How many moles of NH3 would form from the complete reaction of 14. 0 g N2

Answers

Answer 1

The total number of moles NH3 is 1.00 mole, under the condition that the reaction is of 14. 0 g N2.

The given balanced chemical equation for the reaction of nitrogen gas (N2) and hydrogen gas (H2) to form ammonia gas (NH₃) is
N₂(g) + 3H₂(g) → 2NH₃(g)

The molar mass of N₂ is 28.01 g/mol. To evaluate the number of moles of N₂ in 14.0 g of N₂ we divide the mass by the molar mass
Number of moles of N₂ = Mass of N₂ / Molar mass of N₂
Number of moles of N₂ = 14.0 g / 28.01 g/mol
Number of moles of N₂ = 0.4998 mol

Then, the number of moles of NH3 that would form from the complete reaction of 14.0 g N2 can be evaluated
Number of moles of NH₃ = Number of moles of N₂ × (2 moles NH₃ / 1 mole N₂)
Number of moles of NH₃ = 0.4998 mol × (2 mol NH₃ / 1 mol N₂)
Number of moles of NH₃ = 0.9996 mol

Hence, approximately 1.00 mole of NH₃ would form from the complete reaction of 14.0 g N₂.

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Related Questions

In a single covalent bond, _____total electrons are shared (please enter the number of shared electrons).

Answers

In a single covalent bond, two total electrons are shared.

Two total electrons are shared by one covalent bond. One pair of electrons are shared by two atoms in a single covalent connection.

In order to create a stable electron configuration for both atoms, each atom contributes one electron to create a shared pair.

A single covalent bond involves the sharing of one pair of electrons between two atoms.

Common examples of this kind of link between two nonmetals include the bond between the two hydrogen atoms in a molecule of H2 or the bond between the carbon and oxygen atoms in a molecule of CO.

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Determine the mass of carbon dioxide that should be produced in the reaction between 3. 74g of carbon and excess oxygen what is the maximum recent yield if 11. 34g of CO2 is recovers

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The mass of carbon dioxide that should be produced in the reaction is equals to 44.01 g. The maximum Percent yield is 25.8%.

We have a molecule of carbon dioxide. In a reaction between carbon and excess oxygen to form a molecule of carbon dioxide. The reaction [tex]C + O_2 →CO_2[/tex]

Mass of carbon use in reaction =3.74 g

Mass of carbon dioxide that recover

=11.34 g

We have to determine the mass of carbon dioxide. Molar mass of Carbon dioxide= 44.01 g/mol

From the reaction, the one mole of carbon and one mole of oxygen reacts and form one mole of carbon dioxide molecule.

Mass of carbon dioxide = molar mass of carbon dioxide × moles of carbon dioxide

= 44.01 g/mol × 1 mole

= 44.01 g

Now, the maximum recent yield = [tex]\frac{ 11.34}{44.01} [/tex]

= 0.258

Percentage of yield = 25.8%. Hence, required value is 25.8%.

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For the following reaction, if nh3 is used up at a rate of 0. 30mmin, what is the rate of formation of h2? 2nh3→n2 3h2

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The rate of formation of [tex]H_{2}[/tex] is 0.45 min⁻¹

The balanced chemical equation is:

2 [tex]NH_{3}[/tex]→ [tex]N_{2}[/tex] + 3[tex]H_{2}[/tex]

From the equation, we can see that for every 2 moles of [tex]NH_{3}[/tex] consumed, 3 moles of H2 are formed. Therefore, the ratio of the rate of formation of [tex]H_{2}[/tex]to the rate of consumption of [tex]NH_{3}[/tex] is 3/2.

