What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine concentration in ppm

Answer:

The answer is [tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr

[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr

Explanation:

For flash distillation

F = V+L

[tex]\frac{V}{F} + \frac{L}{F} = 1[/tex]

[tex]\frac{F}{V} -\frac{L}{V} = 1[/tex]

Fz = Vy+Lx

Y = [tex]\frac{F}{V}\times Z - \frac{L}{V}\times X[/tex]                  let, [tex]\frac{V}{F} = F[/tex]

[tex]y = \frac{Z}{F} -[ \frac{1}{F} -1]\times X[/tex]

Highlighted reading

F = 299;  [tex]\frac{V}{F}[/tex] = 0.85 ; z = 0.36

y = [tex]\frac{0.36}{0.85} - (-0.15)\times X[/tex]

 = 0.423 + 0.15x ------------(i)

[tex]y^{*}[/tex] = -43.99713[tex]x^{6}[/tex] + 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.469[tex]x^{}[/tex]+ 0.02011

At equilibrium, [tex]y^{*}[/tex] = y

0.423+0.15[tex]x^{}[/tex] = [tex]y^{*}[/tex]

-43.99713[tex]x^{6}[/tex]+ 148.27274[tex]x^{5}[/tex] - 195.46[tex]x^{4}[/tex]+127.99[tex]x^{3}[/tex]-43.3[tex]x^{2}[/tex]+ 7.319[tex]x^{}[/tex]-0.403

F(x) for Newton's Law

Let [tex]x_{0} = 0[/tex]

     [tex]x_{1}[/tex]     = [tex]\frac{0-[{-0.403}]}{7.319}[/tex]

             = 0.055

     [tex]x_{2}[/tex]      = [tex]\frac{{0.055}-{f(0.055)} {{{{{{{}}}}}}}}{f^{'} (0.055)}[/tex]

             = [tex]\frac{{(0.055)}-(-0.11)}{3.59}[/tex]

             = 0.085

    [tex]x^{3}[/tex]    = [tex]\frac{{0.085}-(0.024)}{2.289}[/tex]

           = 0.095

   [tex]x^{4}[/tex]     = [tex]\frac{{0.095}-(-0.0353)}{-1.410}[/tex]

            = 0.07

From This x and y are found from equation (i) and L and V are obtained from [tex]\frac{V}{F}[/tex]  and F values

[tex][\frac{LX_1}{46} + \frac{L(1-x)}{18}]\times0.454[/tex] mol/hr

[tex][\frac{Vy_1}{46}+\frac{V(1-y)}{18}]\times0.454[/tex]mol/hr

The question relates to flash distillation with a focus on an ethanol-water azeotropic mixture. It addresses the limitation of distillation due to the azeotrope formation where the maximum ethanol concentration is about 95.6%. Understanding vapor-liquid equilibrium and distillation curves is essential for this process.

The question is concerned with the principles of flash distillation and how it relates to azeotropic mixtures, specifically an ethanol-water mixture. When dealing with flash distillation, we are interested in how the composition of the mixture changes as it goes through vaporization and condensation cycles. Flash distillation is often used for the separation of components in a liquid mixture that have significantly different volatilities. The crux of this process is that the vapor formed will be richer in the more volatile components, in this case, ethanol.

Regarding the ethanol-water azeotrope, the mixture cannot be distilled beyond a certain concentration of ethanol due to the constant boiling nature at that specific composition. An azeotrope is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. In the case of ethanol and water, the maximum concentration of ethanol that can be obtained by simple distillation is approximately 95.6%. This is because, at this concentration, the mixture behaves as if it were a single component with a unique boiling point, and further separation by boiling alone is not possible without additional chemical agents or advanced separation techniques like extractive distillation.

In chemical engineering, it's also important to consider the vapor-liquid equilibrium and how it is represented by phase diagrams, such as the distillation curve given in this problem. Understanding these diagrams is critical for designing and operating distillation processes, whether in the context of a laboratory setup or an industrial scale operation.

