How many milliliters of a 1:5000 w/v solution of the preservative benzalkonium chloride can be made from 125 mL of a 0.2% w/v solution of the preservative? a. 1250 mL b. 2500 mL C. 625 mL d. 1875 ml

Answers

Answer 1

Final answer:

To determine the volume of a 1:5000 w/v solution that can be made from 125 mL of a 0.2% w/v solution of benzalkonium chloride, we calculate the mass of the solute in the initial solution and divide it by the concentration of the desired dilute solution, giving us 1250 mL.

Explanation:

To calculate the target volume of a 1:5000 w/v solution from a 0.2% w/v solution of benzalkonium chloride, we first need to understand the concentration terms. A 0.2% w/v solution means 0.2 g of solute is present in every 100 mL of solution. For a 1:5000 w/v solution, there is 1 g of solute in 5000 mL of solution, or essentially y, 0.0002 g/mL (1 g / 5000 mL).

We then figure out how much benzalkonium chloride is in the initial 125 mL of the 0.2% solution. Since 0.2% w/v is equivalent to 0.2 g/100 mL, we have 0.2 g in 100 mL, hence in 125 mL, we'll have 0.25 g (since 125 mL is 1.25 times 100 mL).

To work out how much of the 1:5000 solution we can make with 0.25 g, we use the concentration of the 1:5000 solution, which is 0.0002 g/mL. Dividing the total available mass of solute (0.25 g) by the concentration (0.0002 g/mL), we get 1250 mL. Thus, the answer is a. 1250 mL.


Related Questions

What is the preferred electrical charge of a Sodium ion? O a. +1 O O O b. +2 Oco O O d.-1 O e. 2 O

Answers

the answer is a) +1 (Na+)

What are the names of the following compounds: FeCl HNO NaSO SO

Answers

Answer:

FeCl: Ferric Chloride (also called iron chloride), comes from Fe (ferrum, or iron), and Cl (Chlorine)

HNO: Nitroxyl, from N (Nitrogen), and the acidic nature of a radical ending in -yl.

NaSO:  Sodium sulfate, Na (Sodium), S (Sulfur), O (Oxygen).

SO: Sulfur monoxide (Mono-One), O (Oxygen) and S (Sulfur).

Glycerol is a syrupy liquid often used in cosmetics and soaps. A 3.25-L sample of pure glycerol has a mass of 4.10 x 10 g.
What is the density of glycerol in g/cm"? Express your answer in grams per cubic centimeter.

Answers

Explanation:

It is known that density is the amount of mass present in liter of solution or substance.

Mathematically,      Density = [tex]\frac{mass}{volume}[/tex]

It is given that volume is 3.25 L and mass is [tex]4.10 \times 10^{3} g[/tex]. Hence, calculate the density of glycerol as follows.

                     Density = [tex]\frac{mass}{volume}[/tex]

                                   = [tex]\frac{4.10 \times 10^{3} g}{3.25 L}[/tex]

                                    = [tex]1.26 \times 10^{3} g/L[/tex]

As, 1 L = 1000 [tex]cm^{3}[/tex].

So,           [tex]1.26 \times 10^{3} g/L \times \frac{1000 cm^{3}}{1 L}[/tex]

                 = [tex]1260 \times 10^{6} g/cm^{3}[/tex]    

Thus, we can conclude that the density of glycerol is [tex]1260 \times 10^{6} g/cm^{3}[/tex].

In natural convection heat transfer, the parameter
a) Nusselt Number
b) Rayleigh Number
c) Grashof Number
c) Lewis Number

Answers

Answer:

correct answer is option c i.e Grashof Number

Explanation:

The Grashof number is a dimensionless number, which is named after renowned scientist  Franz Grashof. The Grashof quantity is defined as the proportion of the buoyant force to viscous force performing on a fluid in a pace boundary layer.

Its function in natural convection is more or less the same as that of Reynolds's number in compelled convection.

Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas from aqueous glucose and oxygen gas. Calculate the moles of glucose produced by the reaction of 2.40 moles of water. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Answers

Answer: 0.4 moles of glucose are produced by the reaction of 2.40 moles of water.

Explanation:

Photosynthesis is a phenomenon in which green plants containing chlorophyll use sunlight as a source of energy to convert carbon dioxide and water to form glucose and oxygen.

