Which of the following affects the amount by which the freezing point of liquid is lowered by the addition of a solute? More than one answer may be correct.
1.The volume of the solvent.
2.The amount of the compound dissolved.
3. The identity of the chemical species being dissolved.
4. The value of the freezing point for the pure solvent.
5. How soluble the solute is in the solvent.

Based on the Lewis structure of NO2- and VSEPR theory, the central N in NO2- has a trigonal planar electron geometry due to the three electron pairs surrounding it.

The two oxygen atoms are located in the equatorial positions, while the lone pair of electrons occupies the axial position. The lone pair-lone pair repulsion is stronger than the lone pair-bond pair or bond pair-bond pair repulsions. This leads to a compression of the bond angles. Therefore, the estimated bond angle about the central N in NO2- is slightly less than 120°. The bond angle can be affected by various factors such as the electronegativity of the atoms involved and the presence of lone pairs. In the case of NO2-, the presence of a lone pair on the central N leads to a deviation from the ideal 120° bond angle. This is due to the repulsion between the lone pair and the oxygen atoms, causing a decrease in the bond angle. Therefore, the statement that most accurately estimates the bond angle about the central N in NO2- is "it is slightly less than 120°".

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The value of q is 66.4 kJ.

The given chemical equation is:

[tex]\mathrm{N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)} \qquad \Delta H = +66.4 \text{ kJ/mol}[/tex] = 66.4 kJ/mol

Since the enthalpy change (ΔH) is positive, it means that heat is absorbed during the reaction, indicating that this reaction is endothermic.

The amount of heat absorbed (q) can be calculated using the following formula:

q = nΔH

where n is the number of moles of reactant that undergo the reaction.

For this reaction, 1 mole of [tex]\mathrm{N_2}[/tex]reacts, so n = 1.

Therefore, the amount of heat absorbed is:

q = (1 mol) × (66.4 kJ/mol) = 66.4 kJ

So, the value of q is 66.4 kJ.

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To solve this problem, we need to use the solubility product expression for calcium iodate:

Ca(IO3)2(s) ⇌ Ca2+(aq) + 2IO3-(aq)

The solubility product constant, Ksp, is given by:

Ksp = [Ca2+][IO3-]^2

To calculate Ksp for a saturated solution of Ca(IO3)2 in 0.0100 M KIO3, we need to determine the concentrations of Ca2+ and IO3- ions at equilibrium. We assume that Ca(IO3)2 is completely dissociated and that the activity coefficients of the ions are equal to 1. Therefore, we can write:

Ca(IO3)2(s) ⇌ Ca2+(aq) + 2IO3-(aq)

Let's assume that the molar solubility of Ca(IO3)2 in this solution is s. Then, the concentrations of Ca2+ and IO3- ions at equilibrium are [Ca2+] = s and [IO3-] = 2s. Substituting these values into the Ksp expression, we get:

Ksp = [Ca2+][IO3-]^2 = s(2s)^2 = 4s^3

Now, we need to determine the value of s by using the given concentration of KIO3. Since KIO3 is a strong electrolyte, it completely dissociates in water to give K+ and IO3- ions. Therefore, the concentration of IO3- ions in the solution is equal to the concentration of KIO3, which is 0.0100 M. Using this value, we can write the following equilibrium equation for Ca(IO3)2:

Ca(IO3)2(s) ⇌ Ca2+(aq) + 2IO3-(aq)

ICE table:

I: Ca(IO3)2(s) -- -- --

C: -s +s +2s

E: s 2s --

Substituting the equilibrium concentrations into the Ksp expression, we get:

Ksp = 4s^3 = 5.12 × 10^-10

Therefore, the solubility product constant for Ca(IO3)2 in 0.0100 M KIO3 is 5.12 × 10^-10.

To calculate the molar solubility of Ca(IO3)2 in 0.020 M Ca(NO3)2, we can assume that Ca(NO3)2 is completely dissociated and that the activity coefficients of the ions are equal to 1. Therefore, we can write:

Ca(IO3)2(s) ⇌ Ca2+(aq) + 2IO3-(aq)

Let's assume that the molar solubility of Ca(IO3)2 in this solution is s. Then, the concentrations of Ca2+ and IO3- ions at equilibrium are [Ca2+] = 0.020 M and [IO3-] = 2s. Substituting these values into the Ksp expression, we get:

Ksp = [Ca2+][IO3-]^2 = (0.020)(2s)^2 = 0.08s^2

Since Ksp = 2.0 × 10^-4, we can write:

0.08s^2 = 2.0 × 10^-4

Solving for s, we get:

s = 0.0158 M

Therefore, the molar solubility of Ca(IO3)2 in 0.020 M Ca(NO3)2 is 0.0158 M.

