What is the pH of a solution containing 0.049 M of formic acid and 0.055 M of sodium formate?

Answer:

The answer is NaF.

Explanation:

Lattice energy:

It can be define as "the amount of energy released when ions combine to from the ionic solids or energy required to break the compound into its ions"

Example:

Consider the example of ionic solid NaCl. The sodium chloride is formed when ions like Na+ and Cl - are combine. During the formation of NaCl crystals the energy is released which is called lattice energy of sodium chloride.

Dependence of lattice energy:

Lattice energy depend upon following factors:

1. Charges of the ions

2. Sizes of the ions

As the charge of ion increase the lattice energy also increase. There is a direst relationship between them.

The size of the ions and lattice energy have inverse relation. As the size decreases, lattice energy increases. Therefor, the compound NaF consist of smallest cation (Na+) and the smallest anion (F-) have largest lattice energy.

                F= q1 × q2/ r²

As the size of ions decreases the distance between the oppositely charged ions in a ionic compound decreases. They will more tightly packed in a crystal and their center will more closer to each others hence increase the lattice energy.

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.

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 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: 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:

Molarity of silver ion: 0.296 M

Explanation:

Reaction:

AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻

From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.

First, let´s calculate the number of moles of AgCl produced:

1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.

The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:

moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol

Molarity of silver ion = 0.296 M

The molarity of silver ion in the original AgNO3 solution is 0.293 M.

To determine the molarity of silver ion in the original AgNO3 solution when 37.1 mL reacts to yield 1.56 g of AgCl, we first need to calculate the moles of AgCl produced using the molar mass of AgCl.

The molar mass of AgCl is 143.3 g/mol. So, the moles of AgCl formed are calculated as follows:

(1.56 g AgCl) / (143.3 g/mol) = 0.010878 mol AgCl

Since AgNO3 reacts with KCl in a 1:1 mole ratio to produce AgCl, the moles of AgNO3 that reacted is also 0.010878 mol. To find the molarity of AgNO3, we divide the moles by the volume of solution in liters:

(0.010878 mol AgNO3) / (0.0371 L) = 0.293 M

Answer:

2 grams

Explanation:

The percentage in weight/weight is equivalent to the solute mass divided by the solution mass and multiplied by 100. For example, if you have 25g/100g then the percentage will be 25%. In this case you just need to multiply 8*25 and divided it by 100, so the answer is 2 grams of mupirocin per 25 g of the ointment.

Explanation:

The water is hard due to the presence of ions of Mg²⁺. Now in hard water soaps are ineffective . Since in hard water Mg²⁺ ion forms precipitate , which concludes less number of soap molecules are present in the solution and less amount of  froth  which disables the cleansing property of the soap .

Now the experiment , the very last step is the hard water test where some amount of MgCl₂ is added ,

Now this magnesium salt act as a source of Mg²⁺ ion and now the soap action can be determined , whether it is able to form froth or not .

MgCl2 is added to the soap solution during a soap testing experiment to simulate the conditions of hard water. This allows for assessment of the soap's effectiveness as a cleansing agent in hard water, as soaps often form insoluble precipitates when reacting with minerals like magnesium.

In a soap testing experiment, MgCl2 (magnesium chloride) is added to the soap solution as a way to simulate hard water conditions because hard water contains high levels of magnesium and other minerals. One property of soap is its ability to react with ions present in water. In hard water, soaps react with calcium or magnesium ions to form insoluble precipitates, which makes the soap less effective as a cleaning agent. This is notable because soaps form insoluble calcium and magnesium compounds in hard water, while detergents form water-soluble products.

The behavior of soap in this soap-magnesium chloride solution, therefore, would give an indication of its prospective behaviour in hard water conditions.

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Answer:

The energy is 3.559 * 10⁻¹⁹ J

Explanation:

For this problem we need to use Planck's equation, this equation describes the relationship between energy radiated by light (a photon, to be precise) and its frequency. The equation is:

E = h * v

Where E is energy (in Joules), h is Planck's constant (6.626 * 10⁻³⁴ J·s), and v is the frequency (in s⁻¹).

Now we solve the equation, using the data given in the problem:

E = 6.626 * 10⁻³⁴ J·s * 5.371 * 10¹⁴ s⁻¹ = 3.559 * 10⁻⁴⁷ J

Thus the energy is 3.559 * 10⁻¹⁹ J

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:

molar volume of vapor ethanol is [tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]

mass of ethanol is 15430.22 g

Explanation:

By using ideal gas equation

PV = nRT

Where P is pressure , R is gas constant

so, volume is

[tex]V = \frac{RT}{P} = \frac{8.314\times (480+273)}{6000\times 10^3}[/tex]

[tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]

molar volume of vapor ethanol is

[tex]V = 1.043 \times 10^{-3} m^3/mol[/tex]

b)

volume of vessel is given [tex]0.35 m^3[/tex]

therefore total moles of ethanol in given vessel will be

[tex]n =\frac{V_{vessel}}{V}[/tex]

[tex]n =\frac{0.35}{1.043\times10^{-3}}[/tex]

n = 335.44 mol

we know that

mole is given as [/tex]n = \frac{mass}{moleculae weight}[/tex]

weight of ethanol is 46 g/mol

n\times 46 = mass

[tex]335.44\times 46 = 15430.22 g[/tex]

mass of ethanol is 15430.22 g

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 .

