equal volumes and concentrations of hydrochloric acid and ethanoic acid are titrated with sodium hydroxide solutions of the same concentration. which statement is correct?

Answers

Answer 1

Reaching the equivalent point requires the same amount of sodium hydroxide.

What determines a titration's equivalency point?

When the volume of analyte is equal to the concentration of titrant administered, this occurs. With 12.3 mL of 0.2 molL-1 HCl, 30 mL of NaOH are neutralized.

During a titration, what transpires at the equivalence point?

When just enough titrant is injected to completely neutralize the analyte solution during titration, this is known as the equivalence point. Molecules of base equal molecular weights of acid at the equivalence point in an acid-base titration, and the solution is made up entirely of salt and water.

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Related Questions

7. Calculate the kcalorie for each food burned in a calorimeter:
a. 1 cup of popcorn that heats 1250 grams of water from 25oC to 50.8oC
b. A sample of butter that heats 357 grams of water from 25oC to 38.8oC
c. A hamburger contains 37 g of carbohydrate, 19 g of fat, and 24 g of protein.

Answers

33.7Kcal is the heat for 1 cup of popcorn that heats 1250 grams of water from 25°C to 50.8°C.

What is calorimeter?

Calorimeters come in a wide range of designs. A common kind, the bomb calorimeter, essentially comprises of a container within which the reaction occurs and a liquid, such water, that takes in the heat of the action and raises temperature as a result.

The total quantity of heat created may be determined using the measurement of this temperature rise and information about the weight and thermal properties of the container and liquid.

q= m×c×ΔT

  =1250 ×4.18×( 50.80-25)  

  = 1250 ×4.18×  25.8

  =134805J/4cal

  =33701.2cal/1000

  = 33.7Kcal

Therefore, 33.7Kcal is the heat for 1 cup of popcorn that heats 1250 grams of water from 25°C to 50.8°C.

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When an action potential occurs, sodium gates open, resulting in the inside of the cell becoming more negative relative to the outside.

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When a voltage change crosses the threshold, the wattage na+ in the axonal barrier will open. The entry of sodium is responsible for the growing component of the nerve impulse, but the influx of ions also depolarizes nearby axonal regions. When the depolarization approaches the threshold, the nerve impulse descends the axon.

Is depolarization advantageous or detrimental?

Depolarization occurs when an electrical shift occurs in a nerve cell. The majority of cells have a negative charge in comparison to their environment. Through the brief process of depolarization, the cell's internal charge changes from negative to positive.

Does depolarization occur when sodium channels are opened?

Depolarization, a rapid increase in potential that occurs in neurons, is an all-or-nothing process that is started by the opening of sodium within the plasma membrane's ion channels. Repolarization, the ensuing restoration to the resting potential, is facilitated by the activation of na / k channels.

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When 3,3-dimethylbut-1-ene is treated with HBr alone, the major product is 2-bromo-2,3-dimethylbutane. When the same alkene is treated with HBr and peroxide, the sole product is 1-bromo-3,3-dimethylbutane. Explain these results by referring to the mechanisms.

Answers

It yields 2-bromo-2,3-dimethylbutane as the final product. The proton is added to the radical process after the bromine anion. 1-bromo-3,3-dimethylbutane.

Which of the following types of chemicals results from an alkene's hydration?

1 Response. Alcohol is created when alkene and water interact. Electrophilic addition is the reaction at hand. The double bond of the carbon in an alkene with a high electron density is attacked by water.

What do examples of hydration of alkenes mean?

Alkene Hydration Example: ethanol is produced by the hydration of ethylene. An alkene with the condensed structural formula H2C=CH2, ethylene is also known as. The double bond (C=C) in the ethene molecule is the active site.

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calculate the ph of a 0.010 m solution of ch3co2h. the ka of ch3co2h is 1.8 x 10-5. report your answer to two decimal places without units.

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The pH of a 0.010 M solution of CH3COOH (acetic acid) can be calculated using the concentration of the hydrogen ion (H+) and the Ka value of the acid. The answer to two decimal places, the pH of the solution is 4.75.

pH Value is a measure of the acidity or alkalinity of a solution. It is expressed on a scale of 0 to 14, with 7 being neutral. A solution with a pH less than 7 is considered acidic and a solution with a pH greater than 7 is considered alkaline or basic. The pH value is an important factor in many biological, chemical, and environmental processes.

