which of the following are produced in the ozonolysis of the following molecule? the skeletal structure of a molecule with a smiles string of ccccCH3CH2COOH (CH3)2CHCOOH (CH3)2CHCH2COOH CH3CH2CH2COOH CH3CH2CH2CH2COOH

Ionic bonds lead to the formation of crystal lattices, rather than separate, discrete molecules.

Positively charged cations and negatively charged anions are produced when electrons are transported from one atom to another in an ionic bond.

Then, these ions arrange themselves into a three-dimensional array to reduce the system's potential energy. A repeating unit cell can serve as a representation of the final structure, a crystal lattice.

Instead of distinct, discrete molecules, crystal lattices are formed as a result of ionic bonding.

Strong electrostatic interactions between the ions with opposing charges hold the lattice together. Ionic chemicals do not exist as isolated molecules since the lattice permeates the entire crystal.

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NH3 is commonly known as ammonia and H2O is commonly known as water. Molecular compounds are formed by the combination of two or more non-metal elements.

These compounds are also known as covalent compounds as they are held together by covalent bonds. Covalent bonds are formed by the sharing of electrons between two atoms. Molecular compounds are characterized by their low melting and boiling points and are generally poor conductors of electricity.
NH3, which is ammonia, is a colourless gas with a pungent odour. It is used in the production of fertilizers, cleaning agents, and as a refrigerant. NH3 is composed of one nitrogen atom and three hydrogen atoms. It is a basic compound and reacts with acids to form ammonium salts.
H2O, which is water, is a colourless, odourless, and tasteless liquid. It is essential for all forms of life on Earth and is the most common substance on Earth's surface. Water is composed of two hydrogen atoms and one oxygen atom. It is a polar compound, which means it has a positive and negative end. Water is a versatile solvent and is capable of dissolving many substances.
In conclusion, NH3 is commonly known as ammonia and H2O is commonly known as water. These molecular compounds have unique properties and are essential to life on Earth.

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The correct statement concerning galvanic (voltaic) cells and electrolytic cells is that external electric energy must be supplied to an electrolytic cell, while in a galvanic (voltaic) cell, the redox reaction is always spontaneous.

Galvanic cells produce electrical energy from a spontaneous redox reaction, where the anode undergoes oxidation and the cathode undergoes reduction. This flow of electrons generates an electrical current that can be used to power electronic devices. In contrast, electrolytic cells require an external source of electrical energy to drive a non-spontaneous redox reaction. The anode serves as the site of oxidation, and the cathode serves as the site of reduction. By supplying electrical energy, the reaction can proceed in the desired direction, producing a product that would not form spontaneously.
In summary, galvanic cells can serve as a source of electrical energy, while electrolytic cells require external electric energy to drive a non-spontaneous reaction. The cathode is the site of reduction in a galvanic (voltaic) cell, but the cathode is the site of oxidation in an electrolytic cell. Finally, in a galvanic (voltaic) cell, the redox reaction is always spontaneous, while in an electrolytic cell, an external energy source is required to drive the non-spontaneous reaction.

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The hydrogen ion concentration ([h]) for a 0.0473 m solution of Barium hydroxide, [tex]Ba(OH)2[/tex] assuming complete dissociation of the compound is 0.0946 mol/L.

The symbol [h] typically refers to the hydrogen ion concentration in a solution.

The balanced equation for the solution of barium hydroxide is

[tex]Ba(OH)_{2} (s)[/tex] → [tex]Ba_{2}[/tex](aq) + 2OH-(aq)

Here [tex]Ba(OH)_{2}[/tex] dissolves completely. The concentration of barium hydroxide and OH- will be equal to the concentration  [tex]Ba(OH)2[/tex] originally added to the solution

The OH- concentration will be calculated as:

[OH-] = 2 × 0.0473 mol/L

[OH-]= 0.0946 mol/L

[h] = [H+] = [OH-] = 0.0946 mol/L

Therefore, we can conclude that the hydrogen ion concentration ([h]) is 0.0946 mol/L.

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CH₃CH₂OH < CH₃CH₂COOH < ClCH₂CH₂COOH < CH₃CHClCOOH < CH₃CH₂SO₃H. This would be the order of increasing acidity.

