Consider four small molecules, A–D, which have the following binding affinities for a specific enzyme (these numbers are the equilibrium constants Kd for the dissociation of the enzyme/molecule complex). Which binds most tightly to the enzyme? Which binds least tightly?

A) 4.5 μM

B) 13 nM

C) 8.2 pM

D) 6.9 mM

Explanation:

Here it is given that carbon is sample 2 = 25.9 g

No. of moles of carbon will be calculated as follows.

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

                                                 = [tex]\frac{1.47 g}{12.01 g/mol}[/tex]

                                                 = 0.1224 mol

It is also given that mass of hydrogen = 0.123 g

Hence, calculate number of moles of hydrogen as follows.

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

                                                      = [tex]\frac{0.123 g}{1.008 g/mol}[/tex]

                                                      = 0.122 mol

Therefore, [tex]\frac{\text{moles of carbon}}{\text{moles of hydrogen}}[/tex]

                              = [tex]\frac{0.1224 mol}{0.122 mol}[/tex]

                                = 1.003

Therefore, calculate the number of moles of hydrogen as follows.

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

                                          = [tex]\frac{2.17 g}{1.008 g/mol}[/tex]

                                          = 2.1528 mol

Hence, calculate the moles of carbon as follows.

      Moles of carbon = [tex]\text{moles hydrogen} \times \frac{\text{moles of carbon}}{\text{moles hydrogen}} [/tex]

                                   = [tex]2.1528 mol \times 1.003[/tex]

                                  = 2.16 mol

Mass of carbon = moles carbon × molar mass carbon

                           = (2.16 mol) × (12.01 g/mol)

                           = 25.9 g

Thus, we can conclude that 25.9 g of carbon is expected in the sample.

Final answer:

The expected amount of carbon in Sample 2 would be 25.93 grams, calculated using the constant carbon-to-hydrogen mass ratio determined from Sample 1.

Explanation:

The question requires determining how much carbon is expected in Sample 2, given the known amounts of carbon and hydrogen in Sample 1 and the amount of hydrogen in Sample 2. We assume that the compound is the same in both samples, meaning the ratio of carbon to hydrogen must be constant. First, we find the ratio of carbon to hydrogen in Sample 1:

Sample 1: 1.47 g C / 0.123 g H = 11.95 g C/g H

Using this ratio, we calculate the expected amount of carbon in Sample 2:

Expected carbon in Sample 2: 2.17 g H x 11.95 g C/g H = 25.93 g C

Therefore, we would expect Sample 2 to contain 25.93 grams of carbon, assuming it is composed of the same pure compound as Sample 1.

Answer:

Fischer projection is the method used for representing a three-dimensional organic molecule as a two dimensional molecule.  

This method can be used for determining the D- and L- configuration of the organic molecules.

In the Fisher projection of carbohydrate molecule, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the right-hand side, then the carbohydrate is said to have D-configuration.

Whereas, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the left-hand side, then the carbohydrate is said to have L-configuration.

Final answer:

To determine if a carbohydrate is D or L in a Fischer projection, look at the hydroxyl (-OH) group on the penultimate carbon; if it's to the right, it's a D-sugar, and if it's to the left, it's an L-sugar. This classification does not directly relate to the sugar's optical activity but its stereochemistry relative to glyceraldehyde.

Explanation:

How to Determine if a Carbohydrate is D or L Using a Fischer Projection

When examining a Fischer projection of a monosaccharide, you can determine whether it is a D-sugar or an L-sugar by looking at the orientation of the hydroxyl (-OH) group on the penultimate carbon (second-last carbon) in the chain. The rule is straightforward: if the -OH group on this carbon is to the right side of the Fischer projection, the sugar is designated as a D-sugar. Conversely, if the -OH group is to the left side, the sugar is an L-sugar.

This method of classification is based on the relative configuration to glyceraldehyde, where D-glyceraldehyde has the -OH on the right at the chiral center farthest from the carbonyl group, hence all D-sugars follow this pattern. L-sugars are the mirror images (enantiomers) of the D-sugars, with their -OH groups flipped to the opposite side.

It's important to note that the D/L configuration does not directly correlate with the optical activity of the sugar (++/--) but rather describes its stereochemistry relative to glyceraldehyde. The D/L nomenclature is fundamental in distinguishing the stereochemistry of sugars and their derivatives.