In industrial processes, pH values are often monitored and controlled to ensure that reactions occur under optimal conditions. For example, the pH of water used in boilers must be maintained within a specific range to prevent corrosion and scale buildup.

The pH value is a simple but powerful tool for measuring and understanding the acidity or alkalinity of a solution. It provides critical information for a wide range of applications, from health and agriculture to industrial processes and environmental monitoring.

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3. By how many degrees Celcius would the temperature of 3 kg of a
substance change if the substance absorbed 4.328 kJ of energy?
Assume that the specific heat of the substance equals 0.630 J/g*C.

Answers

Answer:

Below

Explanation:

3 kg = 3000 g     and     4.328 kj = 4238 j

 (units have to match units of given specific heat   J/(g°C)  )

4238 j / (3000 g  * x °C) = .630  J / (g °C)

(4238 / .630 ) / 3000 =  x °C

x = 2.24 °C

A solution contains both iron(II) and iron(III) ions. A 50.00-mL sample of the solution is titrated with 35.0 mL of 0.0280 M KMnO4, which oxidizes Fe2+ to Fe3+. The permanganate ion is reduced to manganese(II) ion. The equation for this reaction isAnother 50.00-mL sample of the solution is treated with zinc, which reduces all the Fe31 to Fe21. The equation for this reaction isThe resulting solution is again titrated with 0.0280 M KMnO4; this time 48.0 mL is required. What are the concentrations of Fe2+ and Fe3+ in the solution?

Answers

The concentrations of Fe2+ and Fe3+ in the solution are [Fe2+] = 0.0196 M and [Fe3+] = 0.0269 M, respectively.

Let's assume the initial concentrations of Fe2+ and Fe3+ ions in the solution to be [Fe2+] and [Fe3+], respectively.

The first titration (with 35.0 mL of 0.0280 M KMnO4) oxidizes Fe2+ ions to Fe3+ ions.

n(Fe2+) = (35.0 mL) x (0.0280 M) = 0.980 mol

The second titration (with 50.00 mL of the solution after zinc reduction) gives the number of moles of Fe3+ ions remaining in the solution.

n(Fe3+) = (0.0280 M) x (48.0 mL) = 1.344 mol

Using the mole ratio, we can now find the initial concentrations of Fe2+ and Fe3+ ions in the solution:

[Fe2+] = n(Fe2+) / (50.00 mL) = 0.980 / (50.00 mL) = 0.0196 M

[Fe3+] = n(Fe3+) / (50.00 mL) = 1.344 / (50.00 mL) = 0.0269 M

Titration is a laboratory technique used in chemistry to determine the concentration of an unknown substance. It is a type of analytical method that involves adding a known volume of a standard solution (titrant) to a solution of the unknown substance (analyte) until a reaction is complete and the end-point is reached. The end-point is the point at which a change in color or some other observable change indicates that the reaction is complete.

The end-point is used to determine the exact amount of titrant required to react with the analyte and to calculate the concentration of the unknown substance. The results of a titration are used in various applications such as determining the purity of substances, monitoring the production of chemical products, and determining the effectiveness of certain chemical reactions. Titration is a simple, cost-effective, and widely used technique in the field of analytical chemistry.

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which of the following produce an alkaline / acid h2s (k/a h2s) reaction on tsi and an k/k h2s on lia? a. edwardsiella b. proteus vulgaris c. salmonella d. citrobacter e. providencia f. morganella morganii

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Based on the given information, the microorganisms that produce an alkaline/acid H2S reaction on TSI (Triple Sugar Iron) and a K/K H2S reaction on LIA (Lysine Iron Agar) are given here.

a. Edwardsiella - Alkaline/Acid H2S on TSI and K/K H2S on LIA

c. Salmonella - Alkaline/Acid H2S on TSI and K/K H2S on LIA

d. Citrobacter - Alkaline/Acid H2S on TSI and K/K H2S on LIA

e. Providencia - Alkaline/Acid H2S on TSI and K/K H2S on LIA

f. Morganella morganii - Alkaline/Acid H2S on TSI and K/K H2S on LIA

These microorganisms produce hydrogen sulfide (H2S), which reacts with the iron in the TSI and LIA media to produce a characteristic change in the media's pH. The change in pH can be either alkaline or acidic, depending on the microorganism. A K/K H2S reaction on LIA indicates the presence of H2S in the bacterial colony, while a K/A H2S reaction on TSI indicates both the presence of H2S and an alkaline reaction.