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:

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]

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:

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:

Hydrogen bond is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N-H, O-H, or F-H, and electronegative O,N or F atom. The interaction is written:

                                        A-B . . . B        or           A-H . . . .A

A and B represent O,N or F; A-H is one molecule or part of a molecule and B is a part of another molecule, and the dotted represents the hydrogen bond. So, hydrogen bond is a type of intermolecular forces (attractive forces between molecules, NOT between atoms of the same molecule), and only a few elements can participate on hydrogen bond formation.

Note: dipole-dipole interactions are forces between polar molecules, that is, between molecules that possess dipole moments.  

Final answer:

A hydrogen bond is a weak intermolecular attraction where a partially positive hydrogen atom, bonded to an electronegative atom, is attracted to a lone pair of electrons on a nearby electronegative atom.

Explanation:

A hydrogen bond is a type of intermolecular or intramolecular attractive force. It occurs when a hydrogen atom, covalently bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine, develops a partial positive charge. This partially positive hydrogen atom is then attracted to the lone pair of electrons on a neighboring electronegative atom, often in a different molecule. This attraction is relatively weak compared to covalent bonds but is stronger than other dipole-dipole interactions. In water, hydrogen bonds are a crucial part of the molecule's properties, including its high boiling point and surface tension.

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:

The value of the Prandlt number for this fluid is: 1656.04

Explanation:

As it is stated in the problem : Pr = Cp*mu/k

where:

cp: heat capacity of the fluid

mu: viscosity of the fluid

k: thermal conductivity of the fluid

Now for a given fluid we have

cp= 0.58 J/(g* deg C)

mu=1934 Ibm / (ft * h)

k = 0.28 W/(m * deg C)

If we put these values in the ecuation of the Prandlt number we have:

Pr = (0.58 J/(g* deg C)) × (1934 Ibm / (ft * h) / 0.28 W/(m * deg C)) =

As we can see we have to convert the units so we can operate all the values in the same units of measurement and then cancel them so as to obtain a dimensionless result.

Converting the value of: mu = 1934 Ibm / (ft * h)

1 ft= 0.3048 m

1 h= 3600 s

1 lbm= 453.59 g

mu= 1934 Ibm / (ft * h) × (453,59 g/ lbm) × (1h/3600 s) × (1 ft/0.3048 m) = 799.47 g/ (m *s).

Pr = (0.58 J/(g* deg C)) × (799.47 g/ (m *s) / 0.28 W/(m * deg C)) = 1656.04

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:

I hope you find these examples useful! good luck!

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!

The ΔH value for the combustion of butane is given as 5315 kJ for the reaction as written, which is for 2 moles of butane. To find the enthalpy of combustion per mole, divide this number by 2, yielding -2658 kJ/mol.

The student asked to give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) + 5315 kJ. The ΔH value for the reaction is given as +5315 kJ, indicating the amount of energy released during the combustion of butane.

Now, the enthalpy of combustion is typically expressed on a per mole basis, referring to the amount of heat released when one mole of a substance is burned. Since the reaction above shows the combustion for 2 moles of butane, we can calculate the enthalpy of combustion per mole by dividing the total energy by 2. So, the enthalpy of combustion for one mole of butane would be 5315 kJ / 2 = 2657.5 kJ/mol, which is typically reported in kilojoules per mole (kJ/mol).

Expressed with four significant figures, the enthalpy of combustion per mole of butane would be -2658 kJ/mol (the negative sign indicates that energy is released).

Answer:

Sodium chloride has a crystalline face-centered  cubic structure whereas metallic sodium body-centered  cubic structure.

Explanation:

Hello, atomic arrangements provide the molecules different features and behaviors, since the sodium metal has a body-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-body-centered.svg) the lack of inner atoms, allows the material to be soft and bendable. On the other hand the compacted sodium chloride's face-centered  cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-face-centered.svg), provides it a crystalline structure which is hard and brittle since the atoms are closer.

Best regards!

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.

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.

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:

Option c, Two atomic orbitals combine to form one molecular orbital

Explanation:

Molecular orbitals are formed by linear combination of atomic orbitals.

Some of the important facts of molecular orbital theories are as follows:

So, among given, option c which is 'atomic orbitals combine to form one molecular orbital' is incorrect.

All the statements provided are true with regards to Molecular Orbital Theory, except for 'Two atomic orbitals combine to form one molecular orbital.' In fact, two atomic orbitals combine to form two molecular orbitals, one being a lower-energy bonding molecular orbital and the other one being a higher-energy antibonding orbital.

In the context of Molecular Orbital Theory, three out of the four statements made in the question are correct. Namely, molecular orbitals are indeed formed by the combination of atomic orbitals, and they represent regions of a molecule where electrons are most likely to be found. Additionally, bonding molecular orbitals do have lower energy than the atomic orbitals from which they are formed.

However, the statement suggesting that 'Two atomic orbitals combine to form one molecular orbital' is not accurate. In reality, when two atomic orbitals combine, they form two molecular orbitals: one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital.

This nuanced concept is a cornerstone of MO theory and helps explain a range of phenomena, from the paramagnetism of the oxygen molecule to violations of the octet rule. Understanding this allows us to further appreciate the differences between bonding and antibonding orbitals and their roles in the structure and stability of molecules.

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Answer:

656,25 moles of CH3OH

Explanation:

If the mixture has 7% of weight in methanol (CH3OH), it means that for every 100 kg of the mixture there are 7 kg of methanol.

To solve the problem we just use this relation and convert the kilograms of methanol to gmoles of methanol using the molecuar weight of the methanol (32g/mol):

[tex]300 kgmixture*\frac{7kgCH3OH}{100kg}*\frac{1000g}{1kg}*\frac{1molCH3OH}{32g}=656,25molesCH3OH[/tex]