Answer:

Vw = 2.80907 E-3 m³/Kg

Explanation:

specific volume is the inverse of density:

⇒ Vw = 1 / ρ

∴ water at:

⇒ P = 1400 KPa * ( 1000 Pa / KPa ) = 1400000 Pa

⇒ T = 200°C = 473 K

Water at these conditions is found as saturated steam, specific volume would be:

⇒ Vw = R*T / P

∴ R = 8.3144 Pa.m³/ Kg.K

⇒ Vw = (( 8.3144 ) * ( 473)) / 1400000

⇒ Vw = 2.80907 E-3 m³/Kg

Answer:

Try to remember that 1 nanometer is 1 x 10^9, so calculate 6.23 x 10^-3 m  x 1 x 10^9. Your answer is 6.23 x 10^-3 m = 6.23 x 10^6 nm

Explanation:

Answer:

6.23 x 10^-3 m = 6.23 x 10^6 nm

Explanation:

Explanation:

Nitrogen triiodide (NI₃)

Valence electrons of nitrogen = 5

Valence electrons of iodine = 7

The total number of the valence electrons  = 5 + 3(7) = 26

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is shown in image below.

Nitrogen posses one lone pair and thus the geometry is pyramidal.

The Lewis structure of NI3 is drawn by counting the valence electrons, creating single bonds between nitrogen and iodine, completing octets for iodine atoms, and placing remaining electrons on the nitrogen to complete its octet. This results in a structure where nitrogen is single-bonded to three iodine atoms, each surrounded by three lone pairs, and nitrogen has one lone pair.

To draw the Lewis structure for NI3 (nitrogen triiodide), follow these steps:

Count the total number of valence electrons. Nitrogen has 5 valence electrons, and each iodine has 7 valence electrons, totaling (5 + 3*7) = 26 valence electrons.

Draw a single bond between the nitrogen atom and each iodine atom. This will use up 6 of the valence electrons (2 for each bond).

Complete the octets for the iodine atoms by adding six more electrons to each iodine in the form of electron pairs, using up 18 of the remaining valence electrons.

Place any remaining electrons (2) on the nitrogen atom to complete its octet.

Examine the structure. Each iodine has 8 electrons, and nitrogen has 8 electrons, making the structure complete.

If you follow these steps, the resulting Lewis structure for NI3 will show a central nitrogen atom single-bonded to three iodine atoms, with each iodine atom surrounded by three lone pairs of electrons, and one lone pair of electrons on the nitrogen atom.

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Answer:

Re=8561.79

Friction factor is 0.014

Flow is turbulent flow.

Explanation:

Given that

Diameter ,d=1 in

d=0.0254 m     (1 in =0.0254 m )

Volume flow rate,Q = 2 gpm

We know that

[tex]1\ gpm=7.5\times 10^{-5}\ m^3/s[/tex]

[tex]2\ gpm=2\times 7.5\times 10^{-5}\ m^3/s[/tex]

[tex]2\ gpm=15\times 10^{-5}\ m^3/s[/tex]

We know that

Q= A x V

[tex]A=\dfrac{\pi}{4}\times 0.0254^2\ m^2[/tex]

[tex]A=0.00050\ m^2[/tex]

So

[tex]V=\dfrac{15\times 10^{-5}}{0.00050}[/tex] m/s

V=0.3 m/s

So Reynolds number(Re)

[tex]Re=\dfrac{\rho VD}{\mu }[/tex]

Properties of water at 25 C

[tex]\mu=8.9\times 10^{-4}\ Pa.s[/tex]

[tex]Re=\dfrac{1000\times 0.0254\times 0.3}{8.9\times 10^{-4}}[/tex]

Re=8561.79

Re>4000 ,It means that flow is turbulent flow.

Friction factor

If we assume that pipe is smooth

[tex]f=\dfrac{0.136}{Re^{0.25}}[/tex]

[tex]f=\dfrac{0.136}{8561.79^{0.25}}[/tex]

f=0.014.

Friction factor is 0.014

To identify whether the flow is laminar or turbulent and to determine the friction factor, you should first find the Reynolds number using the given parameters, such as the flow rate of 2.0 gpm and then refer to a Moody chart or similar source.

The Reynolds number (NR) and friction factor of the flow can be determined using the given parameters (Flow rate, pipe diameter, and water temperature). The Reynolds number is an indicator that can reveal whether flow is laminar or turbulent. For flow in a tube of uniform diameter, the Reynolds number is defined as an equation related to the properties of the fluid and the characteristics of the flow.