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The given reactions consist of a variety of elementary and global reactions. Among them, reactions A and F are bimolecular elementary reactions involving the collision of two molecules, reactions C and D are termolecular and unimolecular elementary reactions, respectively.

Reactions B and E are global reactions that occur through a series of elementary steps. Each reaction demonstrates distinct characteristics in terms of the number of molecules involved and the reaction mechanism.

A. Elementary, bimolecular. This is a bimolecular reaction because it involves the collision of two molecules, CO and OH, in a single elementary step.

B. Global. This reaction does not occur in a single step, but rather involves a series of elementary steps that together constitute the global reaction.

C. Elementary, termolecular. This is a termolecular reaction because it involves the collision of three molecules, H, OH, and O₂, in a single elementary step.

D. Elementary, unimolecular. This is a unimolecular reaction because it involves the rearrangement of a single molecule, HOCO, in a single elementary step.

E. Global. This reaction does not occur in a single step, but rather involves a series of elementary steps that together constitute the global reaction.

F. Elementary, bimolecular. This is a bimolecular reaction because it involves the collision of two molecules, OH and HM, in a single elementary step. The resulting product HOM can then undergo further reactions, but these would not be included in the classification of this initial step.

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Classify the following reactions as being either global or elementary. For those identified as elementary, further classify them as unimolecular, bimolecular, or termolecular.

Give reasons for your classification. A. CO + OH CO2+H. B. 2Co +o2->2CO, C. H, +OH+H+O2 D. HOCO H-CO2 E. CH, 20CO, 2H,O F. OH+HM-H,OM.

Water molecules surround solute molecules. Option B

The principle of "like dissolves like" states that non-polar chemicals are soluble in non-polar solvents and polar compounds are soluble in polar solvents.

Water can only dissolve other polar solutes since it is a polar solvent; it cannot dissolve non-polar solutes.

The polar molecules that surround the polar solutes when they are dissolved in water are separated when they do so because of the high dielectric constant of water.

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Missing parts;

Which statement accurately describes part of the dissolving process of a polar solute in water?

Solute molecules repel water molecules.

Water molecules surround solute molecules.

Water molecules are strongly attracted to nonpolar substances.

Solute molecules remain concentrated in one part of the solvent.

If we 800,000 atoms of the radioactive substance. After the 3 half-lives have past, the number of the atoms remain are 100000 atoms.

The initial amount of the radioactive substance = 800,000 atoms

The Number of half lives = 3  half - lives

The amount remaining of the radioactive element after the "n" half lives :

N = [tex]No[/tex][tex](1/2)^{n}[/tex]

Where,

No = the initial amount

n = Number of the half lives

N = 800,000( 1/2 )³

N = 100000 atoms

N is the number of the reaming atoms = 100000 atoms.

Therefore, the number of remaining atoms after the 3 half - lives is 100000 atoms.

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The products of this reaction are Tin (Sn) and Water (2H_{2}O). The reactants, Tin(IV) oxide (SnO_{2}) and Hydrogen gas (2H_{2}), undergo a redox reaction to produce these products.

In the given chemical reaction, SnO_{2} + 2H_{2]} → Sn + 2H_{2}O., the products are Sn (Tin) and 2H_{2}O (Water). Let's analyze the reaction step-by-step.
1. The reactants are SnO_{2} (Tin(IV) oxide) and 2 H_{2}(Hydrogen gas).

Tin is commonly used in the manufacturing of metal alloys, such as bronze and pewter, and can also be found in the production of tinplate for food packaging.
2. The reaction involves the reduction of SnO_{2}  and the oxidation of H_{2}
3. SnO2 loses oxygen and is reduced to Sn (Tin). Meanwhile, H2 gains oxygen and is oxidized to H2O (Water).
4. The balanced chemical equation is: SnO_{2} + 2H_{2]} → Sn + 2H_{2}O.
In summary, the products of this reaction are Tin (Sn) and Water (2H_{2}O). The reactants, Tin(IV) oxide (SnO2) and Hydrogen gas (2 H_{2}), undergo a redox reaction to produce these products.

