In which compound does hydrogen form an ionic bond as a proton?
a) H2O
b) CH3COOH (acetic acid)
c) CH3CH2OH (ethanol)

Answer:

1,43 M

Explanation:

molair mass NaCl = 58,44 g/mol

You need to calculate how many moles 25 g is:

25 g / 58,44 g/mol = 0,4278 mol

0,4278 mol / 0,3 L = 1,43 M

Answer:

hhjcioz xlioyudiyyxyisrupautwtritu regards Roy

Okay, let's solve this step-by-step:

1) AT = -7.0 K (given)

2) C = 410 J/K (given)

3) Mass of reactant = 0.20 mol (given)

4) To convert temperature change (K) to energy change (J): Energy change = Heat capacity x Temperature change

So in this case: Energy change = 410 J/K x -7.0 K = -2870 J

5) To get enthalpy change per mole (AH), we divide the total energy change by the number of moles of reactant:

-2870 J / 0.20 mol = -14350 J/mol

Therefore, AH = -14350 J/mol.

Let me know if you have any other questions!

To determine the enthalpy change (∆H) for the reactant, first use the relationship q = C × ∆T to calculate the heat exchange in the reaction. Then, convert the resulting value from joules to kilojoules. Finally, divide by the number of moles of the reactant to find ∆H. The enthalpy change for the reaction is -14.35 kj/mol.

This chemistry problem involves the use of thermochemical equations and calorimetry principles. Given in the problem, the change in temperature (∆T) is -7.0 K, the heat capacity (C) of the calorimeter and contents is 410 J/K, and a mole of reactant involved is 0.20 mol. Let's use the equation q = C × ∆T to calculate the heat absorbed or released in a reaction where q is the heat gained or lost, C is the calorimeter’s heat capacity, and ∆T is the change in temperature. Hence, the heat exchange (q) = 410 J/K * -7.0 K = -2870 Joules.

This value is negative because it's giving off heat (exothermic). We see that the value obtained is in joules, but we need the output in Kj. 1 Joule is 1x10^-3 Kj, so -2870 Joules is -2.87 Kj. To find ∆H (Enthalpy change), we divide the heat exchanged by the amount of moles. Therefore, ∆H = q/n = -2.87 Kj / 0.20 moles = -14.35 Kj.mol⁻¹. So the enthalpy change for the reaction is -14.35 Kj.mol⁻¹.

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The solubility of silver phosphate, Ag₃PO₄, in pure water is approximately 2.6 x 10⁻⁶ mol/L.

Solubility is the maximum amount of solute that can be dissolved in a given amount of solvent at a particular temperature and pressure, usually expressed in units of grams per liter (g/L) or moles per liter (mol/L).

The solubility of Ag₃PO₄ can be calculated using the Ksp expression;

[tex]K_{sp}[/tex] = [Ag⁺]³ [PO₄³⁻]

Let x be the solubility of Ag₃PO₄ in mol/L. Then, at equilibrium, the concentrations of Ag⁺ and PO₄³⁻ ions will be x mol/L. Therefore;

[tex]K_{sp}[/tex] = (x)³ (x)³ = x⁶

Solving for x, we get;

x = [tex](Ksp)^{(1/6)}[/tex] = (2.6 x 10⁻¹⁸[tex])^{1/6}[/tex]

≈ 2.6 x 10⁻⁶ mol/L

Therefore, the solubility is 2.6 x 10⁻⁶ mol/L.

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The production of titanium metal through the reduction of titanium(IV) chloride with sodium metal involves a redox reaction. Titanium(IV) chloride is a compound that contains titanium in the +4 oxidation state, while sodium metal is an element with a tendency to lose electrons to form Na+ ions.

The reduction of titanium(IV) chloride involves the transfer of electrons from sodium metal to titanium(IV) ions.
The chemical equation for the reaction is:
TiCl4 + 4Na → Ti + 4NaCl
This equation shows that one molecule of titanium(IV) chloride reacts with four atoms of sodium to produce one atom of titanium and four molecules of sodium chloride.
The reduction of titanium(IV) chloride with sodium metal is an example of a single-displacement reaction. In this type of reaction, one element displaces another element in a compound to form a new compound and a different element. In this case, sodium metal displaces the titanium(IV) ion in titanium(IV) chloride to form titanium metal and sodium chloride.
In summary, the production of titanium metal through the reduction of titanium(IV) chloride with sodium metal involves a redox reaction in which titanium(IV) ions are reduced to titanium metal by sodium metal. The chemical equation for the reaction is TiCl4 + 4Na → Ti + 4NaCl, and the reaction is a single-displacement reaction.