To obtain accurate melting points on a Mel-Temp, grind the sample into a fine powder, put the smallest visible amount in a capillary tube, fill the tube with about 3mm of powder, set the voltage to rise quickly to 20 degrees below expected temperature and slow down to 2 degrees per minute, and record the temperature at which liquid first appears and the last crystal disappears.

The four important steps needed to obtain accurate melting points on a Mel-Temp are as follows:

By following these steps, you can ensure that you are determining the melting point accurately.

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Answer:

The extinction coefficient is 4240 M⁻¹cm⁻¹

Explanation:

The Beer-Lambert law for the absorbance of a monochromatic light by a molecule is given by the following mathematical expression:

[tex]A=EbC[/tex]----------------------(1)

Here, E is the extinction coefficient, b is the length of light pathway through the solution and C is the concentration in mol/L. This is a linear equation and fits well with its formula  

y = mx + C-------------------(2)

Here, m is the slope and C is the y intercept. y and x are the two variables. Comparing  equation 1 and 2, Absorbance (A) and molar concentration (C) are the two variables, while the product of extinction coefficient (E) and path length (b) makes the slope of the line. The y-intercept in the given equation is due to the solvent interference.

From this information, we can determine the extinction coefficient by the slope of the line with the following formula.

[tex]m=Eb[/tex]

[tex]E=\frac{2120 (M^{-1} )}{0.5(cm)}[/tex]

[tex]E= 4240 M^{-1}cm^{-1}[/tex]

Answer:The result indicates that the polar component of the mixture fail to separate. In a case like this a third liquid like acid, base or complexing agent is often added to retain more water in the organic solvent.

Explanation:

Answer:

Seems to me that the sample had not been properly adsorbed, did not establish a true equilibrium, and therefore washed through the column rather that separated. The sample should be allowed to stand a short time before elution rather than putting the solvent in immediately.

Explanation:

Answer:

6.104x10^2=610.4

9.5x10^3=9500

Explanation:

To convert from scientific notation (6.104 x10^2, 9.5 x 10^-3) to standard notation, move the decimal point in accordance with the power of 10 (right for positive, left for negative). The converted numbers become 610.4 and 0.0095, respectively.

In scientific notation, the base number is expressed as a decimal between 1 and 10, and tens are represented by powers of 10. In standard notation, we write the number in usual decimal representation.

To convert 6.104 x10^2 to standard notation, move the decimal two places to the right (because the exponent is positive), giving you 610.4.

For 9.5 x 10^-3, move the decimal three places to the left (because the exponent is negative). So, it becomes 0.0095.

Remember that the number of significant figures must remain the same when converting to standard notation.

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The question is incomplete, here is the complete question:

Calculate the volume in liters of a [tex]8.75\times 10^{-5}M[/tex] silver (II) oxide solution that contains 200.g of silver (II) oxide . Round your answer to 3 significant digits.

Answer: The volume of solution is [tex]1.84\times 10^4L[/tex]

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

We are given:

Molarity of solution = [tex]8.75\times 10^{-5}M[/tex]

Given mass of silver (II) oxide = 200. g

Molar mass of silver (II) oxide = 124 g/mol

Putting values in above equation, we get:

[tex]8.75\times 10^{-5}M=\frac{200}{124\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{200}{124\times 8.75\times 10^{-5}}=1.84\times 10^4L[/tex]

Hence, the volume of solution is [tex]1.84\times 10^4L[/tex]

To calculate the volume in liters of a silver(II) oxide solution, you need to know the amount of the solution in milligrams and the density of the solution. Use the formula Volume (in liters) = x mg / (density in mg/L) and round the final answer to the appropriate number of significant digits.

To calculate the volume in liters of a silver(II) oxide solution, we need to know the amount of the solution in milligrams. Let's say the solution contains x mg of silver(II) oxide.