Given that [tex]NH_{3}[/tex] is being consumed at a rate of 0.30 min⁻¹, the rate of formation of [tex]H_{2}[/tex] can be calculated as follows:

Rate of formation of [tex]H_{2}[/tex] = (3/2) × Rate of consumption of [tex]NH_{3}[/tex]

Rate of formation of [tex]H_{2}[/tex] = (3/2) × 0.30 min⁻¹

Rate of formation of [tex]H_{2}[/tex] = 0.45 min⁻¹

Therefore, the rate of formation of [tex]H_{2}[/tex] is 0.45 min⁻¹

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the main processes that happens during the alpha type of radioactivity decay is:
a. an electron is given off
b. an electron is absorbed and turned into a neuron
c. an atom splits into two pieces
d. an alpha particle combines with small atom to make a larger one

Answers

Answer:

The main process that happens during the alpha type of radioactivity decay is (d) an alpha particle combines with a small atom to make a larger one.

Explanation:

The main process that happens during the alpha type of radioactivity decay is (d) an alpha particle combines with a small atom to make a larger one.

During alpha decay, a nucleus emits an alpha particle, which consists of two protons and two neutrons. The emission of the alpha particle reduces the atomic number of the parent atom by 2 and the mass number by 4. Therefore, a new nucleus is formed, which has an atomic number that is 2 less and a mass number that is 4 less than the parent nucleus. This type of decay is commonly observed in heavy nuclei, such as uranium and plutonium.

Which tool likely made these marks?
hammer
file
saw
O screwdriver

Answers

Answer: Saw

Explanation:

Pretty obvious

Answer:

Which tool likely made these marks?

O hammer

O file

O saw

O screwdriver

Explanation:

You're welcome.

How many moles of KClO3 are needed to produce 1039 L of O2according to the following equation?2KClO3→2KCl(s)+3O2(g)

Answers

Approximately 30.9 moles of KClO3 are needed to produce 1039 L of O2 according to the given equation. To determine how many moles of KClO3 are needed to produce 1039 L of O2 according to the equation 2KClO3 → 2KCl(s) + 3O2(g), follow these steps:


Step:1. Determine the stoichiometric ratio between KClO3 and O2 from the balanced equation. In this case, it is 2 moles of KClO3 producing 3 moles of O2.
Step:2. Convert the given volume of O2 (1039 L) to moles using the ideal gas law. Assume standard temperature and pressure (STP) conditions, where 1 mole of any gas occupies 22.4 L.
Moles of O2 = 1039 L / 22.4 L/mol = 46.4 moles (approximately)
Step:3. Using the stoichiometric ratio, calculate the moles of KClO3 needed to produce 46.4 moles of O2.
(2 moles KClO3 / 3 moles O2) x 46.4 moles O2 = 30.9 moles of KClO3 (approximately)

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Half reaction for conversion of sulphite to sulfate

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Overall, this reaction is important in environmental chemistry as it can be used to remove sulphites from water and other solutions.

The half reaction for the conversion of sulphite to sulfate involves the transfer of two electrons and two hydrogen ions. The half reaction can be represented as follows:
[tex]SO_3^{2-} + 2H^+ + 2e^- -- > SO_4^{2-}[/tex]
This half reaction shows that sulphite ([tex]SO_3^{2-}[/tex]) is oxidized to sulfate ([tex]SO_4^{2-}[/tex]) by losing two electrons and two hydrogen ions. This process can occur in the presence of an oxidizing agent, such as hydrogen peroxide or chlorine. The sulfate ion is the final product of the reaction and the sulfur dioxide molecule is released as a gas into the atmosphere.

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g if you need to know the hydroxide ion concentration of an aqueous solution with legal ramifications, which method would be the best method to use to ensure accuracy without any reasonable doubt.

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The best method to determine the hydroxide ion concentration of an aqueous solution with legal ramifications would be to use titration with a standardised acid solution.

Titration with a strong acid and a reliable indicator, followed by careful calculations, is the best method to determine hydroxide ion concentration. This method provides high accuracy and precision, allowing you to confidently determine the hydroxide ion concentration in the aqueous solution.This method is highly accurate and provides precise results. It involves adding the acid solution to the solution of unknown hydroxide ion concentration until the equivalence point is reached, which is indicated by a color change in the solution. The amount of acid solution used can then be used to calculate the hydroxide ion concentration of the original solution with high accuracy and without reasonable doubt.