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In order to identify the carbon atoms that most likely came from acetylene in the synthesis process, we need to understand a bit more about the properties of acetylene and its reactions.

Acetylene is a hydrocarbon with the chemical formula C2H2, meaning it is composed of two carbon atoms and two hydrogen atoms. When acetylene undergoes a reaction known as hydroboration-oxidation, it can be transformed into a variety of organic compounds, including alcohols and aldehydes.

In order to determine which carbon atoms in the starting material most likely came from acetylene, we would need to know more about the specific synthesis process being used. However, it's possible that the carbon atoms that came from acetylene would be the ones directly attached to the carbon-carbon triple bond in the molecule.

Without more information, it's difficult to say for sure which specific carbon atoms in the starting material came from acetylene. However, by understanding the properties and reactions of acetylene, we can make some educated guesses about which atoms might be involved in the synthesis process.

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"The information that should be included on a label for a peroxide-forming chemical is - Date received, date opened, and date to discard".

The information that should be included on a label for a peroxide-forming chemical is the date received, date opened, and date to discard. This information is essential to ensure the safe handling and storage of the chemical.

It is important to keep track of when the chemical was received, as well as when it was opened, in order to determine its shelf life and prevent any potential hazards.

Additionally, including the date to discard on the label ensures that the chemical is not used beyond its safe and effective period.

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The net ionic equations are as follows:

a. [tex]OH^-(aq) + H^+(aq)[/tex] → [tex]H_2O(l)[/tex]

b. [tex]Ca(s) + 2H^+(aq)[/tex] → [tex]Ca^{2+}(aq) + H_2(g)[/tex]

c. [tex]Ca + 2HCl + 2H_2O[/tex] → [tex]Ca(OH)_2 + H_2[/tex]

a. The neutralization of NaOH with HCl can be represented by the following balanced chemical equation:

NaOH + HCl → NaCl + [tex]H_2O[/tex]

To write the net ionic equation, we need to show only the species that are involved in the chemical reaction. The complete ionic equation is:

[tex]Na^+(aq) + OH^-(aq) + H^+(aq) + Cl^-(aq)[/tex] → [tex]Na^+(aq) + Cl^-(aq) + H_2O(l)[/tex]

The net ionic equation is:

[tex]OH^-(aq) + H^+(aq)[/tex] → [tex]H_2O(l)[/tex]

b. The dissolution of calcium in acid can be represented by the following balanced chemical equation:

Ca + 2HCl → [tex]CaCl_2 + H_2[/tex]

To write the net ionic equation, we need to show only the species that are involved in the chemical reaction. The complete ionic equation is:

[tex]Ca(s) + 2H^+(aq) + 2Cl^-(aq)[/tex] → [tex]Ca^{2+}(aq) + 2Cl^-(aq) + H_2(g)[/tex]

The net ionic equation is:

[tex]Ca(s) + 2H^+(aq)[/tex] → [tex]Ca^{2+}(aq) + H_2(g)[/tex]

c. The dissolution of calcium in water can be represented by the following balanced chemical equation:

[tex]Ca + 2H_2O[/tex] → [tex]Ca(OH)_2 + H_2[/tex]

To combine equations a and b so they result in equation c, we can first write the net ionic equation for the neutralization of NaOH with HCl:

[tex]OH^-(aq) + H^+(aq)[/tex] → [tex]H_2O(l)[/tex]

Then, we can substitute [tex]OH^-[/tex] with Ca and [tex]H^+[/tex] with 2HCl from the net ionic equation for the dissolution of calcium in acid:

[tex]Ca(s) + 2H^+(aq)[/tex] → [tex]Ca^{2+}(aq) + H_2(g)[/tex]

The resulting equation is:

[tex]Ca + 2HCl + 2H_2O[/tex] → [tex]Ca(OH)_2 + H_2[/tex]

which represents the dissolution of calcium in water.

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Calcium oxide, CaO, is formed when calcium reacts with oxygen, while calcium sulfide, CaS, is formed when calcium reacts with sulfur.

In the beaker where calcium reacts with oxygen, the compound calcium oxide, CaO, is formed. This is because calcium has a valency of +2 and oxygen has a valency of -2. Therefore, they combine in a 1:1 ratio to form a neutral compound. The chemical formula for calcium oxide is CaO.