To rank the given compounds in order of increasing acidity, we need to consider their molecular structure and the factors that affect acidity. The compounds are:

(a) CH₃CH₂SO₃H
(b) CH₃CH₂OH
(c) CH₃CH₂COOH
(d) CH₃CHClCOOH
(e) ClCH₂CH₂COOH

The factors that influence acidity are the stability of the conjugate base and the presence of electron-withdrawing groups. A more stable conjugate base leads to a stronger acid. Electron-withdrawing groups increase acidity by stabilizing the negative charge on the conjugate base.

(b) CH₃CH₂OH is an alcohol, which generally has a low acidity. This is because the conjugate base, an alkoxide ion, is not very stable due to the negative charge on the oxygen atom.

(c) CH₃CH₂COOH and (d) CH₃CHClCOOH are carboxylic acids, which are more acidic than alcohols. The presence of a carbonyl group stabilizes the conjugate base through resonance.

(d) CH₃CHClCOOH has a chlorine atom, an electron-withdrawing group, which further stabilizes the conjugate base, making it more acidic than (c) CH₃CH₂COOH.

(e) ClCH₂CH₂COOH also has a chlorine atom, but it is further away from the carboxyl group, making its electron-withdrawing effect weaker than in (d).

(a) CH₃CH₂SO₃H is a sulfonic acid, which has a highly stable conjugate base due to resonance and the inductive effect of three oxygen atoms. This makes it the most acidic compound.

In conclusion, the order of increasing acidity is:
(b) CH₃CH₂OH < (c) CH₃CH₂COOH < (e) ClCH₂CH₂COOH < (d) CH₃CHClCOOH < (a) CH₃CH₂SO₃H

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The written formula for Manganese (IV) nitrate is Mn(NO₃)₄.

In this formula, Manganese (Mn) has a +4 charge (indicated by the Roman numeral IV) and Nitrate (NO₃) has a -1 charge.

To create a neutral compound, we need four nitrate ions to balance the +4 charge of Manganese.

Therefore, we write the formula as Mn(NO₃)₄.

Hence, The written formula for Manganese (IV) nitrate is Mn(NO₃)₄, which consists of one Manganese ion with a +4 charge and four nitrate ions, each with a -1 charge, to form a neutral compound.

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Answer:

Explanation:

it's B. 2 HCI: 1 CaCO3 :)

Using crystal field theory, the number of unpaired electrons in [Mn(NH₃)₆]²⁺ is three.

Using crystal field theory, we can determine the number of unpaired electrons in the complex ion [Mn(NH₃)₆]²⁺+ as follows:

The complex ion consists of a central Mn²⁺ ion, which is surrounded by six NH₃ ligands. Mn²⁺ has an electron configuration of [Ar] 3d⁵, meaning it has five electrons in its d orbitals. The NH₃ ligands are considered weak field ligands, meaning they cause a small energy difference between the d orbitals.

In weak field complexes, the electrons preferentially occupy the lower energy orbitals in a way that maximizes the number of unpaired electrons (Hund's rule). In an octahedral complex, such as [Mn(NH₃)₆]²⁺, the d orbitals are split into two groups: the t²g orbitals (dxy, dxz, dyz) and the eg orbitals (dz², dx²-y²). The t²g orbitals are lower in energy than the eg orbitals.

As there are five d electrons in Mn²⁺, they will first fill the t²g orbitals, with one electron each (following Hund's rule). This results in three unpaired electrons in the [Mn(NH₃)₆]²⁺ complex ion.

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The hydrogen ions from the acid react with the hydroxide ions from the base to form water. This reaction is also known as an acid-base neutralization reaction. The resulting solution is neutral, meaning that it has a pH of 7. The correct option is C.

When an acid is neutralized by a base, the resulting solution contains water and a salt.

In a neutral solution, the concentration of hydrogen ions (H⁺) is equal to the concentration of hydroxide ions (OH⁻). This means that the answer is C - always equal to the concentration of hydroxide ions. The equation for this reaction is:

HA + BOH → H₂O + BA

In this equation, HA represents an acid, BOH represents a base, H₂O represents water, and BA represents a salt.

The pH of a solution can be determined by calculating the concentration of hydrogen ions. If the concentration of hydrogen ions is greater than the concentration of hydroxide ions, the solution is acidic and has a pH less than 7. If the concentration of hydroxide ions is greater than the concentration of hydrogen ions, the solution is basic and has a pH greater than 7. In a neutral solution, the concentration of hydrogen ions is equal to the concentration of hydroxide ions, resulting in a pH of 7.

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The correct electron dot structure of water is shown by option B

Each valence electron in the electron dot structure is represented by a dot that is positioned around the element's atomic symbol.