In this case, if the Reynolds number (NR) is below 2000, the flow is considered laminar. If NR is above 3000, the flow becomes turbulent. For values of NR between 2000 and 3000, it may be either or both, depending on factors such as the roughness of the pipe's internal surface and the flow velocity.

The friction factor is a measure of the total resistance created by the force on a fluid as it moves through a pipe. The calculation of this factor also considers the Reynolds number and pipe roughness.

To determine whether your given conditions (2.0 gpm water flow through a 1-inch Schedule 40 steel pipe at 25◦C) will result in laminar or turbulent flow and the corresponding friction factor, please consider calculating the Reynolds number first using the suitable formula and the given conditions and then consulting a Moody chart or a similar source to find the corresponding friction factor.

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Answer : The  value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

[tex]2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)[/tex]

First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].

[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]

[tex]\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}][/tex]

where,

[tex]\Delta H^o[/tex] = enthalpy of reaction = ?

n = number of moles

[tex]\Delta H_f^0[/tex] = standard enthalpy of formation

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

[tex]\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)][/tex]

[tex]\Delta H^o=-1035.6kJ=-1035600J[/tex]

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].

[tex]\Delta S^o=S_f_{product}-S_f_{reactant}[/tex]

[tex]\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}][/tex]

where,

[tex]\Delta S^o[/tex] = entropy of reaction = ?

n = number of moles

[tex]\Delta S_f^0[/tex] = standard entropy of formation

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

[tex]\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)][/tex]

[tex]\Delta S^o=-153J/K[/tex]

Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].

As we know that,

[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]

At room temperature, the temperature is 500 K.

[tex]\Delta G^o=(-1035600J)-(500K\times -153J/K)[/tex]

[tex]\Delta G^o=-959100J=-959.1kJ[/tex]

Therefore, the value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ

Answer:

The number 10,847,100 in Scientific Notation is [tex]1.0847x10^{7}[/tex]

Explanation:

Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.

Answer: The volume of bone for given sample is [tex]30.49cm^3[/tex]

Explanation:

To calculate volume of a substance, we use the equation:

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

Density of bone = [tex]820mg/cm^3=0.82g/cm^3[/tex]      (Conversion factor:  1 g = 1000 mg)

Mass of bone = 25 g

Putting values in above equation, we get:

[tex]0.82g/cm^3=\frac{25g}{\text{Volume of bone}}\\\\\text{Volume of bone}=30.49cm^3[/tex]

Hence, the volume of bone for given sample is [tex]30.49cm^3[/tex]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of [tex]CH_3OH[/tex] = 26.5 g

Molar mass of [tex]CH_3OH[/tex] = 32.04 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

[tex]Moles= \frac{26.5\ g}{32.04\ g/mol}[/tex]

[tex]Moles\ of\ CH_3OH= 0.8271\ moles[/tex]

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

[tex]m=\frac {0.8271\ moles}{0.735\ kg}[/tex]

Molality = 1.13 m

Answer:

E) More neutrons in the nucleus

Explanation:

Isotope -

The atoms which have same number of protons but different number of neutrons.

And since , atomic number = protons number

and , mass number = proton + neutrons .

Hence ,

The atomic number will not change , but the mass number will change .

Hence , the correct option is more neutrons in the nucleus .

Answer:

0.8 mL.

Explanation:

You need to know that the equivalence in mcg to mg is 1000 mcg are 1 mg; so 500 mcg are 0.5 mg. Now you know that there are 0.5 mg per 2 mL, so if you divide all by 2, you will know that you have 0.25 mg per mL. Now you applied a rule of three: ((0.2 mg)(1 ml))/(.25 mg) = 0.8 mL. So the nurse need to administer 0.8 mL to provide 0.2 mg of dose.

Answer:

[tex]\Delta H_{rxn}[/tex] for the given reaction is -890.3 kJ

Explanation:

[tex]\Delta H_{rxn}=\sum n_{i}.\Delta H_{f}(product)_{i}-\sum n_{j}.\Delta H_{f}(reactant)_{j}[/tex]

where [tex]n_{i}[/tex] and [tex]n_{j}[/tex] represents number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence [tex]\Delta H_{rxn}=[1mol\times \Delta H_{f}(CO_{2})_{g}]+[2mol\times \Delta H_{f}(H_{2}O)_{l}]-[1mol\times \Delta H_{f}(CH_{4})_{g}]-[2mol\times \Delta H_{f}(O_{2})_{g}][/tex]

so, [tex]\Delta H_{rxn}=[1mol\times -393.5kJ/mol]+[2mol\times -285.8kJ/mol]-[1mol\times -74.8kJ/mol]+[2mol\times 0kJ/mol]=-890.3 kJ[/tex]

So, Correct answer is -890.3 kJ

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

To convert cm³ or mm³ to m³, one must divide by 1,000,000 or 1,000,000,000 respectively. Thus, 2.7 cm³ is equal to 2.7 x 10^-6 m³ and 2.0 mm³ is equal to 2.0 x 10^-9 m³.