The balanced chemical equation is:

[tex]6CO_2+6H_2O\overset{sunlight}\rightarrow C_6H_{12}O_6+6O_2[/tex]

According to stoichiometry:

6 moles of water produces = 1 mole of glucose

2.40 moles of water produces = [tex]\frac{1}{6}\times 2.4=0.4[/tex] moles of glucose

Thus 0.4 moles of glucose are produced by the reaction of 2.40 moles of water.

The reaction of carboxylic acids with alcohol in the presence of an acid catalyst yields Select one: amides O esters O no reaction occurs O aldehydes

Answers

Answer:

esters

Explanation:

The -OR group of the alcohol replaces the -OH of the carboxylic acid, forming an ester. See attachment for condensed mechanism.

Which statement is true about obligate anaerobes? View Available Hint(s) Which statement is true about obligate anaerobes? They obtain energy by oxidizing ferrous ions. They will use O2 if it is present, but may obtain energy by fermentation if needed. They use O2 for cellular respiration and cannot grow without it. They are poisoned by O2. They live exclusively by cellular respiration or by anaerobic respiration.

Answers

Answer: They are poisoned by O₂

Explanation:

Obligate anaerobes cannot survive in normal concentrations of oxygen. Depending on the species, tolerance varies from 0.5% to 8% oxygen.

Under normal cellular conditions, O₂ turns into O₂⁻ and H₂O₂, toxic to the organism. Obligate anaerobes lack enzymes superoxide dismutase and catalase, capable of turning O₂⁻ and H₂O₂ back into breathable O₂.

The statement that is true about obligate anaerobes is they are poisoned by [tex]O_2[/tex].

Microorganisms known as obligatory anaerobes are incapable of surviving or developing in the presence of oxygen. They cannot detoxify the reactive oxygen species (ROS) created during aerobic respiration because they lack the enzymes catalase and superoxide dismutase. As a result, oxygen is poisonous to them. When cells are exposed to oxygen, toxic byproducts can arise that injure the cells' biological constituents and ultimately cause cell death.

Obligate anaerobes are constrained to anaerobic metabolic pathways, in contrast to facultative anaerobes, which can flip between aerobic and anaerobic metabolism depending on oxygen availability. Typically, they break down organic substances without the need of oxygen through fermentation or anaerobic respiration to produce energy.

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Which of the following is a strong base? КОН b. H2 11 c. NH d. HCI a.

Answers

Answer:

КОН

Explanation:

Potassium hydroxide -

It is an inorganic compound and commonly known as the caustic potash , with molecular formula of KOH .

It is highly soluble in water , and dissociates into its respective ions in water , i.e. , K⁺ and OH⁻

Potassium hydroxide is a very strong base and a colorless solid .

Hence , from the given compound , KOH , is the strongest base among all .

A 36.5 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 25 mg per kg of body weight b.id What is the meaning of the Latin abbreviation b.i.d.? O every other day O as needed O twice daily O once daily How many hours should pass between each administration? number of hours: 413 How many milligrams of amoxicillin should be given at each administration? mass of amoxicillin: 413 Amoxicillin should be stored between 0 °C and 20 °C. Should the amoxicillin be stored in the freezer or the refrigerator?

Answers

Answer:

a) b.i.d: twice daily.

b) 12 hour between each administration.

c) mg amoxicilin/administration = 413.9 mg/administration.

d) should be stored in the refrigerator

Explanation:

mass child = 36.5 Lb * ( 453.592 g/Lb ) =  16556.11 g = 16.556 Kgdosage: 25mg/kg body b.i.d.

∴ b.i.d. : refer to twice a day; so 12 hours will pass between each administration of the medication.

⇒ mg amoxicilin/administration = 25 mg/ kg * 16.556 Kg = 413.9 mg amoxicilin.

Amoxicillin should be stored in the refrigerator, since in this section the temperature is kept within the storage range

How many neutrons in an atom of 30p? O a. 17 O b. 30.974 O c. 15 O d. 14 O e. 16

Answers

Answer:

The correct option is: c. 15

Explanation:

Phosphorous is a chemical element which belongs to the group 15 of the periodic table and has atomic number 15. It is a highly reactive non-metal of the p-group.  

Since, atomic number of an atom is the number of electrons and number of protons for neutral atoms.

So, the number of protons = number of electrons = 15

The atomic mass is obtained by adding the number of neutrons and the protons.

So, number of neutrons + number of protons = 30

So, number of neutrons + 15 = 30

Therefore, the number of neutrons in ³⁰P = 15

A technician tares a 100.0 mL volumetric flask at 150.00 g. After adding sodium chloride to the flask it then weighs 158.84 g. Assuming an error of 0.2 mL in the volumetric volume and 0.005 g in the weight, calculate the molar concentration of sodium chloride and its associated standard deviation.