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move to the safety shower if you spill hazardous chemicals on your skin or clothing . stand under the shower and pull the lever. remove clothing and continue flushing the area for

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Take he safety shower Remove your shirt and any other items of clothes from the area where the chemical was present.

When a dangerous chemical comes into touch with the skin, it has the potential to harm the body. Many dangerous compounds have the potential to enter the body through the skin.

As a precaution, you must remove your lab coat, shirt, and any other items that came into touch with the dangerous chemical once it has already soaked into your lab coat. Doing so will prevent the chemical from entering your body via the skin. Some of these substances can harm the skin directly, thus it is important to avoid them.

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select the correct conversion factor for each step in the following roadmap for a substance with a density of 1 g/ml. mass mass volume

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A conversion factor is a number used to change one set of units to another, by multiplying or dividing. To convert kilograms in grams and litres in millilitres, multiply with 1000 it is also known as unit conversion.

So, when converting from kilogrammes to grammes, we must first multiply by a conversion factor with kilogrammes in the denominator. As a result, kilogrammes will cancel and Graham's will be in the denominator. So the fact that there are 1000 grammes in one kilogramme is the equality that makes this true. So 1000 g and one kilogramme would be our first conversion, and kilogrammes would then cancel. To convert to millimetres, use the density conversion with one gramme in the denominator and one millimetre in the numerator because grammes will cancel.

This will give us a millilitre answer. So the first conversion we'd like to use is 1000 g over one kilogramme. So that goes in the first box, and the second box converts mass and grammes to volume. Do you prefer the one mil leader over the one g . These are the conversion factors you would need to use to complete these two steps from kg to ml.

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Which of the following metals would be oxidized by water and oxygen in an acidic environment but not in basic environment? [answer A, B,C,D] A- Cr B- Hg C- Zn D- Cu'

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The metal that would be oxidized by water and oxygen in an acidic environment but not in basic environment is Zinc (Zn). The answer is C.

Zinc reacts with water and oxygen to form zinc hydroxide and zinc oxide in a basic environment. However, in an acidic environment, zinc ions will be reduced to zinc metal by the hydrogen ions (protons) in the solution, rather than being oxidized. This is because the hydrogen ions are strong enough to reduce the zinc ions, while in a basic environment, the hydroxide ions are not strong enough to reduce the zinc ions and instead they oxidize the zinc to form zinc hydroxide.

On the other side, Chromium (Cr), Mercury (Hg), and Copper (Cu) can be oxidized by water and oxygen in both acidic and basic environments.  

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Consider the reaction:

2A (g) + 3 B (g) → 2 C (g) ΔHrxn = +254.3 kJ

What will be the enthalpy change (in kJ) if 0.812 mol B reacts in excess A?

Answers

Answer:

The enthalpy change (in kJ) if 0.812 mol B reacts with excess A is +202.3 kJ. This can be calculated by first determining the number of moles of A that are needed to react with 0.812 mol B. Since the mole ratio of A to B is 2 : 3, the number of moles of A needed is 0.812 mol B x (2/3) = 0.5413 mol A. Then, the enthalpy change can be calculated as 0.5413 mol A x (254.3 kJ/2 mol A) = 202.3 kJ.

A 500 mL bottle of soda is packaged atdouble atmospheric pressure. If the Henry’s Law constant for CO2 in water is 3.4 x 10-2 mol/(L atm), at 25 °C, how many grams of CO2 are dissolved in the soda before it is opened?
one, please help me? Thanks.

Answers

A 500 mL bottle of soda is packaged at double atmospheric pressure. If the Henry’s Law constant for CO2 in water is 3.4 x 10-2 mol/(L atm), at 25 °C,2.905 grams of CO2 are dissolved in the soda before it is opened.