To convert cubic centimeters (cm³) and cubic millimeters (mm³) to cubic meters (m³), you need to know the unit conversions. One square meter is equal to 1,000,000 cubic centimeters and 1,000,000,000 cubic millimeters.

This means you can convert 2.7 cm³ to meters by dividing by 1,000,000, yielding an answer of 0.0000027 m³ (to two significant figures, or 2.7 x 10^-6 m³).

Similarly, 2.0 mm³ can be converted to meters by dividing by 1,000,000,000, resulting in an answer of 0.000000002 m³ or 2.0 x 10^-9 m³.

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The conversion from cm3 and mm3 to m3 is done by multiplying the original number by 1e-6 for cm3 and 1e-9 for mm3. The results for the examples given are 2.7 cm3 equals 2.7e-6 m3, and 2.0 mm3 equals 2.0e-9 m3.

The task is to convert measurements from one unit (cubic centimeters or cubic millimeters) to another (cubic meters). It's important to understand the relevant conversion factors:

1 cm3 = 1e-6 m3

1 mm3 = 1e-9 m3

Applying these to your examples we get:

2.7 cm3 = 2.7 * 1e-6 m3 = 2.7e-6 m3

2.0 mm3 = 2.0 * 1e-9 m3 = 2.0e-9 m3

So these are the conversions using two significant figures.

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Answer:

N₂ is the limiting reactant

Explanation:

The balanced reaction between N₂ gas and H₂ gas is:

N₂ + 3H₂ → 2NH₃

In order to determine the limiting reactant, we have to calculate the number of moles of each rectant, using their molecular weight:

Lastly, we multiply the number of moles of N₂ by 3, and the number of moles of H₂ by 1; due to the coefficients in the balanced reaction. Whichever number is lower, belongs to the limiting reactant.

N₂ => 10.7

H₂ => 50.0

Thus N₂ is the limiting reactant

Answer:

Because for dilute and aqueous solutions the mass of solvent will be a very close value to the volume of solution.

Explanation:

Molar concentration is defined as:

[tex][M]=\frac{molessolute}{volumesolution}[/tex]

And molal concentration is defined as:

[tex][m]=\frac{molessolute}{kgsolvent}[/tex]

And:

Msolution=Msolute+Msolvent

For dilute solutions, we have small amounts of solute, then we have:

Msolution=Msolute+Msolvent, and as the mass of  solute is very small: Msolution≅Msolvent

If the solution is also aqueous (water as solvent), and considering that the density of water is around 1 gm/cm3 or 1 kg/m3:

Msolvent≅Msolution≅Vsolution

Therefore, if we look to the molar and molal equations, we have the same numerator in both (moles of solute) and nearby numbers for the denominator, giving to the molar and molal concentration close values.  

Answer:

pH of the solution = 4.80

Explanation:

pKa = 8.3

Concentration of tris acid = 8.3

[tex]pH = \frac{1}{2} \times p_{ka} - \frac{1}{2} logC[/tex]

[tex]pH = \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(0.05)[/tex]

     [tex]= \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(5\times 10^{-2})[/tex]

     [tex]= 4.15 - \frac{1}{2}\times (-1.30103)[/tex]

    [tex]= 4.15 - (-0.650515)[/tex]

   [tex]= 4.15 + 0.650515[/tex]

   = 4.80

pH of the solution = 4.80

Answer: [tex]41.4\times 10^{6}[/tex]

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s preceding the first integers are never significant.

All zero’s after the decimal point are always significant.

Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power.  It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.

The engineering notation written in the form:

[tex]a\times 10^b[/tex]

where,

a = the number which is greater than 0 and less than 999

b = an integer multiple of 3

Now converting the given value of 41,405,000 into engineering notation, we get [tex]41.4\times 10^{6}[/tex]

Therefore the scientific notation with 3 significant figures is [tex]41.4\times 10^{6}[/tex]

Infrared spectroscopy is a technique used to identify functional groups in molecules by measuring absorption frequencies. It helps determine the presence or absence of specific groups in a molecule.