Answers

Final answer:

To find the molar concentration of NaCl, subtract the tare weight from the total weight to get the mass of NaCl, calculate moles of NaCl, and divide by the solution volume. To estimate the standard deviation, propagate the errors from the mass and volume measurements according to the rules of error propagation. Specific numerical values for the standard deviation cannot be provided without exact formulas.

Explanation:

The question pertains to calculating the molar concentration of sodium chloride (NaCl) in a solution, and its associated standard deviation, given certain experimental measurements and potential error margins. First, the mass of NaCl added to the solution is found by subtracting the tare weight of the volumetric flask from the total weight after NaCl was added, yielding 8.84 g of NaCl. The molecular weight of NaCl is 58.44 g/mol, which allows determination of the moles of NaCl present.

To find the molar concentration, divide the moles of NaCl by the volume of the solution in liters (assuming the 100.0 mL flask volume as ideal, the error in volume would be considered in calculating the standard deviation, not the concentration itself). Then, to address the error margins, propagate the errors from the mass and volume measurements to estimate the standard deviation of the calculated concentration.

Note: Without specific formulas for error propagation and the exact calculation method for standard deviation provided in the question, a detailed numerical solution including the standard deviation calculation cannot be accurately presented. However, this process typically involves the square root of the sum of squared fractional uncertainties of the measurements involved.

Explain what D and L represent in stereoisomers

Answers

Final answer:

The D and L stereochemical descriptors are used to represent the configuration of stereoisomers in monosaccharides. The D- or L- designation is based on the position of the -OH group on the penultimate carbon in the Fisher projection. The D-configuration is commonly found in nature and only dextrorotary amino acids are used by cells to build proteins.

Explanation:

The D and L stereochemical descriptors are used to represent the configuration of stereoisomers, specifically in the context of monosaccharides or sugars. The designation of D or L is based on the position of the -OH group on the second-last carbon (penultimate C) in the Fisher projection. If the -OH group is on the right side, it is assigned D-configuration, and if it is on the left side, it is assigned L-configuration.

These descriptors do not indicate the rotation of plane polarized light, but purely define the configuration. Enantiomers that are D- and L- pairs have the same common name, with the D- or L- designation indicating their configuration. It's important to note that the D- and L- designation does not always correlate with the dextro/levo rotatory nature of the enantiomers in a polarimeter.

For example, D-glucose and L-glucose are enantiomers with the penultimate C defining D- or L-configuration. The D-configuration is commonly found in nature, and only dextrorotary d amino acids (L amino acids) are used by cells to build polypeptides and proteins.

The highest temperature recorded in the same city during the past year was 304.89 K. What was the temperature in degrees Celsius?

Answers

Final answer:

The temperature of 304.89 Kelvin is equivalent to 31.74°C when converted using the formula C = K - 273.15.

Explanation:

The student has asked about converting the highest temperature recorded in a certain city from Kelvin to degrees Celsius. The formula for converting Kelvin to Celsius is: C = K - 273.15, where C is the temperature in Celsius and K is the temperature in Kelvin. Applying this formula to the given temperature (304.89 K), we get:

C = 304.89 K - 273.15

C = 31.74°C

Hence, the temperature of 304.89 K is equivalent to 31.74°C.

You want to determine the density of a compound but have only tiny crystal, and it would be difficult to measure mass and volume accurately. There is another way to determine density, however called the flotation method. If you placed the crystal in a liquid whose density is precisely that of the substance, it would be suspended in the liquid, neither sinking to the bottom of the beaker nor floating to the surface. However, for such an experiment, you would need to have a liquid with the precise density of the crystal. You can accomplish this by mixing two liquids of different densities to create a liquid having the desired density a a Consider the following: you mix 7.30 mL of CHCI3 (d = 1.492 g/mL) and 8.90 mL of CHBT3 (d = 2.890 g/mL) giving 16.2 mL of solution. What is the density of this mixture? Density = g/mL

Answers

Answer:

2.26g/mL

Explanation:

Given parameters:

Volume of CHCl₃ = 7.3mL

Density of CHCl₃ = 1.492g/mL

Volume of CHBT₃ = 8.9mL

Density of CHBT₃ = 2.89g/mL

Unkown:

Density of the mixture = ?

Solution

Density can be defined as the mass per unit volume of substance. It is usually expressed using the equation below:

    Density = [tex]\frac{mass}{volume}[/tex]

For the given liquids, the volumes are known but we do not know their masses:

To derive the mass, we simply make mass the subject of the formula in the density equation.