The Lewis dot structure, sometimes referred to as the electron dot structure, uses dots to represent the valence electrons of an atom.

Water contains two hydrogen and one oxygen atom with the oxygen atom having two lone pairs of electrons as shown above in the question.

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Mitochondria is a double membrane bound organelle which is found in most eukaryotic organisms. They are found inside the cytoplasm and essentially function as the cells digestive system. Here number 3 represents the mitochondria. The correct option is C.

Mitochondria popularly known as the power house of the cell play an important role in breaking down the nutrients and produce energy rich molecules for the cell. Its size ranges from 0.5 to 1.0 micrometer in diameter.

The mitochondria is a double membraned rod shaped structure which is found both in plants and animals. It comprises of an outer membrane, inner membrane and a gel material called matrix.

Thus the correct option is C.

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Answer: CH4

Explanation: Methane

To create a concise informative plot for each of the terms listed, the following items are required: 1. Reaction, Chemicals or System Being Investigated:
Title: Give a descriptive title that clearly indicates the nature of the investigation.
Axes: Label the x-axis and y-axis with the appropriate variables that are being measured.
Data: Plot the data points on the graph.
Table with Headers Containing Units: Create a table that displays the data collected during the experiment, with headers containing units.
Name or Symbol of the Variable: Clearly identify the variables being measured, either by their name or symbol.

2. Special Conditions of the Experiment:
Title: Include the special conditions being tested in the title.
Axes: Label the x-axis and y-axis with the appropriate variables that are being measured.
Data: Plot the data points on the graph.
Table with Headers Containing Units: Create a table that displays the data collected during the experiment, with headers containing units.
Name or Symbol of the Variable: Clearly identify the variables being measured, either by their name or symbol.

3. Best Fit Line or Curve with an Equation:
Title: Give a descriptive title that clearly indicates the nature of the investigation.
Axes: Label the x-axis and y-axis with the appropriate variables that are being measured.
Data: Plot the data points on the graph.
Best Fit Line or Curve with an Equation: Draw the best fit line or curve through the data points and display the equation on the graph.
Units (if any): Include units on the axes and in the equation.
Name or Symbol of the Variable: Clearly identify the variables being measured, either by their name or symbol.

4. Units (if any):
Title: Give a descriptive title that clearly indicates the nature of the investigation.
Axes: Label the x-axis and y-axis with the appropriate variables that are being measured, including their units.
Data: Plot the data points on the graph.
Table with Headers Containing Units: Create a table that displays the data collected during the experiment, with headers containing units.
Name or Symbol of the Variable: Clearly identify the variables being measured, either by their name or symbol.

5. Table with Headers Containing Units:
Title: Give a descriptive title that clearly indicates the nature of the investigation.
Axes: Label the x-axis and y-axis with the appropriate variables that are being measured.
Table with Headers Containing Units: Create a table that displays the data collected during the experiment, with headers containing units.
Name or Symbol of the Variable: Clearly identify the variables being measured, either by their name or symbol.

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The acidic equilibrium equation for HBrO (hypobromous acid) is: HBrO (aq) ⇌ H⁺ (aq) + BrO⁻ (aq). It represents the reversible dissociation of HBrO into hydronium ions (H⁺) and hypobromite ions (BrO⁻) in aqueous solution.

The acidic equilibrium equation for HBrO (hypobromous acid) can be written as follows:

HBrO (aq) ⇌ H⁺ (aq) + BrO⁻ (aq)

In this equation, HBrO is the acid (in aqueous solution), while H⁺ and BrO⁻ are the conjugate base and acid, respectively. The double arrow (⇌) indicates that the reaction can proceed in either direction, and the (aq) notation indicates that all species are in aqueous solution.

The acidic equilibrium equation for HBrO (hypobromous acid) can be written as follows:

HBrO (aq) ⇌ H⁺ (aq) + BrO⁻ (aq)

In this equation, HBrO is the acid (in aqueous solution), while H⁺ and BrO⁻ are the conjugate base and acid, respectively. The double arrow (⇌) indicates that the reaction can proceed in either direction, and the (aq) notation indicates that all species are in aqueous solution.

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The equilibrium equation for HBrO, an aqueous solution, is HBrO (aq) ↔ H+(aq) + BrO-(aq). This illustrates that HBrO can dissociate into H+ and BrO- ions and vice versa.

