}{HA}=0.01[/tex]

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Answer: The given number in scientific notation is [tex]9.342\times 10^{-8}[/tex]

Explanation:

Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

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.

We are given:

A number having value = 0.000000093425

Converting this into scientific notation, we get:

[tex]\Rightarrow 0.000000093425=9.342\times 10^{-8}[/tex]

Hence, the given number in scientific notation is [tex]9.342\times 10^{-8}[/tex]

Answer:

[tex]9.343\times 10^{-8}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10.

The given number:

0.000000093425 can be written as [tex]9.3425\times 10^{-8}[/tex]

Answer upto 4 significant digits = [tex]9.343\times 10^{-8}[/tex]

Answer:

The density of old trunk is [tex]11.3 g/cm^3[/tex] and it is equal to the density of metal lead.

Explanation:

Mass of the trunk , m= 222 g

Length of the cubic trunk = 2.7 cm

Volume of the cube = [tex]a^3[/tex]

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

[tex]D=\frac{m}{a^3}=\frac{222 g}{(2.7 cm)^3}=11.2787 g/cm^3\approx 11.3 g/cm^3[/tex]

The density of old trunk is [tex]11.3 g/cm^3[/tex] and it is equal to the density of metal lead.

Final answer:

The density of the metal cube is calculated to be approximately 11.28 g/cm³, which closely matches the density of lead (11.3 g/cm³), suggesting that the cube is made of lead.

Explanation:

To determine the type of metal a cube is made of, you need to calculate its density and compare it with the given densities of possible materials. The density is calculated by dividing the mass of the cube by its volume.

The mass of the metal cube is given as 222 g, and the length of each side is 2.7 cm. The volume V of a cube is found by cubing the length of one of its sides: V = 2.7 cm × 2.7 cm × 2.7 cm = 19.683 cm³. The density ρ is then found by dividing the mass m by the volume V: ρ = m/V = 222 g / 19.683 cm³ ≈ 11.28 g/cm³.

Comparing this calculated density with the given densities, the closest match is lead (11.3 g/cm³). Therefore, the metal cube is most likely made of lead.

Answer:

Complete reaction

mole %: 0% N₂; 20% H₂; 80% NH₃

Gram %: 0% N₂; 2,9% H₂; 97,1% NH₃

75% reaction:

mole%: 8,3% N₂; 41,7% H₂; 50% NH₃

Gram: 20% N₂; 7,2% H₂; 72,8% NH₃

Explanation:

The global reaction in the reactor is:

N₂(g) + 3 H₂ (g) → 2 NH₃ (g)

For a complete reaction of 20 moles of N₂(g) you need:

20 moles N₂ ₓ [tex]\frac{3 mol H_2}{1 mol N_2}[/tex] = 60 moles of H₂(g)

So, 10 moles of H₂(g) will stay in the end.

The moles produced of NH₃ are:

20 moles N₂ ₓ [tex]\frac{2 mol NH_3}{1 mol N_2}[/tex] = 40 moles of NH₃(g)

Thus, the final composition in moles is:

0 moles N₂; 10 moles H₂; 40 moles of NH₃

The mole % is:

0% N₂; ¹⁰/₅₀ ₓ 100 = 20% H₂; ⁴⁰/₅₀ ₓ100 = 80% NH₃

In grams you have:

10 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 20,2 g of H₂

40 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 681,2 g of NH₃

0% N₂; [tex]\frac{20,2 g}{701,4}[/tex] ₓ 100 = 2,9% H₂; [tex]\frac{681,2 g}{701,4}[/tex] ₓ100 = 97,1% NH₃

With a 75% of conversion you have that the moles produced of NH₃ are:

40 moles of NH₃(g) × 75% = 30 moles of NH₃

The necessary moles to produce these moles of NH₃ are:

30 moles NH₃ ₓ [tex]\frac{3 mol H_2}{2 mol NH_3}[/tex] = 45 moles of H₂(g)

So, 25 moles of H₂(g) will stay in the end.