   Mass of the liquid = Density of liquid x volume

mass of CHCl₃ = 7.3 x 1.492 = 10.89g

mass of CHBT₃ = 8.9 x 2.89 = 25.72g

Now to calculate the density of the mixture:

  Density = [tex]\frac{mass of CHCl_{3} + mass of CHBT_{3}  }{Volume of CHCl_{3} + volume of CHBT_{3} }[/tex]

Density of mixture = [tex]\frac{10.89 + 25.72}{7.3 + 8.9}[/tex] = 2.26g/mL

Final answer:

To calculate the density of a mixture made from CHCl3 and CHBr3, combine the masses of each component based on their volumes and densities, and then divide by the total volume to get a density of 2.26 g/mL.

Explanation:

To determine the density of a mixture made by combining 7.30 mL of CHCl3 (d = 1.492 g/mL) and 8.90 mL of CHBr3 (d = 2.890 g/mL) yielding a total volume of 16.2 mL, we follow these steps:

Calculate the mass of CHCl3 used: 7.30 mL × 1.492 g/mL = 10.8916 g.

Calculate the mass of CHBr3 used: 8.90 mL × 2.890 g/mL = 25.721 g.

Add the masses of CHCl3 and CHBr3 to find the total mass of the mixture: 10.8916 g + 25.721 g = 36.6126 g.

Use the formula for density = mass / volume to calculate the density of the mixture: 36.6126 g / 16.2 mL = 2.26 g/mL.

Therefore, the density of the mixture is 2.26 g/mL.

What is the focus on the biological organization levels for ecology?

Answers

Answer:

The biology of the ecosystem is always studied from the composition of organisms, population and  community.

Explanation:

In an ecosystem, various processes are sustained and these processes regulate the environment, as the biotic communities in the ecosystem make up the biosphere. Various life processes, their interaction, movement, and successional development ecosystem and the distribution of organisms are the focus of biological organization levels of ecology. First comes the organism ecology level as the researchers study their adaptations for the structure and physiographic behavior. Second is the population ecology level with a group of interbreeding organisms next level is of Community ecology and ecosystem ecology as they study the different species within the area using a three-dimensional model. After these the ecosystem ecology that is an extension of all the levels in the ecosystem. The natural environment can provide all the needed nutrients and life-supporting factors.

A chemist needs to create a series of standard Cu2 (aq) solutions for an absorbance experiment. For the first standard, he uses a pipet to transfer 25.00 mL of a 2.96 M Cu2 (aq) stock solution to a 250.0 mL volumetric flask and adds enough water to dilute to the mark. He then uses a second pipet to transfer 20.00 mL of the second solution to a 100.0 mL volumetric flask and adds enough water to dilute to the mark. Calculate the concentration of the Cu2 (aq) solution in the 100.0 mL volumetric flask.

Answers

Answer:

The concentration of the Cu2 in the 100.0 ml volumetric flask is 0.0592 M

Explanation:

In the first dilution, Cu2 was diluted ten times (25 / 250 = 1/10). Then, this dilution was diluted again, but now five times (20 / 100 = 1/5). In total, the solution was diluted 50 times (1/10 * 1/5 = 1/50). The final concentration will be 2.96 M / 50 = 0.0592 M

The quantity of the solute or the substance present in the solution is called the concentration. The concentration of the [tex]\rm Cu_{2}[/tex] in the volumetric flask is 0.0592 M.

What is concentration?

Concentration is the molarity of the substance and is given as the ratio of the moles of the solute with the volume in litres.

Given,

The volume of [tex]\rm Cu_{2}[/tex] by first pipet = 25 mlVolume of stock solution = 250 ml

The [tex]\rm Cu_{2}[/tex] is diluted ten times at first,

[tex]\dfrac {25}{250}= \dfrac{1}{10}[/tex]

Given,

Volume of [tex]\rm Cu_{2}[/tex] by second pipet = 20 mlVolume of stock solution = 100 ml

The [tex]\rm Cu_{2}[/tex] is diluted five times the second time,

[tex]\dfrac {20}{100}= \dfrac{1}{5}[/tex]

Total dilution of the solution was done 50 times as,

[tex]\dfrac{1}{10}\times \dfrac{1}{5} = \dfrac{1}{50}[/tex]

The final concentration of the solution will be,

[tex]\dfrac{2.96 \;\rm M}{50} = 0.0592 \;\rm M[/tex]

Therefore, the final concentration is 0.0592 M.