What is the Henry's law ?

The weight of a gas dissolved by a liquid is proportional to the pressure of the gas on the liquid, according to Henry's law. The law, first proposed by the English physician and chemist William Henry in 1803, applies only to dilute solutions and low gas pressures.

Given:

Henry's law constant K​​​​​​H = 3.4 × 10^-2 mol/L atm

We need to find mass of carbon dioxide gas inside the soda bottle.

Step 1: Molar mass of CO2

= Sum of atomic masses of all elements present

= 12.01 g/mol C + 2 × 16 g/mol

= 44.01 g/mol

In Step 2 : Solubility of gas

According to Henry's law,

Solubility of gas = K​​​​H × P

Substitute the given values

S = 3.3 × 10 ^-2 mol /lit atm * 4 atm

S = 0.132 mol/lit

In Step 3 : liters of soda drink

Given that 500 ml of soda bottle.

Converting ml into liters.

As 1 lit = 1000 ml

= 500 ml × (1lit/1000 ml)

= 0.5 lit

Step 4: moles of carbon dioxide in soda drink

As solubility = number of moles of carbon dioxide / volume of soda drink in litre

So, number of moles of carbon dioxide = solubility × volume of soda drink in litre

= 0.132 mol/lit × 0.5 lit

= 0.066 mol carbon dioxide.

Step 5 : Mass of carbon dioxide in the soda water

As molar mass = mass / no. of moles

The mass = molar mass of carbon dioxide * no. of moles of carbon dioxide

mass = 44.01 g/mol × 0.066 mol

= 2.9045 g

Thus, 2.905 grams of CO2 are dissolved in the soda before it is opened.

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Matter is anything that takes up space and has mass. Solids, liquids and gases are the three most familiar states, or forms, of matter. _______________. Matter can even turn into new types of matter when exposed to a chemical reaction. (A) The mass of a bowling ball, for example, is much greater than that of a beach ball. (B) Both solids and liquids have set sizes, or amounts, but liquids do not have a set shape. (C) When matter's temperature changes enough, it can change from one state to another. (D) Metals can turn into rust in the rain, and wood can turn into ashes when burned.

Answers

Matter is anything that takes up space and has mass. Solids, liquids and gases are the three most familiar states, or forms, of matter. When matter's temperature changes enough, it can change from one state to another which is therefore denoted as option C.

What is Matter?

This is referred to as any substance that has mass and takes up space by having volume.

In a scenario where temperature is applied to an state of matter it changes from one form to another and an example is solid which when melted will turn to liquid thereby making it the correct choice.

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Kindly help me with the number 2 answer ​

Answers

The number of moles of nitrogen, N in 65 moles of Pb(NO₃)₄ is 260 moles

How do I determine the number of mole of N?

We'll begin by obtaining the number of mole of N in one mole of Pb(NO₃)₄. Details below:

From the formula of Pb(NO₃)₄, we can see that there are 4 moles of N in 1 mole of Pb(NO₃)₄

With the above information, we can determine the number of mole of N in 65 moles of Pb(NO₃)₄. This is illustrated below:

1 mole of Pb(NO₃)₄ contains 4 moles of N

Therefore,

65 mole of Pb(NO₃)₄ will contain = (65 moles × 4 moles) / 1 mole = 260 moles of N

Thus, we can conclude from the above calculation that the number of mole of N is 260 moles

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use electron configurations orbital notation and electron dot structures to represent the formation of an ionic compound from the metal strontium and nonmetal chlorine

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For metal strontium cation - E.C.- [Kr], electron dot structure -  Sr^2+, and for nonmetal chloride anion - E.C. - [Ne] 3s2 3p6, electron dot structure - Cl^-.

The formation of the ionic compound strontium chloride (SrCl2) from the metal strontium (Sr) and nonmetal chlorine (Cl) can be represented using electron configurations, orbital notation, and electron dot structures.