30 moles NH₃ ₓ [tex]\frac{1 mol N_2}{2 mol NH_3}[/tex] = 15 moles of N₂(g)

So, 5 moles of H₂(g) will stay in the end.

Thus, the final composition in moles is:

5 moles N₂; 25 moles H₂; 30 moles of NH₃

The mole % is:

⁵/₆₀ ₓ 100 =8,3% N₂; ²⁵/₆₀ ₓ 100 = 41,7% H₂; ³⁰/₆₀ ₓ100 = 50% NH₃

In grams you have:

5 moles N₂ [tex]\frac{28 g}{1 mol}[/tex] = 140 g of N₂

25 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 50,5 g of H₂

30 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 510,9 g of NH₃

[tex]\frac{140 g}{701,4}[/tex] ₓ 100 =20% N₂; [tex]\frac{ 50,5g}{701,4}[/tex] ₓ 100 = 7,2% H₂; [tex]\frac{510,9 g}{701,4}[/tex] ₓ100 = 72,8% NH₃

I hope it helps!

Answer:

Δ S° > 0

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by the change in heat per change in temperature.

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And  

If the particles are loosely held , the entropy will increase .

From the reaction given in the question ,

Cu₂O (s)  + C (s) → 2 Cu (s) + CO (g)

Considering the states of the reactant and the product ..

The product formed i.e. CO is in gaseous species , hence , the entropy increases .

The freezing point of the aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.09°C

To calculate the freezing point of the solution, we can use the following equation:

ΔTf = Kf * m

where:

ΔTf is the freezing point depression in C

Kf is the molal freezing point constant of water in C·kg/mol (1.86 C·kg/mol)

m is the molality of the solution in mol/kg

The molality of the solution can be calculated from the osmotic pressure using the following equation:

Π = MRT

where:

Π is the osmotic pressure in Pa

M is the molarity of the solution in mol/m³

R is the ideal gas constant in J/mol·K (8.314 J/mol·K)

T is the temperature in K

Substituting the given values into the equation, we get:

M = Π / (RT) = 120 kPa / (8.314 J/mol·K * 300 K) = 0.4918 mol/m³

The molality of the solution is equal to the molarity divided by the density of the solvent. The density of water is 1 kg/L, so the molality of the solution is:

m = M / d = 0.4918 mol/m³ / 1 kg/L = 0.4918 mol/kg

Now that we know the molality of the solution, we can calculate the freezing point depression using the first equation:

ΔTf = Kf * m = 1.86 C·kg/mol * 0.4918 mol/kg = 0.09 °C

Finally, we can calculate the freezing point of the solution by subtracting the freezing point depression from the freezing point of pure water:

Freezing point of solution = freezing point of pure water - freezing point depression

Freezing point of solution =  0 °C-  0.09 °C = -0.09°C

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The freezing point of an aqueous solution of urea at 300 K with an osmotic pressure of 120 kPa is -0.08928 °C.

1. To find the freezing point depression (ΔTf) of the solution, we can use the formula: ΔTf = i x k x m, where:

2. We need to calculate the molality (m) using osmotic pressure (π), which we can find from the formula: π / (i x R x T), given:

3. Since the density of water is approximately 1kg/L, the molality (m) ≈ molarity (M) in this case: m ≈ 0.048 mol/kg

Now, we can find the freezing point depression (ΔTf):ΔTf = 1 x 1.86 °C kg/mol x 0.048 mol/kg = 0.08928 °C

Answer:

Potential energy for n = 6 Bohr orbit electron is -1.21*10⁻¹⁹J

Explanation:

As per the Bohr model, the potential energy of electron in an nth orbit is given as:

[tex]PE_{n} = -\frac{kZe^{2}}{r_{n}}[/tex]

here:

k = Coulomb's constant = 9*10⁹ Nm2/C2

Z = nuclear charge

e = electron charge = 1.6*10⁻¹⁹ C

r(n) = radius of the nth orbit = n²(5.29*10⁻¹¹m)

Substituting for k, Z(= 1), e and r(n) in the above equation gives:

[tex]PE_{6} = -\frac{9*10^{9}Nm2/C2*1*(1.6*10^{-19}C)^{2}}{(6)^{2}*5.29*10^{-11}m}=-1.21*10^{-19}J[/tex]

Answer:

S AgBr = 2.82 E-3 mol/L

Explanation:

∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13

∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²

∴ C NH3(sln) = 1.00 M

from (1) + (2):

                             1 M                     0                       0

                            1 - 2x                   x                        x

∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6

⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6

⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6

⇒ x² = 8.0 E-6*( 1- 2x )

⇒ x² + 1.6 E-5x - 8.0 E-6 = 0

⇒ x = 2.82 E-3 M

⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M

To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation for the formation constant of the silver(I) ammonia complex.

The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH₃)₂, is 1.6 x 107. To calculate the solubility of AgBr in a 1.00 M ammonia solution, we can use the equation:

AgBr(s) → Ag+(aq) + Br-(aq)

Let x be the concentration of AgBr(s). Then, the equilibrium concentrations of Ag+ and Br- would be x and x, respectively.

Using the equation for the formation constant:

Kf = [Ag(NH₃)₂]+ / [Ag+][NH₃]₂

we can substitute the given values and solve for [Ag+] and [NH₃].

The psychical properties do no change the composition of the lithium metal.

The chemical properties will change the composition of the lithium producing other substances.

light enough to float on water - physical property

silvery gray in color - physical property

changes from silvery gray to black when placed in moist air - chemical property

(in air the lithium will react with the oxygen forming the oxide which have a black color)

can be cut with a sharp knife - physical property

in the liquid state, it boils at 1317 °C - physical property

reacts violently with chlorine to form a white solid - chemical property

in the liquid state, it reacts spontaneously with its glass container, producing a hole in the container - chemical property

burns in oxygen with a bright red flame - chemical property

Explanation:

A property that does not bring any change in chemical composition of a substance are known as physical properties.

For example, shape, size, mass, volume, density, hardness etc of a substance are all physical properties.

On the other hand, a property that changes chemical composition of a substance is known as chemical property.

For example, precipitation, reactivity, toxicity etc are chemical property.

Hence, the given properties of lithium are classified as follows.

If you want to achieve a final concentration of 100 µg/mL of epinephrine in a 25 mL solution, when using a stock solution of 1 mg/mL and a pipette that delivers 20 drops/mL, you need to add 50 drops of your stock solution.

Since we are asked to find the number of drops of stock solution required to achieve a final concentration of 100 µg/mL in a 25 mL solution, the first step is to convert the concentration of the stock solution to the same units, µg/mL. Hence, 1 mg/mL is equal to 1000 µg/mL. Further, we know that 1 mL of the stock solution contains 1000 µg of epinephrine, and our pipette delivers 20 drops/mL, so 1 drop of stock solution contains 1000 µg / 20 drops = 50 µg. Thus, if we need a 100 µg/mL concentration in 25 mL, we need a total of 100 µg/mL * 25 mL = 2500 µg of epinephrine. Therefore, to achieve this, we must add 2500 µg / 50 µg/drop = 50 drops of our stock solution. Hence,

50 drops

of the stock solution should be added to achieve the desired concentration.

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To achieve a final concentration of 100 µg/mL of epinephrine in 25 mL of Locke's solution, you need to add 50 drops of the 1 mg/mL stock solution. The dilution factor for one drop of the stock solution in the muscle bath is 500

a) To determine the required number of drops to achieve a specific concentration in Locke's solution, we follow these steps:

b) Dilution Factor

       Volume of one drop:

      1 mL / 20 drops = 0.05 mL/drop

      25 mL / 0.05 mL = 500

The correct dilution factor is 500.