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The major organic product formed when each of the reagents is added to 3,3-dimethylbutene is as follows:

HBr: The major product is 3-bromo-3,3-dimethylbutane.

H₂SO₄ (concentrated sulfuric acid): The major product is 2,3-dimethylbut-2-ene (also known as 2,3-dimethylbutene or diisobutene).

HgSO₄ (mercuric sulfate) followed by H₂O: The major product is 3,3-dimethyl-2-butanol.

Addition of HBr to 3,3-dimethylbutene results in the electrophilic addition of the H-Br bond across the double bond, forming a bromine atom attached to the carbon at the site of the double bond. The major product is 3-bromo-3,3-dimethylbutane.

Concentrated sulfuric acid (H₂SO₄) acts as a dehydrating agent, removing a molecule of water from 3,3-dimethylbutene. This results in the formation of 2,3-dimethylbut-2-ene, where the double bond is shifted to the neighboring carbon atoms.

The addition of mercuric sulfate (HgSO₄) followed by water (H₂O) leads to oxymercuration-demercuration. The mercuric sulfate adds across the double bond to form a mercurinium ion intermediate, which is then reduced by water.

This process results in the formation of 3,3-dimethyl-2-butanol, where the hydroxyl group is added to one of the carbon atoms of the double bond.

It's important to note that these reactions are general representations, and the actual stereochemistry of the products may vary depending on the specific conditions and reaction conditions used.

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There are three unpaired d-electrons in the octahedral high-spin cobalt(iii) complex ion, [CoF6]3- (small ligand field splitting).

In an octahedral high-spin cobalt(iii) complex with small ligand field splitting, the d-electrons occupy the t2g and eg orbitals. As all six ligands are small, they generate a weak ligand field, which results in the energy difference between the t2g and eg orbitals being small, allowing for a high-spin configuration. Cobalt(iii) has five d-electrons, which fill the t2g orbitals first with three electrons, leaving two unpaired electrons in the eg orbitals. Therefore, the complex has three unpaired d-electrons. There are three unpaired d-electrons in [CoF6]3- due to high-spin configuration and weak ligand field splitting, causing a small energy difference between the t2g and eg orbitals.

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(a) Cs is more metallic than Na.
(b) Rb is more metallic than Mg.
(c) N is less metallic than As.


Metallic character refers to the ability of an atom to lose electrons and form positive ions. Elements with more electrons in their outermost shell tend to have higher metallic character.

In pair (a), Cs has a larger atomic radius and more shielding electrons than Na, making it easier for Cs to lose electrons and become a positive ion, indicating higher metallic character.

In pair (b), Rb has a larger atomic radius and more shielding electrons than Mg, making it easier for Rb to lose electrons and become a positive ion, indicating higher metallic character.

In pair (c), As has one more electron than N in the same energy level, leading to a smaller atomic radius and less shielding electrons for As. Therefore, N is less electronegative and has higher metallic character compared to As.


Overall, Cs, Rb, and N have higher metallic character compared to Na, Mg, and As respectively.

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The compounds (ii) CH₃COONa and (iii) Cr(CO) are not organometallic, while compounds (0) Cacz and (iv) B(C₂H₅)₃ are organometallic.

Organometallic compounds contain a direct bond between a carbon atom and a metal atom. In the given compounds, Cacz and B(C₂H₅)₃  fulfill this criterion and are considered organometallic.

Cacz refers to a carbanion complex with a direct carbon-calcium bond. B(C₂H₅)₃, on the other hand, is boron triethyl, where the boron atom is bonded to three ethyl groups. These compounds exhibit unique reactivity and are widely used in organic synthesis and catalysis.

However, CH₃COONa (sodium acetate) and Cr(CO) (chromium carbonyl) do not have direct carbon-metal bonds and, therefore, are not organometallic.

Sodium acetate is a salt composed of sodium ions and acetate ions, while chromium carbonyl consists of chromium and carbon monoxide ligands. These compounds do not possess the characteristic carbon-metal bond found in organometallic compounds.

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The question is asking to determine the substitution of each epoxide carbon in the reactant used. Specifically, it is asking whether the carbons are primary, secondary, tertiary, or quaternary.

The substitution of a carbon is determined by the number of alkyl (or aryl) groups bonded to it.

Therefore, to determine the substitution of each epoxide carbon in the reactant used, we need to know what groups are attached to those carbons. The information provided is not sufficient to answer the question.