1. Electron Configuration:

Strontium (Sr): [Kr] 5s2

Chlorine (Cl): [Ne] 3s2 3p5

2. Orbital Notation:

Strontium (Sr): [Kr] 5s2

Chlorine (Cl): [Ne] 3s2 3p6

3. Electron Dot Structure:

Strontium (Sr): Sr

Chlorine (Cl): Cl

In the formation of an ionic compound, electrons are transferred from the metal (Sr) to the nonmetal (Cl) to form ions with a complete electron configuration. Strontium loses two electrons to form a cation with a 2+ charge, while chlorine gains one electron from each strontium atom to form an anion with a 1- charge.

The resulting electron configurations, orbital notations, and electron dot structures for the ions are as follows:

1. Electron Configuration:

Strontium cation (Sr2+): [Kr]

Chloride anion (Cl-): [Ne] 3s2 3p6

2. Orbital Notation:

Strontium cation (Sr2+): [Kr]

Chloride anion (Cl-): [Ne] 3s2 3p6

3. Electron Dot Structure:

Strontium cation (Sr2+): Sr^2+

Chloride anion (Cl-): Cl^-

These ions can then bond ionically to form the compound SrCl2. The compound is held together by the electrostatic attraction between the positively charged strontium cation and the negatively charged chloride anion.

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A gas occupies 26.00 mL at a temperature of 450C. What is the volume at 1500C?
a.What gas law(s) will you use to answer this question?
b.What is the equation(s)?

Answers

a.The ideal gas law will be used to answer this question.
b.The equation is PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin.

What is Ideal gas?

Ideal gas is a theoretical gas composed of molecules that have negligible volume and do not interact with one another. It is assumed that the molecules are in constant, random motion and no energy is lost to friction. This type of gas is used to describe the behavior of real gases in many situations and is used to calculate properties such as pressure, volume, temperature, and internal energy.

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the society of automotive engineers has established an accepted numerical scale to measure the viscosity of motor oil. for example, sae 40 motor oil has a higher viscosity than an sae 10 oil. rank the following hydrocarbons by their expected viscosity.

Answers

Because its molecules can slide around each other, a liquid has the ability to flow. The resistance to such flow is called viscosity. For organic juices, as the chain increases the viscosity increases due to the bonding that is present.

Therefore, the ranking should be as follows:

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3CH3CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH3

Viscosity is the degree of a fluid's resistance to go with the flow. The viscosity of a fluid is a measure of its resistance to deformation at a given rate. For liquids, it corresponds to the casual concept of "thickness": for instance, syrup has better viscosity than water.

Viscosity quantifies the internal frictional force among adjacent layers of fluid which are in relative movement. as an example, whilst a viscous liquid is compelled via a tube, it flows more quickly close to the tube's axis than close to its walls. Experiments display that some stress (along with a stress difference between the two ends of the tube) is needed to preserve the glide. that is due to the fact a pressure is needed to overcome the friction between the layers of the fluid that are in relative motion.

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given the reaction caco3 cao co2, if 50.0 g of caco3 react to produce 20.0 g of co2 what is the percent yield of co2

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If 50.0 g of CaCO3 react to produce 20.0 g of CO2, the theoretical yield is 2200 gm of CO2 and the percent yield of CO2 is 0.9%.

The theoretical yield of CO2 can be calculated as follows:

= Theoretical yield of CO2

= (50.0 g Caco3) x (1 mol CO2/1 mol Caco3) x (44.0 g CO2/1 mol CO2)

= 2200 g of CO2

The percent yield of CO2 is calculated as:

= Percent yield of CO2

= (Actual yield of CO2 / Theoretical yield of CO2) x 100%

= (20.0 g CO2 / 2200 g CO2) x 100%

= 0.9%

So, the percent yield of CO2 is 0.9%.

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make a prediction about the relative enthalpies of vaporization among the two nliquids you woll test in lab methanol and ethanol which will be higher? lower? explain

Answers

The relative enthalpies of vaporization between methanol (CH3OH) and ethanol (C2H5OH) can be predicted based on the bond strengths in their respective molecules.

Ethanol has stronger bonds compared to methanol, which means that more energy is required to break the bonds and convert the liquid into a gas. As a result, the enthalpy of vaporization for ethanol is likely to be higher compared to methanol.

In other words, it takes more heat energy to vaporize ethanol compared to methanol, and therefore, ethanol has a higher enthalpy of vaporization.