Therefore, you must add 50 drops of the stock epinephrine solution to the muscle bath in Locke's solution to achieve the desired concentration of 100 µg/mL and dilution factor of 500.

Complete question.

Assume that you have a stock solution of epinephrine at a concentration of 1mg/ml. Further assume that there are 20 drops/ml.

a. Calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 ml of Locke's solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL.

b. Compute the dilution factor which describes the extent to which a single drop of stock drug solution is diluted when added to the smooth muscle bath.

Explanation:

Formula for compressibility factor is as follows.

                     z = [tex]\frac{P \times V_{m}}{R \times T}[/tex]

where,     z = compressibility factor for helium = 1.0005

               P = pressure

          [tex]V_{m}[/tex] = molar volume

                R = gas constant = 8.31 J/mol.K

                T = temperature

So, calculate the molar volume as follows.

                [tex]V_{m} = \frac{z \times R \times T}{P}[/tex]

                             = [tex]\frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}[/tex]

                             = 0.0056 [tex]m^{3}/mol[/tex]

As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.

                    [tex]V_{sp} = \frac{V_{m}}{M_{w}}[/tex]

                           = [tex]\frac{0.0056 m^{3}/mol}{4 g/mol}[/tex]

                           = 0.00139 [tex]m^{3}/g[/tex]

                           = 0.00139 [tex]m^{3}/g \times \frac{1 g}{10^{-3}kg}[/tex]

                                 = 1.39 [tex]m^{3}/kg[/tex]

Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 [tex]m^{3}/kg[/tex].

Answer:

The process described in the problem is steady-state.

Explanation:

The steady-state can be described by the conservation of mass. In a bathtub it will be the same as In=Out+ Accumulation. Since the water has already reached the adequate level and the drain is open, the accumulation is equal to zero, and therefore it is at steady state. This also means that In = Out which is also described in the problem: "the water is poured into the bathtub at an equivalent rate of the water being drained".

Answer:

We could use fractional distillation, solvent extraction, partial crystallization, chromatography, and another.

Explanation:

The only information given is that the compounds have similar polarities, which means they form a homogeneous mixture, so, to separate them, we should use some of the processes of separation of a homogeneous mixture.

Without knowing any of the physical properties of A and B (physical state, density, boiling point, for example) it's impossible to determinate the better process of separation. However, we can describe some of the processes that are used to separate homogeneous mixtures: fractional distillation, solvent extraction, partial crystallization, and chromatography, for example.

Answer:

Option d, It is higher than the energy of both reactants and products.

Explanation:

When reactant molecules collide with each other during the course of the reaction, an intermediate state is reached before the formation of product. This intermediate state is also called activated complex.

Activated complex consist energy higher than that of both the reactants and products. Because of possessing higher energy, it is unstable and temporary.

Hence, among given, option d is correct.

Answer:

d. It is higher than the energy of both reactants and products.

Explanation:

The activated complex is a transient state, or intermediate phase, between reagents (weak or not), in which the final product has not yet been formed. This effective shock, like any chemical bond, needs energy. To form this transient state requires an activation energy, a minimum energy for this intermediate phase to occur.

Within this phenomenon, the activation energy must occur by some criteria, which characterizes a chemical bond in an activated complex:

Within these three rules, it is clearer to define this complex as the moment when a molecule of a given atom collides with another molecule of any other atom and breaks the bond established between them.

For it to happen, the energy needs to be high. When this potential energy is present at a high level during the reaction, it is also necessary to have a high energy charge to complete the complex and also the collision of the reactant molecules to form the binding products.

The concentration of the nickel(II) chloride solution is determined by dividing the amount of substance (in umol) by the volume of the solution (in liters). This results in a concentration of 306.5 umol/L.