Now, in the other beaker where calcium reacts with sulfur, we need to look at the valency of sulfur. Sulfur has a valency of -2, which means it requires two electrons to form a stable compound. Calcium, on the other hand, has a valency of +2. Therefore, in order for the compound to be neutral, we need two calcium atoms to combine with one sulfur atom. This gives us the chemical formula CaS, which is calcium sulfide.

In summary, when equal amounts of calcium react with oxygen and sulfur, they form different compounds. Calcium oxide, CaO, is formed when calcium reacts with oxygen, while calcium sulfide, CaS, is formed when calcium reacts with sulfur. The chemical formula of the compound formed depends on the valencies of the elements involved in the reaction.

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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 following that indicates the least pressure is C) 5.5 mmHg.

To compare these pressure values, let's first convert them all to a common unit, such as Pascal (Pa).

A) 1 atm = 101,325 Pa
B) 777 torr = 101,325 * (777/760) = 103,373.6 Pa (approx.)
C) 5.5 mmHg = 101,325 * (5.5/760) = 749.6 Pa (approx.)
D) 100 kPa = 100,000 Pa
E) 12 psi = 12 * 6,894.76 = 82,737.12 Pa (approx.)

Now, we can compare these values to find the least pressure:
A) 101,325 Pa
B) 103,373.6 Pa
C) 749.6 Pa
D) 100,000 Pa
E) 82,737.12 Pa

So, the least pressure is indicated by option C) 5.5 mmHg, which is approximately 749.6 Pa.

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Answer: clockwise north

Explanation:

A high pressure system has higher pressure at its center than the areas around it. Winds blow away from high pressure. Swirling in the opposite direction from a low pressure system, the winds of a high pressure system rotate clockwise north of the equator and counterclockwise south of the equator. This is called anticyclonic flow. Air from higher in the atmosphere sinks down to fill the space left as air is blown outward. On a weather map, you may notice a blue H, denoting the location of a high pressure system.

Students would stand to model the molecules in a solid against the classroom wall. Therefore, the correct option is option C.

According to the context, the term could refer to any ions that meet this requirement. A molecule is a collection of at least two atoms linked together by the attractive forces referred to as chemical bonds.

When speaking of polyatomic ions. Students would stand to model the molecules in a solid against the classroom wall.

Therefore, the correct option is option C.

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Your question is incomplete but most probably your full question was,  

Mrs. Jones' class was learning about how molecules are arranged in solids, liquids, and gases. To model molecules in liquids, she had the students stand one meter apart.

Which ,begin emphasis,best,end emphasis, describes how the students would stand to model the molecules in a solid?

Answer options with 5 options

A.

as far apart as possible

B.

in rows one meter apart

C.

as close as possible

D.

against the classroom wall

E.

five meters apart

At STP, there are 0.0446 moles of oxygen in 1000 mL

n = (V / Vm)

where n is the number of moles, V is the volume of the gas in liters, and Vm is the molar volume of the gas at STP (22.4 L/mol).

Converting the given volume of 1000 mL to liters:

V = 1000 mL = 1 L

Substituting the values into the formula:

n = (1 L) / (22.4 L/mol) = 0.0446 mol

STP stands for Standard Temperature and Pressure, which are standardized conditions used in chemistry to compare and measure the properties of gases. The standard temperature is defined as 0°C (273.15 K), while the standard pressure is 1 atmosphere (atm) or 101.325 kilopascals (kPa).

At STP, one mole of any gas occupies a volume of 22.4 liters. This value is known as the molar volume and is useful in various chemical calculations. Additionally, the ideal gas law, which relates the pressure, volume, and temperature of a gas, is often used at STP conditions to simplify calculations. STP is essential for defining and comparing gas properties, as different gases behave differently under different conditions.

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27 moles of reactants chemically change into 34 moles of products. Option A

One or more chemicals, referred to as reactants, are transformed into one or more new substances, referred to as products, during a chemical reaction. Chemical bonds between atoms, ions, or molecules are broken and created during chemical reactions.

When  we look at the reaction, we can see that the statement that is true about the total number of reactants and the total number of products in the reaction 27 moles of reactants chemically change into 34 moles of products.

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The melting point of tin is 231.9 degrees C, Iron does not burn when exposed to flame, The density of solid silver is 10.49 grams/centimeter³ and Pure water is odorless are Physical property.

Although a physical change happens when substance alters forms without changing its chemical identity, a chemical change is the result of a chemical reaction. If enough energy is provided, many physical changes can be reversed. A different chemical reaction is the only way to undo a chemical change.