The acidic equilibrium equation for HBrO, or Hypobromous acid, starts by indicating its state as aqueous (aq). The equation is HBrO (aq) ↔ H+(aq) + BrO-(aq). This equation shows the acid (HBrO) ionizing into H+ ions and BrO- ions in water. The symbol ↔ is used to show that the reaction can proceed in both directions, hence it's an equilibrium. In other words, HBrO can dissociate into H+ and BrO- and these ions can also recombine to form HBrO. This is a characteristic process in acid-base chemistry which is typically described by the Brønsted-Lowry definition of acids and bases.

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When a ketone and its enol are in equilibrium, under most conditions, the concentration of the enol is slightly higher than the concentration of the ketone. Option A is Correct.

This is because the enol is the less stable tautomer and therefore the equilibrium lies slightly towards the enol form.

The keto-enol tautomerism is the name of this chemical equilibrium. The enol tautomer and keto would quickly reach equilibrium. Additionally, because the keto form is more stable in this equilibrium, it predominates in the combination. Therefore, the enol's concentration would be lower than the keto's.

Even though enol suggests the presence of a double bond and a hydroxyl group, they carry a carbonyl bond. The stabilising components of both the keto and enol tautomers are necessary for the equilibrium of keto-enol tautomerization.

The enol that might have produced 2-pentanone was pent-1-en-2-ol. Equilibrium is not a very good depiction because it substantially favours the ketone.

Little reaction might be seen if the ketone was treated with bromine because the enol level might be too low.

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An effective safety measure when running a new reaction is to start by b. Run the reaction on a small scale.

By conducting the experiment with smaller quantities of reactants, you can minimize potential hazards and more easily monitor the reaction. This allows you to observe any unexpected outcomes or side reactions that may occur.
Keeping the temperature low is another important safety measure. High temperatures can lead to increased reaction rates and the formation of more side products, making the reaction difficult to control. By maintaining a lower temperature, you can better manage the reaction and ensure a safer process.
Running the reaction for a short time is also a useful strategy. By limiting the reaction time, you can quickly identify any issues that may arise, such as excessive heat generation or unexpected byproducts. This enables you to address these issues promptly before they escalate.
While avoiding catalysts may seem like a safe choice, it is essential to understand that catalysts can improve the reaction's efficiency and selectivity. When used correctly, catalysts can make a reaction safer and more controlled. It is crucial to choose an appropriate catalyst and use it in the correct proportions to ensure a safe reaction.
In summary, when running a new reaction, it is essential to implement safety measures such as running the reaction on a small scale, maintaining a low temperature, and limiting the reaction time. Additionally, the proper use of catalysts can enhance reaction safety and control.

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Answer:

D) Oxygen Gas

Explanation:

Early Earth's atmosphere lacked oxygen gas due to the lack of general vegetation on Earth (plants produce O2 gas as a byproduct of photosynthesis) and the large presence of volcanic gases, which are mostly made up of CO2.

1. CO: 3 electron pairs are shared.
2. O₂: Two electron pairs are shared.
3. ClO: 1 electron pair is shared.
4. CN: Three electron pairs are shared.

Here's the information about the shared electron pairs (bonds) in each of the molecules and ions:

1. CO - Carbon Monoxide has a triple bond between C and O, which means 3 electron pairs are shared.
2. O₂ - Oxygen molecule has a double bond between the two O atoms, which means 2 electron pairs are shared.
3. ClO - Chlorine Monoxide has a single bond between Cl and O, which means 1 electron pair is shared.
4. CN - Cyanide ion has a triple bond between C and N, which means 3 electron pairs are shared.

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The mass of 3.15 mol of AgNO₃ is 533.6 grams, the mass of 0.0901 mol of CaCl₂ is 10.02 grams, and the mass of 11.86 mol of H₂S is 404.5 grams.

To calculate the mass in grams of each of the given substances, we need to use the molar mass of each compound. The molar mass of AgNO₃ (silver nitrate) is 169.87 g/mol, the molar mass of CaCl₂ (calcium chloride) is 110.98 g/mol, and the molar mass of H₂S (hydrogen sulfide) is 34.08 g/mol.

To calculate the mass of 3.15 mol of AgNO₃, we can use the following formula:

mass = moles x molar mass
mass = 3.15 mol x 169.87 g/mol
mass = 533.6 g

Therefore, the mass of 3.15 mol of AgNO₃ is 533.6 grams.

Similarly, to calculate the mass of 0.0901 mol of CaCl₂, we can use the formula:

mass = moles x molar mass
mass = 0.0901 mol x 110.98 g/mol
mass = 10.02 g

Therefore, the mass of 0.0901 mol of CaCl₂ is 10.02 grams.