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The hydrogen sulfite or bisulfite ion HS03 can act as either an acid or a base in water. Write two hydrolysis reactions for HSO,- (One acting as an acid and one as a base.

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The hydrogen sulfite or bisulfite ion (HSO3-) can act as an acid or a base in water, depending on the solution's pH. When HSO3- is in an acidic solution, it can act as a base and accept a proton to form the sulfurous acid (H2SO3):
HSO3- + H3O+ → H2SO3 + H2O

In this reaction, the HSO3- ion accepts a proton (H+) from the hydronium ion (H3O+) to form the sulfurous acid (H2SO3). The reaction's forward direction can be driven by increasing the acidity of the solution or by adding more H3O+ ions.On the other hand, when HSO3- is in a basic solution, it can act as an acid and donate a proton to form the sulfite ion (SO32-):HSO3- + OH- → SO32- + H2O.In this reaction, the HSO3- ion donates a proton (H+) to the hydroxide ion (OH-) to form the sulfite ion (SO32-). The reaction's forward direction can be driven by increasing the solution's basicity or by adding more OH- ions.It is important to note that HSO3- is a weak acid, and its hydrolysis reaction can be influenced by various factors such as temperature, pressure, and the presence of other ions in the solution. Additionally, HSO3- is a sulfite ion that is commonly found in food and beverage products as a preservative. People who are sensitive to sulfites may experience allergic reactions after consuming foods or drinks that contain them.

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write the reaction and the ksp expressions for the following slightly soluble salts dissolving in water

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When a slightly soluble salt dissolves in water, it dissociates into its constituent ions, resulting in an equilibrium reaction. The equilibrium constant for this reaction is known as the solubility product constant, Ksp.

Ksp is a measure of the degree of solubility of a salt and is dependent on the ionic concentration of the solution.Let us consider the example of silver chloride (AgCl), which is a slightly soluble salt. When AgCl dissolves in water, it dissociates into its constituent ions, Ag+ and Cl-. This process is represented by the following chemical equation:AgCl(s) ⇌ Ag+(aq) + Cl-(aq)The equilibrium constant expression for this reaction is given by:[tex]Ksp = [Ag^+][Cl^-][/tex]where [[tex]Ag^+[/tex]] and [tex][Cl^-][/tex] represent the ionic concentrations of silver and chloride ions, respectively.Similarly, the Ksp expressions for other slightly soluble salts, such as calcium carbonate (CaCO3), lead(II) iodide (PbI2), and silver sulfate (Ag2SO4), can be written based on their respective dissociation reactions in water.[tex]For CaCO3: CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)\\Ksp = [Ca2+][CO32^-][/tex][tex]For PbI2: PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)\\Ksp = [Pb2+][I^-]^2[/tex][tex]For Ag2SO4: Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)\\Ksp = [Ag+]^2[SO42^-][/tex]These Ksp expressions are useful for determining the solubility of a salt in water and can be used to predict the formation of a precipitate under certain conditions, such as changes in temperature or pH.

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te express your answer in condensed form in order of increasing orbital energy as a string without blank space between orbitals. for example, [he]2s22p2 should be entered as [he]2s^22p^2

Answers

In order of increasing orbital energy, this configuration can be expressed in condensed form as follows:

[Ne]3s²3p²

The electron configuration of sulfur is 1s²2s²2p⁶3s²3p⁴.

To express this configuration in condensed form in order of increasing orbital energy, we can group the electrons by the principal energy level (n) and list them in order of increasing sublevel (s, p, d, f):

[Ne]3s²3p²

Here, [Ne] represents the electron configuration of the noble gas neon, whose completely filled 2s and 2p subshells are included in the core electrons of the sulfur atom. The valence electrons of sulfur are located in the 3s and 3p subshells, which have higher energy levels than the core electrons.