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Which property of metals is explained by the "sea of electrons" within metallic bonding?

A. Malleability

B. All of the Above

C. Electrical conductivity

D. Thermal conductivity

Answers

Answer:

B. All of the Above

Explanation:

Sea of electrons -

This model of the metallic bonding helps in explaining the properties like ,  malleability , ductility ,high electrical conductivity ,  luster ,high thermal conductivity of the metals in solid state .

The metallic bonding is between metal atoms and the ionic bond links a metal and a non - metal together ,

In case of metallic bonding , bulk of metal atoms are joined .

hence from the question , all the given properties are correct .

The pH of blood depends on the [HCO3-/H2CO3] balance. ([H2CO3] is equal to the amount of dissolved CO2). Calculate the bicarbonate (HCO3-) : carbon dioxide ratio for a normal blood pH of 7.40. (the pKa1 of carbonic acid is 6.10 at 37oC, body temperature). (A) 20 : 1 (B) 1.3 :1 (C) 2 : 1 (D) 1 : 20 (E) 1 : 0.01

Answers

Answer: Option (A) is the correct answer.

Explanation:

The given data is as follows.

  pH = 7.40,         [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex]

[tex]pK_{a}[/tex] = 6.10

We have to find [tex]\frac{[HCO_^{-}{3}]}{[CO_{2}]}[/tex] = ?

According to Henderson-Hasselbalch equation,

                 pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]

Hence, putting the given values into the above equation as follows.

                 pH = [tex]pK_{a} + log_{10} \frac{[Salt]}{[Acid]}[/tex]

or,              pH = [tex]pK_{a} + log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]

                 7.40 = 6.10 + [tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex]

             [tex]log_{10} \frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = 1.30

          [tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] = antilog (1.30)    

                                         = 20

Since, it is given that [tex][H_{2}CO_{3}][/tex] = [tex][CO_{2}][/tex].

Therefore, [tex]\frac{[HCO^{-}_{3}]}{[H_{2}CO_{3}]}[/tex] or [tex]\frac{[HCO^{-}_{3}]}{[CO_{2}]}[/tex] = [tex]\frac{20}{1}[/tex]

Thus, we can conclude that the bicarbonate (HCO3-) : carbon dioxide ratio for a normal blood pH of 7.40 is 20:1.

I have a 5 M stock solution of KCI (For a protocol 100 mM KCl is considered "1X" concentration). If I want to make 10 ml of a 4X concentration KCl solution, how much 5 M stock and how much water do I need to add together?

Answers

Answer:

You need 0.8 ml of 5M stock solution and you have to add 9.2 ml of water.

Explanation:

Protocol solution (1X): 100 mM=0.1M

4X: 0.4M

The concentration of a solution is inversely proportional to the volume of a solution, so:

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

where:

M1= 5M stock solution

V1= amount of solution we need to collect

M2=4X solution

V2= 10 ml (volumen of 4X solution)

Therefore:

5M×V1=0.4M×10ml

V1={0.4M}{5M}10ml=0.8ml

[tex]5M*V_{1}=0.4M*10ml\\ V_{1}=\frac{0.4M}{5M}10ml=0.8ml[/tex]

To make a 10 ml solution we have to add 9.2 ml of water because V2 es 10 ml.

According to Wookieepedia, Kylo Ren stands 1.89 m tall and his mass is 89 kg. Calculate his earth-weight in lbm. (2 pts) a. b. Assuming he has a 50% waist to height ratio, calculate his diameter in inches.

Answers

Answer:

a) 6312.12 lbm

b)  D = 11.843 inch

Explanation:

GIVEN DATA:

Height of person = 1.89 m = 74.41 inch

mass of person m = 89 kg = 196.211 lbm

a) person earth weight =mg

where g is acceleration due to gravity = 32.17 ft/sec^2

                                     [tex] = 196.211\times 32.17 = 6312.12 lbm[/tex]

b) waist to height ratio = 0.5

[tex]waist\ to\  height\  ratio = \frac{circumference\ of\ the\ waist}{height\ of\ the\ person}[/tex]

circumference of waist =[tex] height\  of\  person \times waist\ to\ height ratio[/tex]

                                       [tex]= 74.41 inch \times 0.5[/tex]

                                        =37.21 inch

we know that circumference is given as [tex]= 2\pi r\ or\  \pi D= 37.21[/tex]