Finally, to calculate the mass of 11.86 mol of H₂S, we can use the formula:

mass = moles x molar mass
mass = 11.86 mol x 34.08 g/mol
mass = 404.5 g

Therefore, the mass of 11.86 mol of H₂S is 404.5 grams.

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A chemical equation is balanced when E. the number of atoms of each element is the same in reactants and products because in a balanced chemical reaction the amount of reactants is equal to the amount of products.

A balanced chemical equation represents a chemical reaction with the same number of atoms of each element on both sides of the equation. In other words, the total mass and the number of atoms in the reactants are equal to those in the products.

The balance of the equation is achieved by adjusting the coefficients in front of the chemical formulas of the reactants and products to ensure that the same number of atoms of each element is present on both sides of the equation.  Hence, option E is correct.

Balancing the chemical equation is important as it helps to predict the amount of reactants and products that will be used and formed, respectively, and it also ensures that the reaction follows the law of conservation of mass.

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a) The frequency of ultraviolet light is 8.54 * 10^{15} Hz

b) The wavelength of ultraviolet light is 35 nm

To solve for the frequency of ultraviolet light, we can use the equation E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon. Rearranging the equation, we get f = E/h. Plugging in the given values, we get

f = \frac{(5.6610-18 J)}{(6.62607015 * 10^{-34} J s)

f = 8.54 * 10^{15} Hz.
To solve for the wavelength of ultraviolet light, we can use the equation c = λf, where c is the speed of light, λ is the wavelength, and f is the frequency. Rearranging the equation, we get λ = \frac{c}{f}. Plugging in the given values and the speed of light (299,792,458 m/s), we get

λ = \frac{(299,792,458 m/s)}{(8.54 * 10^{15} Hz)

λ= 35 nm (nanometers).
In summary, the frequency of ultraviolet light with a photon energy of 5.6610-18 J is 8.54 * 1015 Hz, and the corresponding wavelength is 35 nm. Ultraviolet light has a shorter wavelength and higher frequency than visible light, making it more energetic and potentially harmful to living organisms.

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Nuclear fission is only experimental however, nuclear fusion is not experimental.

Fission generates energy by breaking heavier atoms, such as uranium, into smaller atoms like iodine, cesium, strontium, xenon, and barium, to mention a few.

Fusion, on the other hand, is the joining of light atoms, such as two hydrogen isotopes.

Both are nuclear reactions that generate energy, but they are not the same. Fusion is the joining of two light nuclei to create a larger nucleus, whereas fission is the breaking of a heavy nucleus into two lighter ones.

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Amphiprotic species are those that can act as both an acid and a base in a chemical reaction. So, among the given options

(a) H₂O - amphiprotic

(b) S₂⁻ - amphiprotic

(d) HSO₄⁻ - amphiprotic

(e) CO₃²⁻ - amphiprotic

(a) H₂O can act as an acid by donating a proton (H⁺) and as a base by accepting a proton. For example, in the reaction with HCl, water acts as a base and accepts a proton to form H₃O⁺ and Cl⁻. The chemical equation for the amphiprotic character of water can be written as:

H₂O + HCl -> H₃O⁺ + Cl⁻

(b) S₂⁻ can also act as an acid and donate a proton, or as a base and accept a proton. For example, in the reaction with HCl, sulfide ion acts as a base and accepts a proton to form HS⁻ and Cl⁻. The chemical equation for the amphiprotic character of S₂⁻ can be written as:

S₂⁻ + HCl -> HS⁻ + Cl⁻

(d) HSO₄⁻ and CO₃²⁻ are also amphiprotic species. HSO₄⁻ can act as an acid and donate a proton or as a base and accept a proton. For example, in the reaction with NH₃, HSO₄⁻ acts as an acid and donates a proton to form NH₄⁺ and SO₄²⁻. The chemical equation for the amphiprotic character of HSO₄⁻ can be written as:

HSO₄⁻ + NH₃ -> NH₄⁺ + SO₄²⁻

(e) CO₃²⁻ can act as an acid and donate two protons or as a base and accept two protons. For example, in the reaction with HCl, CO₃²⁻ acts as a base and accepts two protons to form H₂CO₃ and Cl⁻. The chemical equation for the amphiprotic character of CO32- can be written as:

CO₃²⁻ + 2HCl -> H₂CO₃ + 2Cl⁻

Therefore, only H₂O, S₂⁻, HSO₄⁻, and CO₃²⁻ are amphiprotic species. a, b, d and e are the correct options.