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which statement is true? answer unselected the number of standard atomic orbitals is less than the number of hybrid atomic orbitals. unselected there is no connection between the number of standard atomic orbitals and the number of hybrid atomic orbitals. unselected the number of standard atomic orbitals is greater than the number of the hybrid atomic orbitals. unselected the number of hybrid atomic orbitals made equals the number of standard atomic orbitals used. unselected i don't know yet

Answers

The statement that is true is that the number of hybrid atomic orbitals made equals the number of standard atomic orbitals used. Hybridization is a process that involves the combination of atomic orbitals to form new hybrid orbitals. The number of hybrid orbitals formed is equal to the number of standard atomic orbitals used in the hybridization process. This is because the new hybrid orbitals are a combination of the original orbitals.

For example, in sp hybridization, one s orbital and one p orbital combine to form two sp hybrid orbitals. In this case, two standard atomic orbitals were used to form two hybrid orbitals. Similarly, in sp2 hybridization, one s orbital and two p orbitals combine to form three sp2 hybrid orbitals. In this case, three standard atomic orbitals were used to form three hybrid orbitals.
Therefore, it can be concluded that the number of standard atomic orbitals used in hybridization is directly related to the number of hybrid orbitals formed. This is an important concept in understanding the geometry and bonding of molecules, and is a fundamental concept in chemistry.

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Aldolase shows no activity if it is incubated with iodoacetic acid before fructose-1,6-bisphosphate is added to the reaction mixture. What causes this loss of activity?

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The loss of activity observed in aldolase when it is incubated with iodoacetic acid before fructose-1,6-bisphosphate is added is due to the chemical modification of a key amino acid residue within the enzyme's active site.

iodoacetic acid is a potent alkylating agent that modifies the thiol group of cysteine residues, thereby inhibiting their activity. In aldolase, the specific cysteine residue that is modified by iodoacetic acid is essential for the enzyme's function, as it participates in the formation of the Schiff base intermediate during the catalytic cycle. Thus, the modification of this residue prevents aldolase from binding and catalyzing the cleavage of fructose-1,6-bisphosphate, resulting in the observed loss of activity.

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PART OF WRITTEN EXAMINATION:
The reduction reaction that occurs in an electrochemical reaction:
I Occurs at the anode
II occurs at the cathode
III involves the loss of electrons
IV involves the gain of electrons
A) I only
B) II only
C) I and III only
D) II and IV only

Answers

The reduction reaction that occurs in an electrochemical reaction involves the gain of electrons and occurs at the cathode. Therefore, the correct answer is option D, II and IV only.

The reduction reaction in an electrochemical reaction involves the gain of electrons, which means that option IV is correct. The reduction reaction occurs at the cathode, where the positively charged ions are attracted to the negatively charged electrode and gain electrons. At the same time, oxidation occurs at the anode, where the negatively charged ions are attracted to the positively charged electrode and lose electrons. Therefore, option II is also correct. Options I and III are incorrect because reduction does not occur at the anode and it involves the gain, not the loss of electrons.

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pick two elements in the same group periodic table

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Sodium and potassium have a lot in common chemically because they belong to the same group in the periodic chart.

Elements of group 1

Each have a single valence electron, which they commonly lose in chemical processes to create a positive charge. They both therefore generate molecules with comparable characteristics, such as salts that are soluble in water.

There are differences between sodium and potassium. Potassium is more reactive than sodium and has a higher potential for spontaneous air fires. Furthermore, it dissolves in water more readily than sodium.

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Which of the following processes are exothermic? endothermic? How can you tell? (a) combustion; (b) freezing water; (c) melting ice; (d) boiling water; (e) condensing steam; (f) burning paper.

Answers

Here's a classification of the given processes into exothermic and endothermic categories:

(a) Combustion: Exothermic. Combustion releases heat as chemical bonds are broken and new ones are formed, usually accompanied by the release of energy.