                                                               D = 11.843 inch

1457 (2.2x10^-8) follow up question: Identify the following as acidic, basic or neutral: pH = 4.56 pH = 10.4 [OH-] = 2.4x10-8 Reply Quote Select: All None Message Actions Expand All

Answers

Answer:

(a) pH = 4.56 is acidic

(b) pH = 10.4 is basic

(c) [OH-] = 2.4x10-8, thus [tex]pH = -Log (4.16x10^{-7}) = 6.3[/tex] is acidic

Explanation:

The pH is the mesure of the acidity or bacisity of a solution. It indicates the concentration of protons in a solution and it is define as:

[tex]pH = -Log ([H^{+}])[/tex]

The scale of pH goes from 1 to 14, being 1 the most acid condition and 14 de most basic condition. Also a pH = 7 is a neutral condition.

Therefore, if the pH is:  1 ≤ pH < 7 the solution will be acidic and if the pH is 7 < pH ≤ 14 the solution will be basic.

To answer (c) it is also necessary to consider the water autoionization to calculate the protons concentration as shown bellow

[tex]K_{w} =[H^{+} ][OH^{-}]=10^{-14}[/tex]

[tex][H^{+} ]=\frac{10^{-14}}{[OH^{-}]}[/tex]

For (c) [OH-] = 2.4x10-8

[tex][H^{+} ]=\frac{10^{-14}}{2.4x10^{-8}}=4.16x10^{-7}[/tex]

And using the definition of pH

[tex]pH = -Log (4.16x10^{-7}) = 6.3[/tex]

In a conjugate acid-base pair, the acid typically has one more proton than the base b. one fewer proton than the base. C. two fewer protons than the base. d. the same number of protons as the base. 17 a. TO17 O 0001

Answers

Answer:

Statement (a) is true

Explanation:

Conjugate base of an acid is formed from deprotonation of corresponding acid.For an example, consider an acid e.g. [tex]CH_{3}COOH[/tex] (acetic acid)Acid-base equilibrium for acetic acid in aqueous solution is represented as: [tex]CH_{3}COOH+H_{2}O\rightarrow CH_{3}COO^{-}+H_{3}O^{+}[/tex]Here [tex]CH_{3}COO^{-}[/tex] (acetate ion) is the conjugate base of acetic acid.So, clearly, acetic acid has one more proton as compared to acetate ion

Hence statement (a) is true

Final answer:

In a conjugate acid-base pair, the acid typically has one fewer proton than the base.

Explanation:

In a conjugate acid-base pair, the acid typically has one fewer proton than the base. When a proton (H+) is removed from an acid, it forms its conjugate base which has one less proton. For example, water (H2O) is an acid in the conjugate acid-base pair H2O/OH-, where water (H2O) has one more hydrogen ion (H+) than the hydroxide ion (OH-), which is the conjugate base.

The acid-base pairs can be represented as:

H2O/OH-H3PO4/H2PO4-H2SO4/HSO4-NH4+/NH3

Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents. A sample of impure tin of mass 0.528 g is dissolved in strong acid to give a solution of Sn2+. The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 4.03×10−2 L of the NO3− solution.

Answers

Final answer:

To find the mass percent of chloride in the original dry sample, you can use the formula: Mass percent of chloride = (mass of chloride / mass of original sample) x 100%. Use the volume of AgNO3 solution used in the titration, the solution's molarity, and the molar mass of chloride to calculate the mass of chloride.

Explanation:

The percent by mass of chloride in the original dry sample can be calculated using the following formula:

Mass percent of chloride = (mass of chloride / mass of original sample) x 100%

In this case, the mass of chloride can be determined by multiplying the volume of AgNO3 solution used in the titration (28 mL) by the molarity of the solution (0.1 M) and the molar mass of chloride (35.453 g/mol).

Then, using the mass of chloride and the mass of the original sample (0.200 g), the percent by mass of chloride in the original dry sample can be calculated.

Name two "Storage Polysaccharides" and two "Structural Polysaccharides"

Answers

Answer:

Examples of storage polysaccharides - starch and glycogen and structural polysaccharides - cellulose and chitin

Explanation:

Polysaccharides are the complex carbohydrate polymers, composed of monosaccharide units that are joined together by glycosidic bond.

In other words, polysaccharides are the carbohydrate molecules that give monosaccharides or oligosaccharides on hydrolysis.

The examples of storage polysaccharides are starch and glycogen. The examples of structural polysaccharides are cellulose and chitin.

To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen, an reaction that proceeds according to the following unbalanced equation.