To convert mg to liters, we need to know the density of the solution. Once we have the density, we can use the formula:

Volume (in liters) = x mg / (density in mg/L)

Round the final answer to the appropriate number of significant digits.

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Answer:

Option B and D

Explanation:

The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.

A feasible way to synthesize the product entails the spontaneity of the reaction.

The spontaneity of a reaction is explained by the standard gibb's free energy.

If Delta-G = -ve ( the reaction is spontaneous)

if Delta -G = +ve ( the reaction is non-spontaneous)

if Delta-G = 0 ( the reaction is at equilibrium)

Hence the option (B and D) with negative Gibb's free energy are the reaction that are spontaneous.

Answer: The molality of the solution is 0.953 m

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute

[tex]V_s[/tex] = volume of solution in ml

Given : 0.907 moles of lead(ii)nitrate is dissolved in 1000 ml of the solution.

density of solution= 1.252 g/ml

Thus mass of solution = [tex]Density\times volume=1.252\times 1000 ml=1252g[/tex]

mass of solute =[tex]moles\times {\text {Molar mass}}=0.907mol\times 331g/mol=300g[/tex]

mass of solvent =mass of solution - mass of solute = (1252-300)g= 952 g

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

[tex]Molality=\frac{n\times 1000}{W_s}[/tex]

where,

n = moles of solute  = 0.907 moles

[tex]W_s[/tex]= weight of solvent in g  = 952 g

[tex]Molality = \frac{0.907\times 1000}{952g}=0.953m[/tex]

Thus molality of the solution is 0.953 m

Final answer:

To find the molality of the 0.907 M lead(II) nitrate solution, we first calculate the mass of the solvent in kilograms and then divide the amount in moles of the solute by this mass. The final molality is approximately 0.953 mol/kg.

Explanation:

The question is asking to find the molality of a 0.907 M lead(II) nitrate solution with a density of 1.252 g/cm3. To calculate the molality, we need the number of moles of solute and the mass of the solvent in kilograms.

First, calculate the mass of 1 litre (1000 cm3) of solution using the density:

Mass of solution = Density imes Volume = 1.252 g/cm3 imes 1000 cm3 = 1252 g

Next, the mass of the solute in 1 litre of solution:

Mass of solute = Molarity imes Molar Mass imes Volume = 0.907 mol/L imes 331.2 g/mol imes 1 L = 300.2794 g

Now, find the mass of the solvent (water) by subtracting the mass of solute from the total mass of the solution:

Mass of solvent = Mass of solution - Mass of solute = 1252 g - 300.2794 g = 951.7206 g

Convert mass of solvent to kilograms:

Mass of solvent (kg) = 951.7206 g imes (1 kg / 1000 g) = 0.9517206 kg

Finally, calculate the molality (m):

Molality (m) = Moles of solute / Mass of solvent (kg) = 0.907 mol / 0.9517206 kg = 0.953 m

Therefore, the molality of the lead(II) nitrate solution is approximately 0.953 mol/kg.

Answer:

Here's what I get  

Explanation:

Does an ice cube feel cold to the hand?

The melting of ice is an endothermic process.

If you hold an ice cube, heat flows from your hand to the ice. When your hand loses heat, it feels cold.

In the same way, the endothermic dissolving of ammonium nitrate removes the heat from your hand, so the pack feels cold.

Answer:

ThOD =239.792 mg/L

Explanation:

Theorical Oxigen demand (ThOD):

is the theoretical amount of oxygen

required to oxidize the organic fraction of a

waste up to carbon dioxide and water.

⇒ C sln = 450 mg C6H12O6 / 2 L H2O = 225 mg/L sln

∴ mm C6H12O6 = 180.156 g/mol

balanced reaction:

∴ mol C6H12O6 = 1 mol

⇒ mass C6H12O6 = (180.156 g/mol)( 1 mol) = 180.156 g

∴ the value of ThOD is determined when 180.156 g C6H12O6 consume mass O2 = 6(32) = 192 g Oxygen;  then in a solution of 225 mg/L, you have:

⇒ ThOD = (192/180.156)×225 mg/L

⇒ ThOD = 239.792 mg/L

The theoretical oxygen demand is 0.24 g/L or 240mg/L.