(b) Freezing water: Exothermic. During freezing, water molecules lose energy and form a solid structure, releasing heat in the process.

(c) Melting ice: Endothermic. Melting ice requires the absorption of heat to break the bonds between water molecules in the solid state and convert them into a liquid state.

(d) Boiling water: Endothermic. Boiling water involves the absorption of heat to convert liquid water into water vapor.

(e) Condensing steam: Exothermic. During condensation, water vapor molecules release heat as they transition from the gaseous state to the liquid state.

(f) Burning paper: Exothermic. Burning paper is a form of combustion, where chemical reactions release heat as the paper is broken down.

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CH3COCl + AlCl3 = CH3C+O + AlCl4-. (True or False)

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True. A carbocation (CH3C+O) and an aluminium tetrachloride anion (AlCl4-) are produced as a result of the interaction between acetyl chloride (CH3COCl) and aluminium chloride (AlCl3).

The aluminium chloride serves as a Lewis acid catalyst in the reaction, which follows a Friedel-Crafts acylation mechanism. An acylium ion (CH3CO+) and an AlCl4- anion are created when the acetyl chloride interacts with the aluminium chloride. The carbocation (CH3C+O) is then created by a rearrangement of the acylium ion. By serving as a counter ion, the AlCl4- anion stabilises the carbocation. As a result, the chemical equation provided is accurate. True. A carbocation and an aluminium tetrachloride anion are created in this Friedel-Crafts acylation reaction between acetyl chloride and aluminium chloride.

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In the Galvanic Series which element is listed as the most active?
A) zinc
B) copper
C) steel
D) magnesium
E) carbon

Answers

The Galvanic Series, the most active element is the one that is most likely to corrode or oxidize when in contact with other elements. In this case, D) Magnesium the Galvanic Series is a list of metals and alloys arranged according to their relative corrosion potentials in a given environment.

This series helps in predicting the corrosion behavior of a metal when in contact with another metal. The elements towards the top of the series are more active, meaning they have a higher tendency to corrode. Here's a brief explanation of the terms you mentioned Galvanic This term refers to the generation of electrical energy from a chemical reaction between two different metals or metal alloys. In the context of the Galvanic Series, it refers to the potential difference that drives the corrosion process. Element An element is a substance made up of atoms with the same atomic number, or the same number of protons in the nucleus. Elements are the fundamental building blocks of matter and cannot be broken down into simpler substances through ordinary chemical processes. In conclusion, magnesium is the most active element in the Galvanic Series and has the highest tendency to corrode when in contact with other elements.

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which of the following is the tetrahedral intermediate in the acid-catalyzed fischer esterification reaction of acetic acid, ch3co2h, and ethanol, ch3ch2oh?

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The tetrahedral intermediate in the acid-catalyzed Fischer esterification reaction of acetic acid and ethanol is formed when the carbonyl carbon of acetic acid undergoes nucleophilic attack by the oxygen of ethanol.

This intermediate then undergoes a dehydration reaction to form the ester product.

Aldehydes and ketones undergo a lot of nucleophilic addition reactions that are catalysed by an acid or base. Acids promote the production of a protonated carbonyl group, which catalyses hydration.

it is more vulnerable to an assault by a nucleophile. As a result, an intermediate hemiacetal is created, which can later be protonated and attacked by a different nucleophile to create a completely substituted acetal. In general, acid catalysis increases the carbonyl group's reactivity in nucleophilic addition processes.

Acids catalyse the hydration of carbonyl oxygen by protonating it, which increases its electrophilicity and susceptibility to nucleophilic attack. As a result, a tetrahedral intermediate is created, which subsequently proceeds through proton transfer to create the final hydrated product.