The melting point of tin is 231.9 degrees C, Iron does not burn when exposed to flame, The density of solid silver is 10.49 grams/centimeter³ and Pure water is odorless are Physical property. Sodium can react with water to form sodium hydroxide and hydrogen is Chemical property.

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The sample of gas that occupies 24m³ of volume at 175°K will have a volume of 30.5m³ at 400.0 K.

We can use the combined gas law to solve this problem, which relates the pressure, volume, and temperature of a gas in a closed system. The combined gas law is expressed as,

(P₁ x V₁)/T₁ = (P₂ x V₂)/T₂,

We can assume that the pressure of the gas is constant, as the problem does not provide information about any changes in pressure. Therefore, we can simplify the combined gas law to,

(V₁/T₁) = (V₂/T₂)

The starting volume, V₁, is 24 m3 and the initial temperature, T₁, is 175°C.

T(K) = T(°C) + 273.15

So, the initial temperature in kelvin is,

T₁ = 175°C + 273.15 = 448.15 K

We are asked to find the final volume, V₂, when the temperature is 400.0 K. Therefore, we can rearrange the equation to solve for V₂,

V₂ = (V₁/T₁) × T₂

Substituting the values we know,

V₂ = (24 m³ / 448.15 K) × 400.0 K

V₂ = 1.27 × 24 m³

V₂ = 30.5 m³

Therefore, the gas would occupy 30.5 m³ at 400.0 K.

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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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For a concentration of PBDEs in herring gull eggs, doubling time for PBDEs in this source is 0.778 yrs. The concentration in 2010 is equals to 44.16 ppm.

We have a concentration of PBDEs in herring gull eggs from the Great Lakes was about 1100 ppb in 1990.

Concentration in 2000 = 7000 ppm

= 7000 × 10³ ppb

We have to doubling time for PBDEs in this source and concentration in 2010.

Time = 10 yr

The rate constant is [tex]k = \frac{ 2.303 }{t} \frac{ log[a]}{log[a - x]}[/tex],

Plugging the values, [tex]k = \frac{ 2.303 }{10} log( \frac{ 7000× 10³ }{1100})[/tex]

= 0.875/yr

Doubling time = 0.69/k

= 0.778 yrs.

Now, the we determine the concentration in 2010. Let the concentration be equal x.

[tex]0.778 = \frac{ 2.303 }{10} log(\frac{ x}{7000 })[/tex]

=> [tex]\frac{ x}{7000 \: ppm} = 10^{3.8}[/tex]

=> x = 44.16

Hence, required value is 44.16 ppm.

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The pH of the solution is 5.

A solution's pH, which measures how acidic it is, is determined by taking the negative logarithm of the concentration of hydrogen ions (H+) at base 10. The concentration of hydronium ions, or [H3O+], is utilized instead because hydrogen ion concentrations are typically relatively low.

The pH can be calculated using the following formula:

pH = -log[H3O+]

Substituting the given value:

pH = -log(1.00 × 10^-5) = -(-5) = 5

Therefore, the pH of the solution is 5.

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In solution 4 of part b, both the hydronium concentration and hydroxide concentration are 1 M due to the neutralization reaction between HCl and NaOH. The initial concentrations of H₃O⁺ and OH⁻ were 0.1 M, but they reacted to form water and resulted in equal final concentrations of 1 M for both ions.

To calculate the hydronium and hydroxide concentrations for solution 4 in part b, we need to first understand the equation for the reaction that occurred.

The equation given is: HCl + NaOH -> NaCl + H₂O

This tells us that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water.

Based on this equation, we know that the initial concentration of hydroxide ions (OH-) in solution 4 is equal to the initial concentration of sodium hydroxide (NaOH), which was given as 0.1 M.

Next, we need to determine the concentration of hydronium ions (H₃O⁺). Since HCl is a strong acid, it completely dissociates in water to produce H₃O⁺ and Cl⁻ ions. Therefore, we can assume that the initial concentration of H₃O⁺ is equal to the initial concentration of HCl, which was also given as 0.1 M.

Now, let's consider what happens when the HCl and NaOH are mixed together. They react to form NaCl and water, which means that the concentrations of H₃O⁺ and OH⁻ will change.

From the equation, we can see that the reaction consumes one mole of HCl and one mole of NaOH. This means that the final concentration of HCl and NaOH will both be zero.

To determine the final concentration of OH-, we need to use the fact that the reaction produces one mole of water for every mole of NaOH that reacts. Therefore, the final concentration of OH⁻ will be equal to the initial concentration of NaOH (0.1 M) divided by the volume of the solution.