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The analytical chemist is performing a titration to determine the concentration of an acetic acid solution.

The chemist will add a known concentration of sodium hydroxide solution to the acetic acid solution until the equivalence point is reached, at which all of the acetic acid will have reacted with the sodium hydroxide.

By measuring the volume of sodium hydroxide solution required to reach the equivalence point, the chemist can calculate the concentration of the acetic acid solution.

In this specific titration, the acetic acid solution is 1.100 M and the sodium hydroxide solution is 0.7600 M, and the volume of the acetic acid solution used in the titration is 229.6 mL.

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If u mean in terms of concentration, it will be 'Dilute'

One that is almost or entirely ionized in water is considered to be a strong acid or alkali. Nitric acid, sulfuric acid, and hydrochloric acid are a few examples. Weak solutions are not ionized in water.

Bases provide a similar problem: a strong base is one that is completely ionized in solution. A weak base is one that is less than 100% ionized in solution.

The fundamental elements that do not completely ionize in water are known as weak bases. Ammonia is one substance that is a weak base. A portion of NH3 that dissolves in water separates into ammonium cation.

An acid that partially separates into its ions in water or an aqueous solution is referred to as a weak acid. On the other hand, in water, a strong acid completely dissociates into its ions. While the conjugate acid of a weak base is also a weak acid, the conjugate base of a weak acid is also a weak solution.

Thus, Weak solutions are not ionized in water.

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In most solutions containing a strong or weak acid, the autoionization of water can be neglected when calculating [H3O+].

The reason for this is related to the relative concentrations of H3O+ ions produced by the acid and the autoionization process of water.

When an acid is added to water, it donates a proton (H+) to the water molecules, which form hydronium ions (H3O+). In the case of a strong acid,

the dissociation is nearly complete, leading to a high concentration of H3O+ ions. For weak acids, the dissociation is partial, but it still contributes a significant amount of H3O+ ions to the solution.

On the other hand, the autoionization of water is a self-ionization process where two water molecules interact, with one donating a proton to the other,

forming a hydronium ion (H3O+) and a hydroxide ion (OH-). However, the equilibrium constant for this process, Kw, is

very small (approximately 1.0 x 10^-14 at 25°C), which means that the concentration of H3O+ ions produced by water's autoionization is extremely low.

Since the concentration of H3O+ ions contributed by the acid is much greater than that produced by the autoionization of water,

it is reasonable to neglect the autoionization of water when calculating [H3O+]. This simplifies the calculations and provides an accurate enough estimation of the hydronium ion concentration in most acid solutions.

In summary, the autoionization of water can be neglected when calculating [H3O+] in solutions containing strong or weak acids due to the significantly higher concentration of H3O+ ions contributed by the acid.

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Answer: things arnent always as they seem

Explanation:

Answer:

500100 benswuer kers olá Marilene

The mass of H₂O is 44,680 g for the given reaction 2H₂ + O₂ → 2H₂O. Therefore, the correct option is option A.

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter. It essentially refers to a body of matter's resistance to changing its speed or location in response to the force that is applied. The change caused by either an applied force is smaller the more mass a body has.

The kilogramme, which is defined approximately equal to 6.6 × 10⁻³⁴ joule second in terms of Planck's constant, is the unit of mass inside the Worldwide System of Units (SI). A joule is equivalent to one kilogramme multiplied by one square metre per second. The mass of H₂O is 44,680 g for the given reaction 2H₂ + O₂ → 2H₂O.

Therefore, the correct option is option A.

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Boiling point = Normal boiling point + ΔT = 373 K + (3.71 g/180.16 g/mol) * (0.512 K/m) / (0.087 kg) = 374.12 K.

To calculate the boiling point of the solution, we'll first find the molality (m) of fructose.

Molality is defined as moles of solute per kilogram of solvent.

1. Calculate moles of fructose: (3.71 g) / (180.16 g/mol) = 0.0206 mol
2. Convert grams of water to kilograms: 87 g = 0.087 kg
3. Calculate molality: (0.0206 mol) / (0.087 kg) = 0.237 m

Next, we'll use the molality and the Kbp (0.512 K/m) to find the change in boiling point (ΔT).

4. Calculate ΔT: (0.237 m) * (0.512 K/m) = 0.121 K

Finally, add ΔT to the normal boiling point (373 K).

5. Boiling point = 373 K + 0.121 K = 374.12 K

The boiling point of the solution is 374.12 K, or approximately 101.0°C.

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The boiling point of the solution would be 100.34°C.