To calculate the theoretical oxygen demand of a solution containing 450mg of glucose (C6H12O6) in 2 L of distilled water, we need to start by determining the stoichiometry of the complete combustion of glucose, which can be represented by the balanced chemical equation:

C6H12O6 + 6O2 → 6CO2 + 6H2O

From the equation, we can see that 1 mole of glucose requires 6 moles of oxygen to completely react. Therefore, to calculate the oxygen demand, we first need to find the number of moles of glucose present in the solution:

The molar mass of glucose (C6H12O6) is approximately 180 g/mol. So, 450 mg of glucose is equal to 0.450 g or 0.450/180 = 0.0025 moles of glucose.

Using the stoichiometry from the equation, the moles of oxygen required is 0.0025 moles of glucose  imes 6 moles of oxygen/mole of glucose = 0.015 moles of oxygen.

Now, the molar mass of oxygen (O2) is approximately 32 g/mol, hence 0.015 moles  imes 32 g/mol = 0.48 grams of oxygen. Since the solution volume is 2 L, we need to find the amount per liter, which gives us 0.48 g / 2 L = 0.24 g/L or 240 mg/L of theoretical oxygen demand.

Answer:

This is due to the physical properties of the sample, since it affects the volume dispensed.

Explanation:

For example, in the case of very dense samples, selected samples to adhere to the surface of the tip, dispensing more slowly. In contrast, ethanol samples are less viscous and more volatile and are dispensed more rapidly. Some of the ways to minimize these inconveniences are the use of ultra low retention pipette tips, since they have a hydrophobic plastic additive that prevents the liquid from adhering to the inside of the tip.

Another way is to use the reverse pipetting.

Explanation:

As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.

Since, the given reaction is exothermic in nature so, when we increase the temperature then reaction will shift in the direction opposing the increase in temperature. Hence, the equilibrium will shift on left side when we increase the temperature.

On the other hand, when we decrease the temperature then the equilibrium will shift on the right side.

The correct classifications for the actions based on the given are as follows:

a. Increase the temperature: This causes a rightward shift in the direction of the net reaction.

b. Decrease the temperature: This causes a leftward shift in the direction of the net reaction.

For a chemical reaction at equilibrium, Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

The reaction in question has a standard enthalpy change of 156 kJ, which indicates that the reaction is exothermic (releases heat).

a. Increase the temperature:

When the temperature is increased for an exothermic reaction, the system tries to absorb the excess heat to restore equilibrium. Since the forward reaction releases heat, the system will favor the reverse reaction to absorb heat. However, because the forward reaction is exothermic and already releases heat, increasing the temperature actually provides energy to the system, which drives the reaction forward to consume the added energy. Therefore, increasing the temperature causes a rightward shift in the direction of the net reaction.

b. Decrease the temperature:

Conversely, when the temperature is decreased, the system tries to release heat to restore equilibrium. For an exothermic reaction, the forward reaction releases heat, so decreasing the temperature will favor the forward reaction to release heat and restore equilibrium. Therefore, decreasing the temperature causes a leftward shift in the direction of the net reaction, favoring the products since the forward reaction is exothermic and will release heat.

In summary, for an exothermic reaction, increasing the temperature will shift the equilibrium to the right (towards the products), and decreasing the temperature will shift the equilibrium to the left (towards the reactants).

Answer:

CH3CH2CH(OH)CH3

Explanation:

This reaction follows the Markonikovs rule which states that the addition of an acid yo an assymetric alkene, the proton given off by the acid goes to the Carbon attached/with more hydrogen substituents and the rest of the acid ion goes to the Carbon with more alkyl substituents.