If we assume that the volume of the solution is 100 mL (as stated in the question), then the final concentration of OH- will be:

[OH⁻] = 0.1 M / 0.1 L = 1 M

Finally, we can use the fact that the concentration of H₃O⁺ and OH⁻ must be equal in a neutral solution to determine the final concentration of H₃O⁺.

Since the final concentration of OH⁻ is 1 M, we know that the final concentration of H₃O⁺ must also be 1 M.

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The answer C less active (or more noble) than hydrogen. This is because copper has a lower tendency to lose electrons and form cations compared to hydrogen. In other words, copper is a relatively stable element that is not as easily oxidized as hydrogen.

They can be seen in the electrochemical series, which ranks elements according to their tendency to undergo oxidation or reduction reactions. Hydrogen is located higher up on the series, indicating that it is more reactive and has a greater tendency to lose electrons and form cations. On the other hand, copper is located lower down on the series, indicating that it is less reactive and has a lower tendency to undergo oxidation. It is worth noting that copper can still undergo oxidation reactions under certain conditions. For example, when exposed to air and moisture, copper can slowly react to form copper oxide. Additionally, copper can be used as an anode in certain electrochemical cells, indicating that it is more anodic than some other metals. However, in general, copper is considered to be a relatively stable and unreactive element, particularly compared to hydrogen.

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Answer:

true

Explanation:

The assertion stating that the term vapour is used to represent the gaseous state of a substance which is otherwise liquid at room temperature is true, but the reason is not a correct explanation.

Vapour is commonly used to describe the gaseous state of a substance that is present at a liquid or solid state at room temperature and atmospheric pressure. When a liquid or solid substance evaporates, vapours are formed that can be inhaled.

Even though the ammonia gas can be referred to as ammonia vapour,  the reason does not explain why vapour is used to describe the gaseous state of substances that are typically present in liquid or solid state at room temperature.

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The vapor pressure of water at 81°C is 3.61 atm.

To calculate the vapor pressure of water at 81°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its enthalpy of vaporization and temperature.

The equation is:

ln(P2/P1) = (ΔHvap/R) x (1/T1 - 1/T2)

where:P1 = vapor pressure at temperature T1

P2 = vapor pressure at temperature T2

ΔHvap = heat of vaporization

R = gas constant (8.314 J/mol K)

We can use the equation to solve for P2, given that we know P1, ΔHvap, T1, and T2. We will assume that the heat of vaporization of water is constant over this temperature range.

First, we need to convert the temperatures to Kelvin:

T1 = 25°C + 273.15 = 298.15 K

T2 = 81°C + 273.15 = 354.15 K

Next, we can plug in the values we know:

ln(P2/3.13x10-²) = (4.39x10⁴ J/mol / 8.314 J/mol K) x (1/298.15 K - 1/354.15 K)

Simplifying the right-hand side of the equation:

ln(P2/3.13x10-² atm) = 19.7

Finally, solving for P2:

P2/3.13x10-² atm = e¹⁹·⁷

P2 = 3.61 atm

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To determine if ethylene glycol n-butyl ether-water is one or two-phase system, we need to compare the tie-line length and the lever rule calculation. A single stable liquid phase with a mass fraction of 30% ethylene glycol n-butyl ether and 70% water is present at equilibrium.

Assuming that the system follows Raoult's law and that the activity coefficients are equal to one, we can use the following equation to calculate the tie-line length:

Tie-line length = (x2 - x1) / (y1 - x1)

where x1 and x2 are the mole fractions of water in the liquid phase and vapor phase, respectively, at the two-phase boundary, and y1 is the mole fraction of water in the liquid phase at the point where the tie-line intersects the tie-line axis.

Using the given data, we can calculate the mole fraction of water in the liquid phase at the two-phase boundary as follows:

x1 = 0.3 (given)

x2 = 1 - x1 = 0.7

y1 = x1 = 0.3 (assuming the vapor phase is pure water)

Substituting these values into the tie-line equation, we get:

Tie-line length = (0.7 - 0.3) / (0.3 - 0.3) = 0

Since the tie-line length is zero, the system is a single-phase liquid, and there is no two-phase region. Therefore, the answer is that the system is one stable liquid phase at equilibrium, with a mass fraction of water of 70% and a mass fraction of ethylene glycol n-butyl ether of 30%.

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