This prediction can be confirmed by experimentally measuring the enthalpies of vaporization for both liquids, which can be done using calorimetry or other methods. However, the prediction is based on the strength of the bonds in each molecule, which makes it a reasonable estimate of the relative enthalpies of vaporization between methanol and ethanol.

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Consider a hypothetical reaction in which A and B are reactants and C and D are products. If 23g of A completely reacts with 27g of B to produce 11g of C, how many grams of D will be produced?

Answers

According to law of conservation of mass, if  23 g of A reactant  completely reacts with 27 g of B to produce 11 g of C,39 grams of D product will be produced.

What is law of conservation of mass?

According to law of conservation of mass, it is evident that mass is neither created nor destroyed rather it is restored at the end of a chemical reaction .

Law of conservation of mass and energy are related as mass and energy are directly proportional which is indicated by the equation E=mc².Concept of conservation of mass is widely used in field of chemistry, fluid dynamics.

Law needs to be modified in accordance with laws of quantum mechanics under the principle of mass and energy equivalence.This law was proposed by Antoine Lavoisier in the year 1789.

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A sample of ethanol containing 0.35 mol is cooled from 125.0cC to 62.0oC. Given the data below, select all the options that correctly reflect how to calculate the total heat change for the process.Boiling point for ethanol at 1 atm = 78.5oC; ΔHovap = 40.5 kJ/mol; Cliq = 1130 J/mol⋅oC; and sgas = 65.9 J/mol⋅oC

Answers

The total heat change for the process is : ΔH = -45. 6 kJ

The total heat change for the process can be calculated using the following equation:

Where m is the mass in moles, and ΔHovap is the enthalpy of vaporization.

In this case, the total heat change can be calculated as follows:

ΔH = -40. 5 kJ / mol + 0. 35 mol(1130 J/ mol⋅ oC)(62. 0oC - 125. 0oC) + 0. 35 mol(65. 9 J/ mol ⋅ oC)ln(1/1)

ΔH = -45. 6 kJ

Ethanol is an organic compound, also known as ethyl alcohol or grain alcohol. It is a clear, colorless liquid with a characteristic pleasant odor and a burning taste. It is highly flammable, and is usually derived from crops such as corn, wheat, or sugar cane. It is commonly used as a fuel, a solvent, and an ingredient in many alcoholic beverages.

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what is the final concentration when 275mL of water is added to 155mL of 0.130M KCl?

Answers

Answer:

Below

Explanation:

.130 * .155   = initial number of moles of KCL

then add .275 liter of water bringing total volume to .275 + .155 = .430 Liter

moles of KCl divided by volume = concentration

  .130*.155 / (.430) = .047 M

Which of the following is a type of crystalline solid? Select all that apply.
Select all that apply:
O metallic
O ionic
O covalent network
O molecular

Answers

Crystalline solids often fall into one of four categories. These categories include network covalent solids, ionic solids, metallic solids, and molecular solids.

Which one of the following describes crystalline structure?

In contrast to its non-crystalline form, a crystalline structure comprises highly tightly packed atoms, leading to a high density of the material.

How many different kinds of crystalline solids exist?

Four categories of crystalline solids exist in solid ions Electrostatic attractions keep positive and negative ions together as they develop. They stand out in the solid state due to their extremely high melting temperatures, brittleness, and poor conductivity.

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which molecules have dipole-dipole forces? check all that apply. which molecules have dipole-dipole forces?check all that apply. ch3cl ccl4 hcl

Answers

All three molecules [tex]CH_3Cl, HCl, CCl_4[/tex] have dipole-dipole forces acting on them.

[tex]CH_3Cl[/tex] or chloromethane- Carbon provides the positive end of the dipole and Cl being more electronegative forms the negative end of the dipole. The bond is slightly polar in nature.

Dipole-dipole forces are formed in [tex]HCl[/tex] or hydrochloric acid because the chlorine atom holds a slightly value of AH charge whereas the charge on hydrogen is slightly value of AH. Because of this force of attraction, a small dipole-dipole force acts between two HCl molecules.

Intermolecular forces of attraction between atoms of [tex]CCl_4[/tex] or carbon tetrachloride is also known as the London Dispersion Force of Attraction. Attraction takes place between temporary dipoles in the structure.