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The specifics of when a month starts differ from calendar to calendar; some rely on new, entirety, or crescent moons, while others do complex computations.

In contrast with solar calendars, which annual cycles are only based on the solar year, a lunar calendar includes a calendar built on the monthly phases of the Moon's aspects (synodic months, lunations). The Islamic calendar is the most extensively used lunar calendar.

A lunisolar calendar, in which the lunar months come into alignment to the solar year by some process of intercalation, such as by inserting a leap month, is distinguished from a strictly lunar calendar. The specifics of when a month starts differ from calendar to calendar; some rely on new, entirety, or crescent moons, while others do complex computations.

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Molecules are two or more atoms held together by chemical bonds. Therefore the correct option is option A.

A molecule is created when two or more atoms bind to one another. It doesn't matter whether the atoms are from the same element (like O2) or from different elements (like H2O, which has two hydrogen atoms and one oxygen atom). Covalent, ionic, or metallic chemical bonds can hold the atoms of a molecule together.

The smallest units of a compound that still have their chemical properties are called molecules. Chemical processes can separate them into individual atoms, but physical processes like heating or freezing cannot. Therefore the correct option is option A.

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Answer:

3, 876.9 ~ 3, 877

Explanation:

4.5 kg = 4500 g....... convert to gram

n =m/M , n= no of mol

m= mass of sub.

M= molar mass

so n= 4500g / 26g/mol

n= 173.076 mol

then 1 mol = 22.4 L ............... at STP

at STP 173.076 mol = X

then when u solve X by

X = (173.076 mol × 22.4 L)/ 1 mol

X = 3, 876.9 L at STP which is aproximate to

3, 877L at STP

Answer:

NUCLEAR FISSION

Explanation:

According to Einstein's equation, E = mc2, where

m

is the mass and

c

is the speed of light states that mass can be converted to energy and vice-versa. In nuclear fission, the mass of reactants is more than mass of the products. The difference in mass is called the mass defect. This mass is converted into energy

To calculate the joules of heat energy required to raise the temperature of 15 grams of gold from 22°C to 85°C, you can use the formula: Q = mcΔT where Q is the heat energy in joules, m is the mass in grams, c is the specific heat, and ΔT is the change in temperature.

Given: m = 15 grams c = 0.129 J/(g x °C) Initial temperature = 22°C Final temperature = 85°C First, find the change in temperature (ΔT): ΔT = Final temperature - Initial temperature ΔT = 85°C - 22°C ΔT = 63°C Now, plug the values into the formula: Q = (15 g) x (0.129 J/(g x °C)) x (63°C) Q = 121.635 J Since the answer should be in whole joules, you can round it to the nearest whole number: Q ≈ 122 J None of the given options match the calculated answer. However, option A (121 J) is closest to the calculated value of 122 J, so it can be considered as the best available answer among the provided options.

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The answers are 1. the value of ΔH° for the combustion of one mole of methane is -801 kJ/mol and 2. the value of ΔE° for the combustion of one mole of methane is also -801 kJ/mol.

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Boric acid is a monoprotic and Lewis acid. B(OH)[tex]_3[/tex] + H[tex]_2[/tex]O ⇌ [BO(OH)[tex]_2[/tex]]− + H[tex]_3[/tex]O+ and HBO[tex]_2[/tex] + H[tex]_2[/tex]O ⇌ [BO[tex]_2[/tex]]− + H[tex]_3[/tex]O+ are the reactions for ionisation of boric acid.

In particular, orthoboric acid is a boron, oxygen, plus hydrogen chemical having the formula B(OH)3. Trihydroxidoboron, hydrogen orthoborate, and boracic acid are other names for it. It occurs naturally as the substance known as sassolite and is typically found as colourless crystals.

A white powder that dissolves in water. It is a weak acid that can react using alcohols to produce borate esters as well as a variety of borate anions and salts. Boric acid is a monoprotic and Lewis acid. B(OH)[tex]_3[/tex] + H[tex]_2[/tex]O ⇌ [BO(OH)[tex]_2[/tex]]− + H[tex]_3[/tex]O+ and HBO[tex]_2[/tex] + H[tex]_2[/tex]O ⇌ [BO[tex]_2[/tex]]− + H[tex]_3[/tex]O+ are the reactions for ionisation of boric acid.

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