Infrared spectroscopy is a technique used to identify different functional groups in molecules based on their characteristic absorption frequencies. It involves passing infrared light through a sample and measuring the frequencies that are absorbed. Different functional groups have different absorption frequencies, allowing scientists to determine the presence or absence of specific groups in a molecule. For example, if a sample absorbs in the carbonyl frequency range but not in the alkyne range, it can be inferred that the molecule contains a carbonyl group but not an alkyne.

Answer: The moles of carbon dioxide gas formed is 12.33 moles.

Explanation:

We are given:

Moles of oxygen gas = 12.33 moles

The chemical equation for the reaction of carbon and oxygen to produce carbon dioxide follows:

[tex]C+O_2\rightarrow CO_2[/tex]

By stoichiometry of the reaction;

1 mole of oxygen gas produces 1 mole of carbon dioxide gas.

So, 12.33 moles of oxygen gas will produce = [tex]\frac{1}{1}\times 12.33=12.33mol[/tex] of carbon dioxide gas.

Hence, the moles of carbon dioxide gas formed is 12.33 moles.

Answer:

W = 1.8 KJ

Explanation:

turbine adiabatically: Q = 0

∴ Lost work:

∴ To = 300 K

ideal gas:

⇒ S2 - S1 = (9/2)*(8.314) Ln (383.5/494.8) - (8.314) Ln (1.7/6.3)

⇒ S2 - S1 = 1.36 J/mol.K

entropy generated in the sourroundings:

∴ Q = ΔU = Cv*ΔT

∴ Cv = 3/2*R = 12.5 J/mol.K.....ideal gas

∴ ΔT = 494.8 - 383.5 = 111.3 K

⇒ Q = 12.5 * 111.3 = 1391.25 J/mol

⇒ ΔSo = 1391.25/300 = 4.638 J/mol.K

⇒ W = ( 300 K ) * [ 1.36 J/mol.K + 4.638 J/mol.K ]

⇒ W = 1799.25 J/mol * 1 mol * ( KJ/1000J )

⇒ W = 1.799 KJ ≅ 1.8 KJ

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

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

where:

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

[tex]v(t) = -\frac{d[A]}{dt} = k [A]^{n}[/tex]

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law:                                    rate = k

Concentration-time Equation:   [A]=[A]o - kt

where

For first-order kinetics, we have:

Rate Law:                                        rate= k[A]

Concentration -Time Equation:      ln[A]=ln[A]o - kt

where:

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

Answer:

the "d" option

Explanation:

 Cómo se pronuncia

To calculate the flow I need the volume and time to comply with the formula:

S = V / t

V = volume

t = time

In this problem I have volume and density as data.

There is no way to calculate the time so the correct answer is the "d" option

Answer:

1. Percentage by weight = 0.5023 = 50.23 %

2. molar fraction =0.153

Explanation:

We know that

Molar mass of HClO4 = 100.46 g/mol

So the mass of 5 Moles= 5 x 100.46

       Mass (m)= 5 x 100.46 = 502.3 g

Lets assume that aqueous solution of HClO4  and the density of solution is equal to density of water.

Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.

We know that

Mass = density x volume

Mass of 1000 ml  solution = 1 x 1000 =1000     ( density = 1 gm/ml)

            m'=1000 g

1.

Percentage by weight = 502.3 /1000

Percentage by weight = 0.5023 = 50.23 %

2.

We know that

molar mass of water = 18 g/mol

mass of water in 1000 ml = 1000 - 502.3 g=497.9 g

So moles of water = 497.7 /18 mole

moles of water = 27.65 moles

So molar fraction = 5/(5+27.65)

molar fraction =0.153

Explanation:

For an isothermal process equation will be as follows.

                W = nRT ln[tex]\frac{P_{1}}{P_{2}}[/tex]

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

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

                                           = [tex]\frac{10000 g/s}{18 g/mol}[/tex]

                                           = 555.55 mol/s

                                           = 556 mol/s (approx)

As T = [tex]50^{o}C[/tex] or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

                  W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

                      = [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}[/tex]    

                     = [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303[/tex]    

                     = -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

                     [tex]3440193.809 J/s \times \frac{0.001 kJ}{1 J}[/tex]

                          = 3440.193 kJ/s

                          = 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

The mass of the insoluble solid is determined by first calculating the volume of the liquid solvent, then using that information to find the volume of the solid, and finally multiplying the volume of the solid by its density.