The reaction under goes 2 steps,

The first step called Oxymercuration involves the dissociation of Hg(OAc)2 into HgOAc+ and OAc-. The double bond in 1-butene (high electron site) bonds to an electron deficient HgOAc+ thereby forming an unstable complex, the lone pairs of oxygen in the water molecule (H2O) is attracted and a conplex is further formed with a proton on the now bonded H2O molecule.

To stabilise this, the proton(H+) is removed and then Deoxymercuration occurs. Deoxymercuration is simply removing the HgOAc+ molecule from the compound. This is done by the further reaction with the mixture of NaBH4 in alkaline medium; this is a string reducing reagent (reducting reagent - addition of excess Hydrogen) to form a secondary alcohol(Butan-2-ol)

Below us the mechanism in the attachment, i hope this was helpful.

Answer: 7.68×10^14 J

Explanation:

The magnitude of the charge q1=q2 = 1.6×10^-19 C

Proportionality constant K = 9×10^9 Nm²/C²

Da= 2×10^-10m

Db= 3 ×10^-15m

W= k q1q2 ( 1/Db -1/Da)

W= 9×10^9 (1.69×10^-19)²(1/3.0×10^-15 - 1/2×10^-10)

W= 7.68 ×10^-14J

Therefore the amount of work requires is 7.68 ×10^-14J

Answer:

Fracking involves the ______drilling to extract oil and natural gas from the rock. Select one: a. vertical b. lateral c. horizontal

The answer is option C (horizontal)

Explanation:

Fracking commonly known as Hydraulic fracturing is an oil and gas production technique that has enabled the extraction of oil and gas from rock (shale rock).

A fracturing fluid typically water with additives is pumped under high pressure into the drilling pipe to widen fractures in the rock or to create new ones and and allow trapped gas or crude oil to flow through a pipe to a wellhead at the surface, deep holes are drilled down into the shale rock. These fractures are held open by an injected proppant, typically sand and release hydrocarbons. Fracking takes about three to five days. Since shale reserves are typically distributed horizontally rather than vertically, horizontal directional drilling method is employed.

Some of the impacts of Fracking on the environment are:

Fracking, or hydraulic fracturing, primarily involves horizontal drilling. Starting vertically, the drill transitions to a horizontal path to enhance contact with gas or oil-bearing rocks, thereby improving extraction efficiency.

Fracking, also known as hydraulic fracturing, involves the horizontal drilling to extract oil and natural gas from the rock. The process begins with a vertical hole that reaches the desired depth and then transitions into lateral or horizontal drilling. This allows for increased contact with the oil or gas-bearing rock, improving the extraction process and its efficiency.

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Final answer:

Filtration devices used in different situations include membrane filtration, gravity filtration, microfiltration, vacuum filtration, activated carbon filtration, and simple filtration.

Explanation:

For A, to remove powdered decolorizing charcoal from 20mL of solution, you can use membrane filtration with a syringe filter. It allows you to push the solution through by depressing the syringe's plunger.

For B, to collect crystals obtained from crystallizing a substance from about 1 mL of solution, you can use gravity filtration. It involves pouring the solution through a funnel with filter paper to separate the crystals.

For C, to remove a very small amount of dirt from 1 mL of liquid, you can use microfiltration. It uses a membrane filter with a pore size small enough to capture the dirt particles.

For D, to isolate 2.0 g of crystals from about 50 mL of solution after performing a crystallization, you can use vacuum filtration. It involves connecting a filtration unit to a vacuum to draw the solution through the filter.

For E, to remove dissolved colored impurities from about 3 mL of solution, you can use activated carbon filtration. It utilizes activated carbon to adsorb the impurities.

For F, to remove solid impurities from 5 mL of liquid at room temperature, you can use simple filtration. It involves pouring the liquid through a filter to separate the solid impurities.

Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

When heating a flammable or volatile solvent for recrystallization, one should not do the heating using an open flame or heat near any open flame in the laboratory.

They are liquids that can catch flame easily either directly or through their vapors.

This means that such liquids should not be heated using an open flame or near any open flame to prevent fire accidents.

Heating near an open flame may cause the vapor to catch fire. Heating in a stoppered flask is not ideal because the main aim of heating is to allow the vapor to escape.

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The molecularity of each reaction is determined by the number of molecules or atoms participating as reactants. The first is bimolecular, the second is termolecular, and the third is unimolecular.

Certainly, here's a more detailed explanation for each reaction:

1. The reaction F(g) + H2(g) → HF(g) + H(g) is bimolecular in nature. Bimolecular reactions involve two molecules or atoms as reactants that come together to form products. In this case, one molecule of fluorine (F2) is reacting with one molecule of hydrogen gas (H2) to produce one molecule of hydrogen fluoride (HF) and one hydrogen atom (H). The balanced equation reflects this stoichiometry: 1 F2 + 1 H2 → 2 HF + 1 H.

2. The reaction H2(g) + 2I(g) → 2HI(g) is termolecular. Termolecular reactions are relatively rare and involve three molecules or atoms as reactants. In this reaction, one molecule of hydrogen gas (H2) is reacting with two molecules of iodine gas (I2) to form two molecules of hydrogen iodide (HI). The balanced equation represents this as 1 H2 + 2 I2 → 2 HI. The presence of three molecules as reactants in this equation makes it termolecular.

3. NO2Cl(g) → NO2(g) + Cl(g) is a unimolecular reaction. Unimolecular reactions involve only one molecule or species as a reactant that decomposes or rearranges to form products. In this reaction, a single molecule of nitrogen dioxide chloride (NO2Cl) is decomposing into one molecule of nitrogen dioxide (NO2) and one chlorine atom (Cl). The equation 1 NO2Cl → 1 NO2 + 1 Cl illustrates the unimolecular nature of this reaction.

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Answer:

d = 3.11 g /cm³

Explanation:

Volume -

The volume of a cuboidal object can be calculated by multiplying the length , width and height ,

Hence , Volume ( V ) = l * w * h ,

From the question ,

l = 8.2 cm

w = 3.8 cm

h = 7 cm

Hence, the volume is ,

V = l * w * h

V = 8.2 cm * 3.8 cm * 7 cm

V = 218.12 cm³

Density -

Density of a substance is given as the mass divided by the volume ,

Hence,

d = m / V

From the question,

m = 0.68 kg

since ,

1 kg = 1000g

m = 0.68 * 1000 = 680 g

V = 218.12 cm³ ( calculated above )

Hence , the value of density can be calculated by using the above equation,

d = m / V

d = 680 g / 218.12 cm³

d = 3.11 g /cm³

Answer: The new water level of the cylinder is 24.16 mL

Explanation:

To calculate the volume of water displaced by silver, we use the equation:

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

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

[tex]10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL[/tex]

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

Answer:

0.370 micromoles of magnesium fluoride the chemist has added to the flask.

Explanation:

[tex]Molarity=\frac{moles}{\text{Volume of solution(L)}}[/tex]

Moles of  magnesium fluoride  = n

Volume of the solution = 380.0 mL = 0.380 L (1 mL = 0.001 L)

Molarity of the solution = [tex]9.75\times 10^{-4} mM=9.75\times 10^{-7} M[/tex]

(1 mM = 0.001 M)

[tex]9.75\times 10^{-7} M=\frac{n}{0.380 L}[/tex]

[tex]n=3.705\times 10^{-7} mol[/tex]

1 mole = [tex]10^6[/tex] micro mole

[tex]n=3.705\times 10^{-7} \times 10^6 \mu mol=0.3705 \mu mol[/tex]

0.370 micromoles of magnesium fluoride the chemist has added to the flask.

The pH of the solution is 3,76

Why?