The concentration of a solution is determined by the formula c = n/V, where 'c' is the concentration, 'n' is the amount of substance (in moles), and 'V' is the volume of the solution (in liters).

In this case, the chemist has used 61.3 umol of nickel(II) chloride (NiCl2), and the total volume of the solution is 200 ml. We need to convert this volume to liters, giving us 0.2 L. Now, using the formula, we calculate: c = 61.3 umol / 0.2 L, which equals 306.5 umol/L.

Make sure to always convert the units appropriately before you substitute the values into the formula.

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Answer: The chemical formula of Iron (III) sulfate is [tex]Fe_2(SO_4)_3[/tex]

Explanation:

Iron is the 26th element of periodic table having electronic configuration of [tex][Ar]3d^64s^2[/tex].

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

Sulfate ion is a polyatomic ion having chemical formula of [tex]SO_4^{2-}[/tex]

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

So, the chemical formula for Iron (III) sulfate is [tex]Fe_2(SO_4)_3[/tex]

Answer:

Mutarotation refers to the change in the optical rotation or optical activity of a solution due to the change in the equilibrium of the two anomers. It depends upon the optical activity and ratio of the anomeric forms in the solution.

To measure the optical rotation of a given solution, a polarimeter can be used and thus the ratio of the anomeric forms can be calculated.

Answer:

The right answer is option A: The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.

Explanation:

In a heat exchanger, when the exchange area is reduced, and at the same condition for the rest of the variables, the amount of heat exchanged by the two fluids is reduced.

In the case of the hot fluid, it will rise at a higher temperature, compared to the temperature at which it would come out with a larger exchange area.

In the case of the hot fluid, it will come out at a lower temperature, compared to the temperature at which it would come out with a larger exchange area.

The correct answer is option A.

Answer:

a- The outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases.

Explanation:

Hello,

In heat transfer processes, specific units such as heat exchangers are useful to either heat up a cold stream with a hot one or vise-versa. However, the heat transfer efficiency increases as surface area increases, it means that the higher the surface area the higher the heat exchanger's efficiency, therefore, if the surface area is reduced, the efficiency is reduced as well, due to the lack of more space for the heat transfer.

In such a way, the outer temperature of the cold fluid will be lower (less heating up) and the outer one of the hot fluid will be higher (less cooling down), thus, the outlet temperature of the cold stream decreases and the outlet temperature of the hot stream increases (a option).

Best regards.

Answer:

0.0375 moles

Explanation:

Given that:

Molar mass of monopotassium phosphate = 136.09 g/mol

Given that volume = 250.0 mL

Also,

[tex]1\ mL=10^{-3}\ L[/tex]

So, Volume = 250 / 1000 L = 0.25 L

Molarity = 0.15 M

Considering:

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

[tex]0.15=\frac{Moles\ of\ solute}{0.25}[/tex]

Thus, moles of monopotassium phosphate needed = 0.0375 moles

Answer:

a) 213.3 mg/L

b) 62.61 mg/L

c) 0.0225 mg/L

Explanation:

Theoretical oxygen demand (ThOD)is essentially the amount of oxygen required for the complete degradation of a given compound into the final oxidized products

a) Given:

Concentration of acetic acid,[tex][CH3COOH][/tex] = 200 mg/L

[tex]CH3COOH + 2O2 \rightarrow 2CO2 + 2H2O[/tex]

[tex]ThOD = \frac{mass\ O2}{mass\ CH3COOH} * conc. CH3COOH[/tex]

Based on the reaction stoichiometry:

mass of [tex]CH3COOH[/tex] = 60 g

mass of [tex]O2[/tex]= 2(32) = 64 g

[tex]ThOD = \frac{64 g}{60 g} * 200mg/L = 213.3 mg/L[/tex]

b) Given:

Concentration of ethanol, [tex][C2H5OH][/tex] = 30 mg/L

[tex]C2H5OH + 3O2 \rightarrow 2CO2 + 3H2O[/tex]