Fex(CO)y + O2 --> Fe2O3 + CO2

If you burn 1.959 g. of Fex(CO)y and obtain 0.799 g. of Fe2O3 and 2.200 g. of CO2, what is the empirical formula of Fex(CO)y?

Answers

Answer:

The empirical formula is: Fe(CO)₅

Explanation:

According the global reaction:

Feₓ(CO)y + O₂ → Fe₂O₃ + CO₂

You should calculate Fe₂O₃ and CO₂ moles, thus:

0,799 Fe₂O₃ grams  × [tex]\frac{1 mole}{159.69 Fe2O3 g}[/tex] = 5,00×10⁻³ Fe₂O₃ moles

2,200 CO₂ grams  × [tex]\frac{1 mole}{44,01 CO2 g}[/tex] = 5,00×10⁻²CO₂ moles

The ratio between Fe₂O₃ moles and CO₂ moles is 1:10. Thus ratio between x and y must be 1:5 because Fe₂O₃ has 2 irons but CO₂ has just one carbon.

Assuming the formula is Fe₁(CO)₅ the molecular weight is 195,9 g/mol. Thus:

1,959 Fe(CO)₅ grams  × [tex]\frac{1 mole}{195,9 Fe(CO)5 g}[/tex] = 1,00×10⁻² Fe(CO)₅ moles

Thus, assuming 1,00×10⁻² moles as basis for calculation, the global reaction is:

1 Fe(CO)₅ + ¹³/₂O₂ → ¹/₂ Fe₂O₃ + 5 CO₂

With this balanced equation the moles produced have sense, thus, the empirical formula is: Fe(CO)₅

I hope it helps!

Final answer:

The empirical formula of the compound Fex(CO)y, formed from 1.959 g of the substance producing 0.799 g of Fe²O³ and 2.200 g of CO², is Fe(CO)⁵.

Explanation:

To determine the empirical formula of the compound Fex(CO)y, we must first find the moles of iron (Fe) and carbon monoxide (CO) in the compound. Given that 0.799 g of Fe²O³ and 2.200 g of CO² were produced, we can calculate the number of moles of Fe and C using their molar masses (Fe: 55.85 g/mol, C: 12.01 g/mol, O: 16.00 g/mol).

From Fe²O³, the mass of Fe is 0.799 g x (2 mol Fe / 159.69 g Fe²O³) = 0.0100 mol Fe.
From CO², the mass of C is 2.200 g x (1 mol C / 44.01 g CO²) = 0.0500 mol C.

To find the mole ratio, we use the smallest number of moles as a divisor. Here, it is 0.0100 mol Fe. The ratio of Fe to C in the compound is 0.0100 mol Fe / 0.0100 mol = 1 Fe to 0.0500 mol C / 0.0100 mol = 5 CO.

Therefore, the empirical formula of the compound is Fe(CO)5.

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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
reactor. What is the excess reactant?

Answers

Answer:

H₂ gas

Explanation:

The reaction between nitrogen gas and hydrogen gas forms ammonia (the Haber-Bosch process):

N₂ + 3H₂ ⇒ 2NH₃

The excess reactant can be found by comparing the moles of nitrogen and hydrogen. The molar mass of N₂ is 28.00 g/mol and the molar mass of H₂ is 2.02 g/mol.

(100 kg N₂)(1000g/kg)(mol/28.00g) = 3570 mol

(100 kg H₂)(1000g/kg)(mol/2.02g) = 49500 mol

The molar ratio between the reactant N₂ and H₂ is 1N₂:3H₂. The moles of nitrogen required to react with H₂ is:

(49500 mol H₂)(1N₂ / 3H₂) = 16500 mol

The amount of nitrogen required is more than what is available, so nitrogen is the limiting reagent and hydrogen is the excess reagent.

which one have less friction loss ?
1- venturi
2- orrifice
3- nozzle
4- all the same

Answers

Answer:

1- venturi

Explanation:

Venturi

In venturi tube the  friction is less as compare to the nozzle and the orifice .

Nozzle have a medium friction loss where as orifice have a very high friction loss of , approximately 75 - 80% loss .

Venturi tube have a  convergent and a divergent section which helps to reduce the friction loss as compared to the orifice.

The value of the Discharge coefficient in venturimeter is Cd = 0.98 but orifice have discharge coefficient Cd = 0.68 .