To solve this problem we have to apply the Henderson-Hasselbach equation, which is used whenever we need to calculate the pH of a solution of an acid HA (Lactic Acid) and its conjugate base A⁻ (Sodium Lactate)

We can use either moles or concentrations for this equation. In this case, we are going to use the moles, and we are going to take the pKa of lactic acid as 3,86:

[tex]pH=pKa+log(\frac{nA^- }{nHA} )=3,86+log(\frac{(0,04moles)}{(0,05mL)*(1.0M)} )\\\\pH=3,76[/tex]

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Final answer:

To calculate the pH of a buffer solution containing lactic acid and its conjugate base sodium lactate, the Henderson-Hasselbalch equation is used. The calculated pH for the buffer solution is approximately 3.76.

Explanation:

The question involves calculating the pH of a solution containing a weak acid and its conjugate base, which is a typical buffer solution scenario. We recognize it as a buffer solution because it contains both the weak acid (lactic acid, HC3H5O3) and its conjugate base (sodium lactate, NaC3H5O3). The Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), can be used to calculate the pH of a buffer solution where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.

Firstly, we need to determine the concentrations of the acid and the base. The concentration of the lactic acid remains at 1.0 M since the added volume is negligible, while the amount of sodium lactate added is given as 0.040 moles. Since the volume is 50.0 mL (or 0.050 L), the concentration of sodium lactate will be 0.040 moles/0.050 L = 0.80 M.

Using this information and the known pKa of lactic acid, which is approximately 3.86, we can now use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:

pH = pKa + log([A-]/[HA])
pH = 3.86 + log(0.80/1.0)
pH = 3.86 + log(0.80)
pH ≈ 3.86 + (-0.09691)
pH ≈ 3.76

Therefore, the pH of the buffer solution is approximately 3.76.

Answer:

it same mass as water

Explanation:

Final answer:

The steam resulting from boiled water will have the same mass as the water from which it was boiled, adhering to the law of conservation of mass, as long as the system is closed and no mass is lost due to external factors.

Explanation:

When you boil a sample of water, the steam that results will have the same mass as the water if you ignore any minor mass loss due to gases dissolved in water escaping or small amounts of water that may stick to the container. This is because the mass of a substance remains constant during a phase change; only the state of the substance changes from liquid to gas. This is in accordance with the law of conservation of mass.

The mass of a substance does not change as it undergoes physical transformations; thus, when you boil water, all of the mass is converted to steam provided the system is closed and no mass escapes. In an open system, there might be a loss of mass due to external factors, but the boiling process itself does not add or remove mass from the water.

Answer:

The activation energy for the reaction is, 1.90682  KJ/mol.

Explanation:

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at 225 K = [tex]0.391 s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at 525 K = [tex]0.700 s^{-1}[/tex]

[tex]Ea[/tex] = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

[tex]T_1=225 K, T_2=525 K[/tex]

Now put all the given values in this formula, we get

[tex]\log (\frac{0.700 s^{-1}}{0.391 s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{225 K}-\frac{1}{525 K}][/tex]

[tex]Ea=1,906.82 J/mole=1.90682 KJ/mol[/tex]

Therefore, the activation energy for the reaction is, 1.90682  KJ/mol.

Answer:

2.1 × 10⁻¹ M

2.0 × 10⁻¹ m

Explanation:

Molarity

The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

3.9 g × (1 mol/93.13 g) = 0.042 mol

The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M

Molality

The moles of solute are 0.042 mol.

The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

200 mL × 1.05 g/mL = 210 g = 0.210 kg

The molality of aniline is:

m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m

Explanation:

It is known that the ionic strength of a solution, I, is represented as follows.

          [tex]I = \frac{1}{2} \sum m_{i}z^{2}_{i}[/tex]

where,  [tex]m_{i}[/tex] = the concentration of the ion,

             [tex]z_{i}[/tex] = the charge of the ion

Now, for 1 M NaCl solution, the ionic strength will be calculated as follows.