To calculate the boiling point elevation, we can use the formula:

ΔTb = Kbp x molality

where ΔTb is the boiling point elevation, Kbp is the boiling point elevation constant of the solvent, and molality is the concentration of the solution in terms of moles of solute per kilogram of solvent.

First, we need to calculate the molality of the solution. We know the mass of fructose (3.71 g) and the mass of water (87 g). We can convert the mass of fructose to moles by dividing by its molar mass:

moles of fructose = 3.71 g / 180.16 g/mol = 0.0206 mol

Then, we can calculate the molality:

molality = moles of fructose / mass of water in kg

molality = 0.0206 mol / 0.087 kg = 0.237 mol/kg

Now we can calculate the boiling point elevation:

ΔTb = Kbp x molality

ΔTb = 0.512 K/m x 0.237 mol/kg = 0.1216 K

Finally, we can calculate the boiling point of the solution:

Boiling point of solution = normal boiling point of solvent + ΔTb

Boiling point of solution = 373 K + 0.1216 K = 373.12 K

We can convert the boiling point to Celsius by subtracting 273.15:

Boiling point of solution = 373.12 K - 273.15 = 100.34°C

Therefore, the boiling point of the solution is 100.34°C.

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Answer:

Potential energy

Explanation:

To determine the volume of gaseous water collected use the balanced chemical equation and the concept of stoichiometry. The balanced equation for the reaction is 2H2(g) + O2(g) → 2H2O(g).

From the equation, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.

Given that we have 8.6 L of H2 and 5.3 L of O2 at STP (Standard Temperature and Pressure), we can use the ideal gas law to convert the volumes to moles.

At STP, 1 mole of any ideal gas occupies 22.4 L.

Moles of H2 = 8.6 L / 22.4 L/mol = 0.3846 mol

Moles of O2 = 5.3 L / 22.4 L/mol = 0.2366 mol

Based on the balanced equation, we can see that the ratio of moles of H2O to O2 is 2:1. Therefore, the moles of H2O produced will be half the moles of O2.

Moles of H2O = 0.2366 mol / 2 = 0.1183 mol

Now, to convert the moles of H2O to volume, we use the ideal gas law again: Volume of H2O = Moles of H2O × 22.4 L/mol = 0.1183 mol × 22.4 L/mol = 2.647 L. Therefore, the volume of gaseous water collected is approximately 2.647 liters.

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Answer:

-368.62 kJ/mol

Explanation:

[(2 x - 470.11 kJ/mol) ( 0kJ/mol)] - [ 2 x -285.8 kJ/mol) (2x 0 kj/mol)] = -368.62 kJ/mol

The equilibrium concentration of OAc- in the 500 mL of 0.100 M solution of HOAc(aq) with a Ka(HOAc) of 1.79 x 10-5 will be approximately 0.00134 M..

To find the [OAc-] at equilibrium, we need to use the Ka expression and an ICE (Initial, Change, Equilibrium) table. The Ka expression for the dissociation of acetic acid (HOAc) is Ka = [H+][OAc-]/[HOAc]. Initially, [HOAc] = 0.100 M, [H+] = 0, and [OAc-] = 0. During the dissociation, [HOAc] will decrease by x, [H+] will increase by x, and [OAc-] will increase by x.

At equilibrium:
Ka = [H+][OAc-]/[HOAc]
1.79 x 10-5 = (x)(x)/(0.100-x)

We can assume that x is small compared to 0.100, so we can simplify the equation to:
1.79 x 10-5 = (x^2)/0.100

Now, solve for x:
x^2 = 1.79 x 10-5 * 0.100
x^2 = 1.79 x 10-6
x ≈ 0.00134

Since x represents the change in [H+] and [OAc-], the equilibrium concentration of OAc- is approximately 0.00134 M.

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There are approximately 5.16 x10^{21}representative particles in 8.56 x 10^{-3} mol of sodium chloride.

To determine the number of representative particles in a given amount of substance, we need to use Avogadro's number, which is approximately 6.022 x 10^{23} representative particles per mole.

Given that there are 8.56 x10^{-3}mol of sodium chloride, we can calculate the number of representative particles as follows:

Number of representative particles = amount in moles × Avogadro's number

Number of representative particles = 8.56 x10^{-3}mol × 6.022 x10^{23}particles/mol

Number of representative particles ≈ 5.16 x10^{21}particles

Therefore, there are approximately 5.16 x[tex]10^{21}[/tex]representative particles in 8.56 x10^{-3} mol of sodium chloride. This calculation is based on the understanding that one mole of any substance contains Avogadro's number of particles, which is a fundamental concept in chemistry.