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stars are formed from the reaction of an alcohol and a carboxylic acid. identify the alcohol and carboxylic acid combination necessary to make each pictured ester. ester with benzene ring attached to carbonyl carbon and a four carbon chain attached to the singly-bonded oxygen choose... ester with a two carbon chain containing the carbonyl carbon and a phenyl ring attached to the singly-bonded oxygen choose... ester with a three carbon chain containing the carbonyl and a two carbon chain attached to the singly-bonded oxygen choose... ester with a four carbon chain containing the carbonyl carbon and a one carbon chain attached to the singly-bonded oxygen choose...

Answers

For the ester with a benzene ring attached to the carbonyl carbon and a four-carbon chain attached to the singly-bonded oxygen, the alcohol and carboxylic acid combination necessary is benzyl alcohol and hexanoic acid.

For the ester with a two-carbon chain containing the carbonyl carbon and a phenyl ring attached to the singly-bonded oxygen, the alcohol and carboxylic acid combination necessary is phenol and acetic acid.

For the ester with a three-carbon chain containing the carbonyl and a two-carbon chain attached to the singly-bonded oxygen, the alcohol and carboxylic acid combination necessary is propanol and butyric acid.

For the ester with a four-carbon chain containing the carbonyl carbon and a one-carbon chain attached to the singly-bonded oxygen, the alcohol and carboxylic acid combination necessary is methanol and pentanoic acid.

Alcohol and carboxylic acid combinations to make each of the given esters based on their structural formulae.

Ester with benzene ring attached to carbonyl carbon and a four carbon chain attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

Ester with a two carbon chain containing the carbonyl carbon and a phenyl ring attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

Ester with a three carbon chain containing the carbonyl and a two carbon chain attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

Ester with a four carbon chain containing the carbonyl carbon and a one carbon chain attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-methyl-2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-methyl-3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

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Vertical sashes should be closed except when
- Measuring the airflow of a hood
- Access to equipment inside the hood is necessary
- There is some chemical reaction occurring inside the hood
- One expects an explosion

Answers

Vertical sashes in a fume hood are an important safety feature that help to contain hazardous materials and protect the user. Typically, these sashes should be closed at all times except when certain circumstances arise. For instance, they may need to be opened to measure the airflow of a hood.

Which is essential for ensuring proper ventilation and preventing dangerous buildup of fumes or vapors. Similarly, if there is a need to access equipment inside the hood, the sashes may be opened temporarily. In some cases, if there is a chemical reaction occurring inside the hood, the sashes may need to be opened slightly to allow for proper ventilation. Finally, if there is an expectation of an explosion, the Vertical sashes should be opened to minimize the risk of injury. In general, it is important to follow proper safety procedures and guidelines when working with fume hood to ensure the safety of both the user and the surrounding environment.

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Using data from Appendix D in the textbook, calculate [OH−] and pH for each of the following solutions. A) 0. 15 M NaBrO B) 8. 2×10−2 M NaHS. C) A mixture that is 0. 13 M in NaNO2 and 0. 25 M in Ca(NO2)2

Answers

A) NaBrO: [OH⁻] = pH = 2.63 × 10⁻³ M, 10.58

B) NaHS: [OH] = 4.07 × 10⁻⁴ M, pH = 13.39

C) Mixture of NaNO₂ and Ca(NO₂)₂: [OH⁻] = 1.59 × 10⁻¹³ M, pH = 10.80.

A) NaBrO is a salt of a weak acid (HBrO) and a strong base (NaOH), so it undergoes hydrolysis. Using the equilibrium constant expression for the hydrolysis reaction of BrO⁻:

BrO⁻(aq) + H₂O(l) ⇌ HBrO(aq) + OH⁻(aq)

Kb = [HBrO][OH⁻] ÷ [BrO⁻]

we can find Kb from the pKa of HBrO:

Kb = Kw/Ka = 1.0 × 10⁻¹⁴ ÷ 2.3 × 10⁻⁹ = 4.35 × 10⁻⁶

[HBrO] = [OH⁻] = √(Kb[BrO⁻]) = 2.63 × 10⁻³ M

pH = 14 - pOH = 10.58

B) NaHS is a salt of a weak acid (H₂S) and a strong base (NaOH), so it undergoes hydrolysis. Using the equilibrium constant expression for the hydrolysis reaction of HS-:

HS⁻(aq) + H₂O(l) ⇌ H₂S(aq) + OH⁻(aq)

Kb = [H₂S][OH⁻] ÷ [HS⁻]

we can find Kb from the pKa of H₂S:

Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.2 × 10⁻⁷ = 8.33 × 10⁻⁸

[OH⁻] = √(Kb[HS⁻]) = 4.07 × 10⁻⁴ M

pH = 14 - pOH = 13.39

C) NaNO₂ and Ca(NO₂)₂ do not undergo hydrolysis, so we can find the [OH⁻] and pH of the solution by assuming that the total concentration of NO₂⁻ is the sum of the concentrations of NaNO₂ and Ca(NO₂)₂.

[NO₂⁻] = [NaNO₂] + 2[Ca(NO₂)₂] = 0.13 M + 2(0.25 M) = 0.63 M

[OH⁻] = Kw/[H₃O⁺] = Kw ÷ [NO2-] = 1.0 × 10⁻¹⁴ ÷ 0.63 = 1.59 × 10⁻¹³ M

pH = 14 - pOH = 10.80

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true/false. An addition polymer is formed when two monomers containing bonds react in the presence of a(n) initiator.

Answers

True, An addition polymerization is a process in which monomers containing carbon-carbon double bonds (such as ethene) react with an initiator (such as a radical) to form a polymer with a long chain of repeating units.

In an addition polymerization, two monomers containing double or triple bonds react in the presence of an initiator, which helps initiate the reaction.

                                             The initiator can break the double or triple bonds, allowing the monomers to form new single bonds and create a polymer chain.

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____ energy is the minimum amount of energy that colliding molecules must possess in order for a chemical reaction to occur. a) collision b) activation c) bond

Answers

Answer: The answer is b activation have a great day

Explanation:

b) Activation energy is the minimum amount of energy that colliding molecules must possess in order to undergo a chemical reaction.

Without this minimum energy, the chemical reaction cannot proceed, and the molecules will simply bounce off each other. The S.I. unit of activation energy is joules (J) or kilojoules per mole(KJ/mol) .There are two factors on which activation energy depends and the factors are the nature of reactants and the effect of the catalysts. There are two types of catalysts : positive catalyst decreases the activation energy and the negative catalyst increases the activation energy. The correct answer is b) activation energy.

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How many moles of Cl in one mole of the CaCl2?

Answers

One mole of CaCl₂contains 2 moles of chloride ions.

Calcium chloride (CaCl₂) is a salt that consists of one calcium ion (Ca2+) and two chloride ions (Cl-). Therefore, one mole of CaCl₂ contains two moles of chloride ions (2 Cl-).

To calculate the number of moles of Cl- in one mole of CaCl₂, we can use the formula:

moles of Cl- = 2 x moles of CaCl₂

Since one mole of CaCl₂ contains 1 mol of calcium ion and 2 moles of chloride ions, the total number of moles in one mole of CaCl₂is:

1 + 2 = 3 moles

So, the number of moles of Cl- in one mole of CaCl₂ is:

moles of Cl- = 2 x moles of CaCl₂ = 2 x 1 = 2 moles

Therefore, one mole ofCaCl₂ contains 2 moles of chloride ions.

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AlCl4- + H+ = AlCl3 + HCl. (True or False)

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False. The equation AlCl4- + 2H+ = AlCl3 + H2O + HCl is balanced. This is due to the fact that two H+ ions are required to balance the -1 charge of AlCl4- and the charges on either side of the equation.