           I = [tex]\frac{1}{2} [(1 \times 1^{2}) + (1 \times 1^{2})][/tex]

             = 1.00

And, for 1 M [tex](NH_{4})_{2}SO_{4}[/tex], concentration of [tex][NH_{4}][/tex] = 2 M and [tex][SO_{4}][/tex] = 1 M.

             I = [tex]\frac{1}{2} [(2 \times 1^{2}) + ((1 \times 2^{2})][/tex]

               = 3.00

As the ionic strength of 1 M [tex](NH_{4})_{2}SO_{4}[/tex], is different  therefore it is a serious error.

Using 1.0 M ammonium sulfate instead of 1 M NaCl is a serious error in the biological preparation due to the differences in dissociation and specific properties of the compounds.

The student's choice to use 1.0 M ammonium sulfate instead of 1 M NaCl is a serious error because they are not equivalent compounds. Ammonium sulfate dissociates into two ions (NH4+ and SO42-), while NaCl only dissociates into two ions (Na+ and Cl-). Therefore, the ionic strength of the solution will not be the same.

In addition, the presence of different ions in ammonium sulfate will affect the specific properties and behavior of the solution compared to NaCl. For example, the presence of ammonium ions can affect the pH and the precipitation of certain compounds.

It is important to use the correct reagents in a biological preparation to maintain the desired conditions and ensure accurate results.

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Answer:

0.4g

Explanation:

1.0% (w/v%) = 1 g of agarose 100 ml of Tris-Acetate-EDTA,  this is the buffer that agarose is run with

the amount of agarose for 40 ml = 1 /100 × 40 ml = 0.4 g  

Final answer:

To prepare 40 mL of a 1.0% (w/v %) agarose gel, measure and mix 0.4 grams of agarose powder with 40 mL of buffer solution in an Erlenmeyer flask.

Explanation:

If you are asked to make 40 mL of a 1.0% (w/v %) agarose gel, you will add 0.4 grams of agarose to the 40 mL of buffer. To made this calculation, you need to understand the meaning of w/v %, which stands for weight/volume percentage. It is defined as the mass of a solute (in this case agarose powder) divided by the volume of the solution, and then multiplied by 100 to get the percentage.

For a 1.0% (w/v %) agarose gel, this means that you need 1 gram of agarose powder for every 100 mL of solution. Therefore, for 40 mL of solution, you simply use the proportion:

1 gram/100 mL = X grams/40 mL

Solving for X gives you:

X = (1 gram/100 mL) * 40 mL = 0.4 grams

Conclusion: To prepare a 40 mL agarose gel at 1.0% w/v concentration, weigh out 0.4 grams of agarose powder using an electronic scale, and then blend it with the appropriate buffer in an Erlenmeyer flask.

Answer:

b.  CI (1255 kJ/mol), Ge (784 kJ/mol), and K (418 kJ-mol)

Explanation:

In general, the first ionization energy for a given period increases as we go from left to right in the periodic table (there are some exeptions as with every rule), and the first ionization energy decreases a we go down in the periodic table. The reason for this are:

1. As we move from left to right in a group, the effective nuclear charge increases which makes it harder to remove the electron.

2. As we increase the period in going top to bottom in the periodic table we are adding another shell farther away from the nucleus, making it easier to remove the electron.

Given the three values for the first ionization energy in kJ/mol : 418, 784 and 1255, we expect the highest value to correspond to Cl which belongs to period 3 (K and Ge belong to period 4).

Now comparing K and Ge which belong to period 4, Ge will have a higher effective nuclear charge than K .

So the match will be Cl (1255 kJ/mol), Ge (784) and K(418)

Correct answer is b.

Note: there could be some confusion since the value of 784 was misplaced in the question statement, but we can deduce that in this question we are asked to match the values for the atoms.

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x =  1.053 g / 0.0405  mol = 26 g/mol