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Answer:

The measurement between to points is known as the distance

it is tecnically magnesium,

Answer:

barium is the answer

The balanced chemical equation represents the stoichiometry of the reaction and ensures that the number of atoms of each element on both sides of the equation is equal, which is necessary for the conservation of mass during a chemical reaction.

The balanced chemical equation for the reaction of dichromate ion [(Cr2O7)2-] with hydrogen peroxide (H2O2) and hydrogen ion (H+) to produce chromate ion (CrO5) and water (H2O) can be written as:

2(Cr2O7)2- + 8H2O2 + 12H+ → 4CrO5 + 16H2O

In this reaction, the dichromate ion is reduced to chromate ion and hydrogen peroxide is oxidized to water. The reaction takes place in an acidic medium, which provides hydrogen ions to protonate the peroxide ion and facilitate the reduction of dichromate ion. The balanced equation shows that two molecules of dichromate ion, eight molecules of hydrogen peroxide, and twelve hydrogen ions are required to produce four molecules of chromate ion and sixteen molecules of water.

Overall, the balanced chemical equation represents the stoichiometry of the reaction and ensures that the number of atoms of each element on both sides of the equation is equal, which is necessary for the conservation of mass during a chemical reaction.

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The substance for which the addition of HCl to its solution will have no effect on its solubility is [tex]PbF_2[/tex](s) (option b).

The addition of HCl to a solution can affect the solubility of some slightly soluble substances by reacting with them to form a more soluble compound. The solubility of a substance may increase or decrease depending on the nature of the reaction.

a. AgBr(s) - The addition of HCl to a solution of AgBr will decrease its solubility because AgBr will react with HCl to form a more soluble compound, silver chloride (AgCl).

b. [tex]PbF_2[/tex](s) - The addition of HCl to a solution of [tex]PbF_2[/tex] will have no effect on its solubility because [tex]PbF_2[/tex] is insoluble in water and does not react with HCl.

c. [tex]MgCO_3[/tex](s) - The addition of HCl to a solution of [tex]MgCO_3[/tex] will decrease its solubility because [tex]MgCO_3[/tex] will react with HCl to form a more soluble compound, magnesium chloride ([tex]MgCl_2[/tex]), and carbon dioxide ([tex]CO_2[/tex]).

d. [tex]Cu(OH)_2[/tex](s) - The addition of HCl to a solution of [tex]Cu(OH)_2[/tex] will decrease its solubility because [tex]Cu(OH)_2[/tex] will react with HCl to form a more soluble compound, copper chloride ([tex]CuCl_2[/tex]), and water ([tex]H_2O[/tex]).

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The answer was 25

Do according to bodmas rule

Answer:

Seafloor Ages

They found that the youngest rocks on the seafloor were at the mid-ocean ridges. The rocks get older with distance from the ridge crest.

hope this helped

Answer:

Rate = k [X]⁻¹ [Z]²

Explanation:

[X] [Y] [Z] initial rate M M M M · s −1

Exp 1 0.30 0.20 0.35      0.210

Exp 2 0.60 0.10 0.70      0.420

Exp 3 0.60 0.20 0.70     0.420

Exp 4 0.60 0.40 0.35      0.105

In Experiment 2 and 3 where the concentrations of Y and Z were constant, doubling the concentration of Y had no effect on the rate of the reaction. This means, that the rate of the reaction is zero order with respect to Y.

In experiment 3 and 4, dividing the concentration of Z by 2, causes the rate of the reaction to decrease by 4. This means the rate of the reaction is second order with respect to Z.

In experiment 1 and 4, doubling the concentration of X, causes the rate of the reaction to decrease by half. This means that X has an order of -1 with respect to the rate of the reaction.

The rate expression is given as;

Rate = k [X]⁻¹[Y]⁰[Z]²

Rate = k [X]⁻¹ [Z]²

Answer:

El volumen final del sistema es 1.2L

Explanation:

La ley de Charles establece que el incremento de la temperatura de un gas produce un incremento en el volumen directamente proporcional cuando la presión permanece constante. La ecuación es:

V₁/T₁ = V₂/T₂

Donde V es volumen y T temperatura absoluta de un gas en 1, el estado inicial y 2, su estado final.

Reemplazando:

V₁ = 1.3L

T₁ = 50°C + 273.15K = 323.15K

V₂ = Incógnita

T₂ = 300K

1.3L/323.15K = V₂/300K

1.2L = V₂