To determine the mass of the solid, we first calculate the volume of the liquid solvent. The volume of a substance can be calculated using the formula volume = mass/density. Thus, the volume of the liquid solvent is 26.6 g / 0.865 g/mL = approximately 30.8 mL.

Now, knowing that the total volume of the solid and liquid together is 91.0 mL, we can find out the volume of the solid as follows: 91.0 mL (total volume) - 30.8 mL (volume of the liquid) = 60.2 mL.

We can then use the given density of the solid (1.75 g/mL) to calculate its mass. According to the formula mass = volume x density, the mass of the solid is therefore 60.2 mL x 1.75 g/mL = 105.35g.

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Answer:

9 L

Explanation:

According to the question , the given reaction is -

2NO(g) + O₂(g)------->2NO₂(g)

Since ,

At STP ,

One mole of a gas occupies the volume of 22.4 L.

Hence , as given in the question -

9 L of NO , i.e .

22.4 L = 1 mol

1 L = 1 / 22.4 mol

9 L = 1 / 22.4  * 9 L = 0.40 mol

From the chemical reaction ,

The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .

Hence ,  

2 moles of NO will produce 2 moles of NO₂.

Therefore ,

0.40 mol of NO will produce 0.40 mol of NO₂.

Hence , the volume of NO₂ can be calculated as -

1 mol = 22.4 L

0.40 mol = 0.40 * 22.4 L = 9 L

To find the volume of NO2 produced, calculate the number of moles of NO reacted using the ideal gas equation. Convert moles to volume using the molar volume of gases at STP.

To find the volume of NO2 produced, we first need to determine the number of moles of NO reacted. Using the ideal gas equation, we can calculate the number of moles:

n = PV/RT = (1 atm)(9 L)/(0.0821 L atm/(mol K))(273 K) = 0.408 mol

Since the reaction has a 2:2 mole ratio, 0.408 mol of NO will produce 0.408 mol of NO2. To convert moles to volume at STP, we can use the molar volume of gases, which is 22.4 L/mol:

Volume of NO2 = 0.408 mol * 22.4 L/mol = 9.1392 L

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Answer:

Explanation:

The metallic properties of steel would strongly not favor its use in construction of aircraft.

Steel is an alloy of carbon and iron. It is denser and generally weighs more than equivalent amount of aluminum. This would imply that the weight of aircrafts would be much more heavier than usual. Weight is a very key component in construction of aircraft. The lighter the mass, the faster propellers can move the craft and lesser amount of energy used. The weight would be a major problem.

Although alloying iron and carbon improves the resistant of the steel to corrosion, it still cannot be compared to that of aluminium. We would build airplanes that would not be durable for so long and that can readily rust on frequent usage.

If airplanes were made of steel, they would be heavier, requiring more fuel and more powerful engines, reducing efficiency and payload capacity. Aluminum's lightness, toughness, and resistance to corrosion are pivotal for modern airplane performance and cannot be matched by steel without compromising efficiency and increasing operating costs.

If airplanes had to be made of steel instead of aluminum, there would be significant consequences for the aviation industry. Steel is much heavier than aluminum, which means that airplanes would be considerably heavier as well. This increase in weight would lead to a need for more powerful engines to provide the necessary lift, which could result in higher fuel consumption and lower efficiency. Moreover, the added weight would decrease the payload capacity of the airplane, as the plane's own structure would constitute a larger portion of its maximum takeoff weight.

The major advantages of aluminum in airplane construction include its lightness, toughness, and resistance to corrosion. These properties are crucial for ensuring that airplanes are able to carry substantial payloads, achieve adequate fuel efficiency, and maintain structural integrity over time. The use of aluminum, a metal that meets these critical needs, is therefore one of the key reasons why modern airplanes are able to achieve the performance levels that they do today. Utilizing steel, despite its strength, would compromise these benefits and could alter the economics and environmental impact of air travel.

Answer:

a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.

b) The greatest total amount is of 1-aminobutane.

c) The least total amount is of the conjugate acid.

Explanation:

The equilibrium in water of 1-aminobutane is:

CH₃(CH₂)₃NH₃⁺ ⇄ CH₃(CH₂)₃NH₂ + H⁺

a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.

b) The equlibrium constant is: K = 1,66x10⁻¹¹.

That means you will have 1-aminobutane:Conjugate acid in a ratio of 6x10¹⁰ : 1 .

The greatest total amount is of 1-aminobutane

c) Thus, The least total amount is of the conjugate acid.

I hope it helps!