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Complete Question:

which molecules have dipole-dipole forces among the given options?

1) [tex]CH_{3}Cl[/tex]

2) [tex]CCl_4[/tex]

3) [tex]HCl[/tex]

check all that apply.

In terms of MO theory, which of species in each of the following pairs will most likely be the one to gain an electron? Explain your rationale.
a. CN or NO (Assuming the order of MO energy level diagram of NO follows that of N2)
b. (O2)2+ or (N2)2+

Answers

a. NO (nitrosyl) is more likely to gain an electron compared to CN (cyanide). This is because the electron density of the nitrogen atom in NO is higher, due to the presence of the unpaired electron in its lowest unoccupied molecular orbital (LUMO), which makes it a more attractive site for accepting electrons.

b. (N2)2+ is more likely to gain an electron compared to (O2)2+. This is because nitrogen atoms have a higher electronegativity compared to oxygen atoms, which means nitrogen attracts electrons more strongly. As a result, the electron density in the LUMO of N2 is higher, making it a more attractive site for accepting electrons.

complete the mechanism for the reaction of butanone with nabh4 nabh 4 followed by the addition of aqueous acid.

Answers

Explanation:

The reaction mechanism is as follows:

1. Butanone reacts with NABH4 in an aprotic solvent to form an intermediate alkoxide.

2. The alkoxide reacts with aqueous acid to form the corresponding alcohol.

3. The alcohol is then protonated by the acid to form the final product.

write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of fructose

Answers

Chemical equation for the reaction whose enthalpy change is the standard enthalpy in the formation of fructose is:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = ΔHf° (fructose)

The standard enthalpy of formation (ΔHf°) of fructose is the change in enthalpy that occurs when one mole of fructose is formed from its elements in their standard state. The standard state for elements is defined as having a pressure of 1 bar (100 kPa) and a temperature of 298 K. The chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of fructose is:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = ΔHf° (fructose)

In this equation, glucose (C6H12O6) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The change in enthalpy (ΔH) represents the standard enthalpy of formation of fructose (ΔHf°). It is important to note that the standard enthalpy of formation is a measure of the energy change associated with the formation of one mole of a substance from its elements in their standard state. The value of ΔHf° is used in thermochemistry to calculate the enthalpy change of reactions, and to determine the heat of combustion of a substance.

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5. A covalent bond occurs when two atoms_______valence______

Answers

Answer:

Valence shell electrons

Explanation:

⁷ 168 g of iron reacts with excess oxygen to form 232 g of an oxide of iron. iron + oxygen H ratio iron oxide a. Calculate the mass of oxygen in the iron oxide. b. Calculate the males of oxygen in the oxide. c. Calculate the moles of iron in the iron oxide. d. Calculate the ratio of the iron to oxygen to give the formula of this oxide of iron. formula ​

Answers

The mass of the oxygen that reacted is 70.4 g.

What is the reaction of iron and oxygen?

The reaction between iron and oxygen is called rusting and is a oxidation process. The balanced chemical equation for the reaction is:

4 Fe + 3 O2 -> 2 Fe2O3

Number of moles of iron = 168 g/56 g/mol = 3 moles

Number of moles of iron III oxide = 232 g /160 g/mol

= 1.45 moles

The mass of the oxygen that reacted is;

If 3 moles of oxygen produces 2 moles of oxide

x moles of oxygen would produce 1.45 moles ofn the oxde

x = 3 * 1.45/2

= 2.2 moles

Mass of the oxygen = 2.2 moles * 32 g/mol

= 70.4 g

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which of the following products would be produced from hydroboration/oxidation of 1-pentyne? e only a and b only b and e only a only

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The products of the hydroboration/oxidation of 1-pentyne are A and B only. A is the primary product, which is 2-methyl-2-pentanol, and B is the secondary product, which is 2-ethyl-1-pentanol.

When 1-pentyne is subjected to hydroboration/oxidation, a boron atom is added to the double bond, forming a new tertiary carbon center. The boron atom is then oxidized to form a hydroxyl group, which is the primary product (A).

The secondary product (B) is formed when the tertiary carbon center undergoes an E2 elimination, resulting in the formation of an ethyl group.

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