A piece of an unknown metal has a volume of 16.6 mL and a mass of 190.1 grams. The density of the metal is g/mL A piece of the same metal with a mass of 94.6 grams would have a volume of ml. Submit Answer

Answers

Answer: The density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL

Explanation:

To calculate the density of unknown metal, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]       ......(1)

Volume of unknown metal = 16.6 mL

Mass of unknown metal = 190.1 g

Putting values in equation 1, we get:

[tex]\text{Density of unknown metal}=\frac{190.1g}{16.6mL}\\\\\text{Density of unknown metal}=11.45g/mL[/tex]

The density of the metal remains the same.

Now, calculating the volume of unknown metal, using equation 1, we get:

Density of unknown metal = 11.45 /mL

Mass of unknown metal = 94.6 g

Putting values in above equation, we get:

[tex]11.45g/mL=\frac{94.6g}{\text{Volume of unknown metal}}\\\\\text{Volume of unknown metal}=8.26mL[/tex]

Hence, the density of the metal is 11.45 g/mL and the volume occupied by 94.6 grams is 8.26 mL

Calculate the mass of 1.0 L of helium (He), 1.0 L of chlorine gas (Cl2), and 1.0 L of air (79% N2, 21% O2 by volume) at 25°C and 1 atm total pressure. Explain why a balloon filled with helium rises and why leaks of chlorine gas can be dangerous.

Answers

To calculate the mass we use the following formulas:

PV=nRT     (1)

and

n = m / M    (2)

where:

P - pressure (atm)

V - volume (L)

n - moles

R - gas constant = 0.082 (L × atm) / (mol × K)

T - temperature (°K) (25°C + 273 = 298°K)

m - mass (g)

M - molecular mass (g/mole)

Now we rewrite equation (1):

n = PV / RT

And replace n with m / M from equation (2):

m / M = PV / RT

m = (P × V × M) / (R ×T)

1 L of He will have a mass of:

m = (1 × 1 × 4) / (0.082 × 298) = 0.1637 g

1 L of Cl₂ will have a mass of:

m = (1 × 1 × 71) / (0.082 × 298) = 2.9055 g

1.0 L of air will contain 0.79 L of N₂ and 0.21 L of O₂

0.79 L of N₂ will have a mass of:

m = (1 × 0.79 × 28) / (0.082 × 298) = 0.9052 g

0.21 L of O₂ will have a mass of:

m = (1 × 0.21 × 32) / (0.082 × 298) = 0.2750 g

mass of air = mass of N₂ + mass of O₂

mass of air = 0.9052 + 0.2750 = 1.1802 g

A balloon filed with helium will rise because as you see 1 L of helium is lighter than 1 L of air.

Chlorine gas is dangerous because chlorine is very toxic for human life and more of that is heavier than the air so will diffuse very hard from the area where the leak appeared.

write and the integrated rate laws hor zeroth-first- second-order rate laws.

Answers

Explanation:

The integrated rate law for the zeroth order reaction is:

[tex][A]=-kt+[A]_0[/tex]

The integrated rate law for the first order reaction is:

[tex][A]=[A]_0e^{-kt}[/tex]

The integrated rate law for the second order reaction is:

[tex]\frac{1}{[A]}=kt+\frac{1}{[A]_0}[/tex]

Where,

[tex][A][/tex] is the active concentration of A at time t

[tex][A]_0[/tex] is the active initial concentration of A

t is the time

k is the rate constant

Answer:

- 0th: [tex]C_A=C_{A0}-kt[/tex]

- 1st: [tex]C_A=C_{A0}exp(-kt)[/tex]

- 2nd: [tex]\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]

Explanation:

Hello,

For the ideal reaction A→B:

- Zeroth order rate law: in this case, we assume that the concentration of the reactants is not included in the rate law, therefore the integrated rate law is:

[tex]\frac{dC_A}{dt}=-k\\ \int\limits^{C_A}_{C_{A0}} {} \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\C_A-C_{A0}=-kt\\C_A=C_{A0}-kt[/tex]

- First order rate law: in this case, we assume that the concentration of the reactant is included lineally in the rate law, therefore the integrated rate law is:

[tex]\frac{dC_A}{dt}=-kC_A\\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A} } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\ln(\frac{C_{A}}{C_{A0}} )=-kt\\C_A=C_{A0}exp(-kt)[/tex]

- Second order rate law: in this case, we assume that the concentration of the reactant is squared in the rate law, therefore the integrated rate law is

[tex]\frac{dC_A}{dt}=-kC_A^{2} \\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A^{2} } } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\-\frac{1}{C_A}+\frac{1}{C_{A0}}=-kt\\\frac{1}{C_A}=kt+\frac{1}{C_{A0}}[/tex]

Best regards.

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