Due to differences in the amount of atoms of each element on the two sides, the above equation is not balanced. We must add coefficients to the reactants and products in order to balance it. We discover that two H+ ions are required to balance the charge on the AlCl4- ion after doing this. In addition, water (H2O) and HCl are also produced together with AlCl3. AlCl4-+2H+=AlCl3+H2O+HCl is the reaction's balanced equation. False. Because there are not an equal amount of atoms on both sides, the above equation is unbalanced. AlCl4- + 2H+ = AlCl3 + H2O + HCl is the balanced equation because two H+ ions are required to balance the charge on the AlCl4- ion and because the reaction also yields water and HCl.

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over long time scales the solubility-temperature feedback (below), can affect climate. an increase in atmospheric co2 concentrations increases the greenhouse effect and causes temperatures to rise. as ocean temperatures rise, the solubility of co2 decreases. as a result, co2 is released from the oceans to the atmosphere, strengthening the greenhouse effect and causing temperatures to rise further. what type of feedback is this?

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This type of feedback is known as a positive feedback loop. In this case, an initial increase in atmospheric CO2 concentrations leads to a rise in temperatures, which causes a decrease in CO2 solubility in the oceans. As a result, more CO2 is released from the oceans into the atmosphere, further strengthening the greenhouse effect and leading to even higher temperatures. The process amplifies the initial effect, which is characteristic of a positive feedback loop.

1)how many red blood cells could you line up across the grain of sand?


2) How many red blood cells could you line up across the diameter of a penny (0. 02 m)?


Please I need helpppppp

Answers

Penny #2439
Sand grain #244

Hope this helps :)

Brainlist pls

what is the molarity of a solution that contains 32.00g sodium chlorid and 275g of water (i will give brainliest)

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The molarity of the solution that contains 32.00 g of sodium chloride and 275 g of water is 1.994 M.

To calculate the molarity of a solution, we first need to determine the number of moles of the solute (sodium chloride, NaCl) and the volume of the solution (in liters).

Given; Mass of NaCl = 32.00 g

Mass of water = 275 g

Calculate the number of moles of NaCl.

To find the number of moles of NaCl, we can use its molar mass, which is the sum of the atomic masses of sodium (Na) and chlorine (Cl).

Molar mass of NaCl=Atomic mass of Na + Atomic mass of Cl

= 22.99 g/mol (for Na) + 35.45 g/mol (for Cl)

= 58.44 g/mol

Number of moles of NaCl = Mass of NaCl/Molar mass of NaCl

= 32.00 g / 58.44 g/mol

= 0.5477 mol

Now, we can convert the mass of water to volume in liters.

The mass of water needs to be converted to volume in liters in order to calculate the molarity of the solution. This can be done using the density of water, which is approximately 1 g/mL or 1 g/cm³.

Density of water= 1 g/mL or 1 g/cm³

Mass of water = 275 g

Volume of water = Mass of water/Density of water

= 275 g / 1 g/mL

= 275 mL (since 1 mL = 1 cm³)

Converting mL to L;

Volume of water = 275 mL / 1000 mL/L

= 0.275 L

Now, we can Calculate the molarity of the solution.

Molarity (M) is termed as the number of moles of solute per liter of solution.

Molarity (M) = Number of moles of solute/Volume of solution (in liters)

Plugging in the values;

Molarity (M) = 0.5477 mol / 0.275 L

= 1.994 M

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Among the compounds, water, 1-butyne,2-butyne and ethane , which compounds are stronger acids than ammonia 1-butyne and ethane water and ethane 2-butyne and 1-butyne water and 1-butyne

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Among the compounds water, 1-butyne, 2-butyne, and ethane, the compounds that are stronger acids than ammonia are water and 1-butyne. This is because water can act as both an acid and a base, while 1-butyne, with its acidic terminal alkyne hydrogen, can donate a proton more readily than ammonia.

Ethane and 2-butyne are not acidic and cannot act as acids. In water, the hydrogen ion (H⁺) is readily released, making it a stronger acid than ammonia. Similarly, 1-butyne has a terminal alkyne group that can release a proton (H⁺) and is therefore a stronger acid than ammonia.

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