[tex]ThOD = \frac{mass\ O2}{mass\ C2H5OH} * conc. C2H5OH[/tex]

Based on the reaction stoichiometry:

mass of [tex]C2H5OH[/tex] = 46 g

mass of [tex]O2[/tex]= 3(32) = 96 g

[tex]ThOD = \frac{96 g}{46 g} * 30mg/L = 62.61 mg/L[/tex]

c) Given:

Concentration of sucrose, [tex][C12H22O11][/tex] = 50 mg/L

[tex]C12H22O11 + 12O2 \rightarrow 12CO2 + 11H2O[/tex]

[tex]ThOD = \frac{mass\ O2}{mass\ C22H22O11} * conc. C12H22O11[/tex]

Based on the reaction stoichiometry:

mass of [tex]C12H22O11[/tex] = 342 g

mass of [tex]O2[/tex]= 12(32) = 384 g

[tex]ThOD = \frac{384 g}{342 g} * 50mg/L = 0.0225 mg/L[/tex]

To calculate the number of moles of solute particles in a solution, multiply the molarity by the volume of the solution. In this case, the result is 0.0070 mol. Taking the LOG (base 10) of the result gives -2.15.

To determine the number of moles of solute particles in a solution, you need to consider the molarity and volume of the solution. In this case, the molarity is 0.887 M and the volume is 7.94 mL. To calculate the number of moles, you can use the formula:

Moles = Molarity x Volume

Converting mL to L:

7.94 mL = 7.94/1000 L = 0.00794 L

Now, substitute the values into the formula:

Moles = 0.887 M x 0.00794 L = 0.0070 mol

Taking the LOG (base 10) of 0.0070, we get -2.15 (rounded to 2 decimal places).

Answer:

The rabbit population will reach 500 after 10 months.

Explanation:

According to the given data:

The initial number of rabbit's equals 2.

Number of rabbit's after 2 months =2x3= 6

Number of rabbit's after 4 months = 6x3=18

Number of rabbit's after 6 months = 18x3=54

Number of rabbit's after 8 months = 54x3=162

Thus we can see that the number of rabbit's form a Geometric series with common ratio =3 and initial term = 2

Now the general term of a geometric series with first term 'a' and common ratio 'r' is given by

[tex]T_{n+1}=ar^{n}[/tex]

Thus we need to find when the term becomes 500 thus using the given data we have  

[tex]500=2\cdot 3^{n}\\\\3^{n}=250\\\\(n)log_3(3)=log_3(250)\\\\(n)=5.025\\\\[/tex]

Thus the fifth term (excluding the start term) will have a rabbit count of 500 now since each term has a time difference of 2 months thus sixth term will occur after [tex]5\times 2=10months[/tex]

Answer:

To make 500 mL of a solution of PPP 1% w/v you need 5g

Explanation:

A percent w/v solution is calculated with the following formula using the gram as the base measure of weight (w):

% w/v = g of solute/ mL of solution × 100

A 1% w/v means that you have 1 g of PPP per 100 mL of solution

Thus, if you need to make 500 mL of solution you will need 5 g of PPP.

I hope it helps!

To prepare a 1% w/v solution of phenolphthalein di-phosphate (PPP) for 500ml, you need to weigh out 5 grams of PPP and dissolve it in enough water to make the total solution volume up to 500 ml.

To prepare a 1% w/v solution of phenolphthalein di-phosphate (PPP) for 500ml, you first need to understand that 1% w/v means you have 1 gram of solute for every 100 milliliters of solution. Then, for 500 milliliters, you would need 5 grams of PPP. The molar mass of PPP (478 g/mol) is only used to calculate moles from grams or vice versa and is not needed to prepare a percent weight/volume solution. Since you have enough PPP (25g), you can weigh out 5 grams of PPP and dissolve it in water to make a total volume of 500ml to achieve a 1% w/v concentration.

Steps to prepare the solution: