Answering questions by analyzing specific observations and using them to come to a more general understanding is known as __________



inductive reasoning

empiricism

skepticism

deductive reasoning

Answer:

The answer is "solubility is how much solvent will dissolve in solute."

Explanation:

Final answer:

Solubility is the amount of solute that can dissolve in a given quantity of solvent at a certain temperature and pressure, resulting in a saturated solution.

Explanation:

Solubility is defined as the maximum amount of a substance (solute) that can dissolve in a given amount of solvent at a specified temperature and pressure, leading to a saturated solution. It is often expressed in grams of solute dissolved in 100 grams of solvent.

The concept of solubility is essential in chemistry as it helps predict the behavior of substances in various environments and is a key factor in reactions like precipitation.

For instance, when a solution's concentration exceeds the substance's solubility, the excess will precipitate out of the solution. Meanwhile, substances with low solubility tend to precipitate readily under most conditions.

The phrase 'like dissolves like' is a useful heuristic that indicates that polar solvents will typically dissolve polar solutes, while nonpolar solvents will dissolve nonpolar solutes. This is key when trying to determine if a solute will be soluble in a given solvent.

Answer:

1.5 mol/L.

Explanation:

∵ M = no. of moles of solute/V of the solution.

∴ M = (3.0 mol)/(2.0 L) = 1.5 mol/L.

The molarity of the given silver chloride solution is 1.5 M, calculated by dividing the number of moles of solute by the liters of solution. To find the concentration of chlorine, one must calculate the moles from the mass of each chloride compound, convert to grams, and divide by the solution volume in liters.

The molarity of a solution is defined as the number of moles of solute per liter of solution. For the first question provided with 3.0 moles of silver chloride (AgCl) dissolved to make a 2.0-liter solution, the molarity (M) can be calculated using the formula M = moles of solute / liters of solution. Therefore, the molarity is 1.5 M, because M = 3.0 moles / 2.0 liters.

For the second question, to calculate the concentration of chlorine in grams per liter, first determine the moles of chlorine in each compound, then convert the total moles to grams and divide by the volume of solution in liters. We can find the molar mass of CaCl₂, MgCl₂, and MnO₂, calculate how many moles of Cl are in each using their respective weights, and sum them up to find the total chlorine content. The grams of chlorine per liter can then be calculated by converting the total grams of chlorine to the volume of the solution.

Answer:

1) Add more sulfate ions.

2) 1 M K₂SO₄.

Explanation:

1) Which of the following actions would shift this reaction toward solid barium sulfate?

Add more sulfate ions will increase the concentration of the products side, so the reaction will be shifted towards the reactants side to relieve the stress and attain the equilibrium again.

So, the right choice is: Add more sulfate ions.

2) In which of the following would barium sulfate be least soluble?  

The answer is 1 M K₂SO₄, it has a common ion effect of SO₄⁻ ions that increase the concentration of the products side, so the reaction will be shifted towards the reactants side to relieve the stress and attain the equilibrium again.

So, the solubility will decrease.

1. b) Adding more sulfate ions shifts the reaction towards solid barium sulfate.

2. Barium sulfate is least soluble in a c) 1 M K₂SO₄ solution.

1. Le Châtelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to relieve that stress.

For the reaction:

To shift this reaction towards solid barium sulfate, one must consider the factors that affect the equilibrium. Using Le Châtelier's principle:

Hence, adding more sulfate ions would shift the reaction towards the solid barium sulfate. The correct option is b).

2. Solubility Consideration

Barium sulfate would be least soluble in a 1 M K₂SO₄ solution. This is because K₂SO₄ provides additional SO₄²⁻ ions, which increases the common ion effect, thus decreasing the solubility of BaSO₄. The correct option is c).

Question: Le Châtelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to relieve that stress. This is true for all types of equilibrium, including the dissolution of salts that are only sparingly soluble.

Here is the chemical reaction that represents solid barium sulfate dissolving in water:

BaSO₄(s) ⇌ Ba₂+(aq) + SO₄²⁻(aq), with Ksp = 1.1×10⁻¹⁰

1. Which of the following actions would shift this reaction toward solid barium sulfate?

a) Add more barium sulfate.

b) Add more sulfate ions.

c) Remove sulfate ions.

d) Remove barium ions.

2. In which of the following would barium sulfate be least soluble?

a) pure water

b) 1 M NaNO3

c) 1 M K2SO4

Depends on the gas and grams of solute. If you were able to give me a gas i would be able to help you but i cannot without the gas or chemical makeup.

No. When water first begins to cool down, it contracts. However, as it gets colder and eventually freezes, it begins to expand.

You can test this by freezing water in a water bottle: when you take it out of the freezer, the cap might have popped off or cracks may have formed in the sides of the bottle.

Answer: Water expands when frozen, not contracts.

Older people had no money to buy stuff

Answer:

The recent graduate invests taking risks and betting on profitability.

The retired man protects the estate by sacrificing the gain.

Explanation:

Answer: hydrogen (H⁺) or hydronium (H₃O⁺) and hydroxide (OH⁻)

Explanation:

1) Arrhenius definitions

The chemist Svante Arrhenius, winner of the Noble Prize in chemistry by 1903, defined acids and bases based on the ability of such substances to give ions in water solutons.

2) Acids

Arrhenius defined an acid as a substance that dissociates in water to form hydrogen ions (H⁺), also called protons. This process is called protonation (formation of protons). Due to its small size and high activity, H⁺ does not exist in that form, but it forms H₃O⁺ ions in aqueous solution.

3) Bases

Arrhenius defined a base as a substance that dissociates in water to form hydroxide (OH⁻) ions.

4) Examples

That was the first definitions of acids and bases and is restricted to water solutions and does not include many compounds (like NH₃) which nowadays are identified as acids or bases.

Some examples of Arrhenius acids and bases are:

According to the Arrhenius definition, acids release hydrogen ions (H⁺) and bases release hydroxide ions (OH⁻) in aqueous solutions. These two ions are central to identifying Arrhenius acids and bases.

According to the Arrhenius definition, acids and bases are classified based on the ions they release in an aqueous solution. An Arrhenius acid is a compound that produces hydrogen ions (H⁺), while an Arrhenius base is a compound that produces hydroxide ions (OH⁻). For example, HCl dissociates in water to form H⁺ ions and Cl⁻ ions, categorizing it as an Arrhenius acid. Conversely, NaOH dissociates to form Na⁺ ions and OH- ions, making it an Arrhenius base.

The central ions in these definitions are thus hydrogen ions (H⁺) for acids and hydroxide ions (OH⁻) for bases.

The answer is that you have to write this in your own words.

To help you on how to get started, here's what you need to do:

Just say what planet you want to go and why you want to go there. Then, say what you wish to learn on that planet.

Template: My space mission would be going to planet _______. I would like to go there because ___________________. I hope to learn that ________________.

Our space mission involves studying Mars' geological features and potential for life. This will enhance our understanding of the possibility of life beyond our solar system and specifically detail the environment on Mars. The mission might also reveal evidence of past or present life.

Our space mission will aim to explore the unique geological features and potential habitability of Mars. Mars is a planet of particular interest due to evidence suggesting its past was warmer and wetter, hinting at the possibility of life beyond our solar system.

We wish to explore this aspect of the solar system because it could fundamentally alter our understanding of life's existence in the universe. The mission will involve the use of hi-tech equipment to identify potential biomarkers, signs of past or present life.

Through this mission, we hope to learn more about the environment on Mars, detail its capacity for sustaining life, and possibly find evidence of life itself. Our findings will not only influence future Mars missions but also add a significant chapter to our understanding of life in the universe.

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Answer:

From the chemical formula of a compound you can learn about its composition:

Also, with the ratios and the atomic masses of each kind of atom, you can obtain the mass composition (%) of the compound.

Explanation:

Using indexes (subscripts), the chemical formula of a compound shows  the elements (kind of atoms) in the compound and the relative proportions of those elements.

The following example will show you how you learn that information from the chemical formula.

A very familiar compound is water and its chemical formula (also very familiar) is H₂O.

From that, you learn that water (a chemical compound):

        * hydrogen, H: 2 × 1.008 g/ mol = 2.016 g/,ol

        * oxygen, O: 1 × 15.999 g/mol = 15.999 g/mol

        * Molar mass of water: 2.016 g/mol + 15.999 g/mol = 18.015 g/ mol

        * H: (2.016 g/mol / 18.015 g/mol) × 100 = 11.19%

        * O: (15.999 g/mol / 18.015 g/mol) × 100 = 88.81%

Answer:

(a) The amounts of reactants decrease.

(b) The amounts of reactants decrease.

(c) The amounts of reactants decrease.

(d) the amounts of reactants increase.

Explanation:

(a) C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) ΔH°rxn = − 151 kJ.

So, the reaction can be represented as:

C₂H₂(g) + H₂O(g) ⇌ CH₃CHO(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

(b) CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) ΔH°rxn = − 451 kJ.

So, the reaction can be represented as:

CH₃CH₂OH(l) + O₂ (g) ⇌ CH₃CO₂H(l) + H₂O(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amount of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

(c) 2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) (exothermic).

So, the reaction can be represented as:

2C₂H₄(g) + O₂(g) ⇌ 2CH₃CHO(g) + heat.

As the temperature is decreases, it is like that decreasing the concentration of products side, that shifts the reaction towards the right side (products side) to attain the equilibrium again. So, the amounts of reactants decrease.

So, the right choice is reaction is: The amounts of reactants decrease.

(d) N₂O₄(g) ⇌ 2NO₂(g) (endothermic).

N₂O₄(g) + heat ⇌ 2NO₂(g)

As the temperature is decreased, it is like that decreasing the concentration of reactants side, that shifts the reaction towards the lift side (reactants side) to attain the equilibrium again. So, the amounts of reactants increase.

So, the right choice is reaction is: the amounts of reactants increase.

Copper nitrate and nitric oxide are produced in this reaction.

The digestion of copper wire by nitric acid is a redox reaction producing copper(II) nitrate, nitrogen dioxide, and water. The resulting copper salts are blue in solution.

The digestion of solid copper wire by nitric acid is a type of redox reaction.

When copper metal reacts with concentrated nitric acid, it forms aqueous copper(II) nitrate, accompanies by the release of brown fumes of nitrogen dioxide (NO₂) and water. This chemical reaction is influenced by factors such as concentration of the acid, the metal's activity, and temperature. With concentrated nitric acid, copper generally reduces the acid to nitrogen dioxide.

In a redox reaction, there is a transfer of electrons between the reacting species, with one substance being oxidized (losing electrons) and the other being reduced (gaining electrons).

For the reaction of copper with concentrated nitric acid, the balanced chemical equation would reflect the formation of Cu(NO₃)₂ and NO₂, which can be represented as follows after balancing:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l).

Copper salts, such as Cu(NO₃)₂, are characterized by a blue color in solution, which indicates the presence of copper(II) ions that have emerged from the copper metal.

Answer:

The pH of the solution is 6.35.

Explanation:

Concentration of hydroxide ions = [tex][OH^-]=2.20\times 10^{-8} M[/tex]

pH + pOH = 14

pH = 14 - pOH

[tex]pOH=-\log[OH^-][/tex]

[tex]pOH=-\log[2.20\times 10^{-8} M]=7.65[/tex]

[tex]pH=14-pOH=14-7.65=6.35[/tex]

The pH of the solution is 6.35.

The pH of the solution with a [tex]\( 2.20 \times 10^{-8} \)[/tex] M hydroxide ion concentration is approximately [tex]\( 6.34 \)[/tex].

To find the pH of a solution with a given hydroxide ion concentration [tex](\( [\text{OH}^-] \))[/tex], we can use the relationship between [tex]\( [\text{OH}^-] \)[/tex] and [tex]\( \text{pOH} \)[/tex], and then convert [tex]\( \text{pOH} \)[/tex] to pH.

The relationship between [tex]\( [\text{OH}^-] \)[/tex] and [tex]\( \text{pOH} \)[/tex] is:

[tex]\[ \text{pOH} = -\log([\text{OH}^-]) \][/tex]

Given [tex]\( [\text{OH}^-] = 2.20 \times 10^{-8} \) M[/tex], we can calculate [tex]\( \text{pOH} \)[/tex]:

[tex]\[ \text{pOH} = -\log(2.20 \times 10^{-8}) \][/tex]

[tex]\[ \text{pOH} = -\log(2.20) - \log(10^{-8}) \][/tex]

[tex]\[ \text{pOH} = -(\log(2.20) + (-8 \times \log(10))) \][/tex]

[tex]\[ \text{pOH} = -(0.3424 - 8 \times 1) \][/tex]

[tex]\[ \text{pOH} = -(0.3424 - 8) \][/tex]

[tex]\[ \text{pOH} = -(7.6576) \][/tex]

[tex]\[ \text{pOH} = -(-7.6576) \][/tex]

[tex]\[ \text{pOH} = 7.6576 \][/tex]

Now, we can convert [tex]\( \text{pOH} \)[/tex] to pH using the relationship:

[tex]\[ \text{pH} + \text{pOH} = 14 \][/tex]

[tex]\[ \text{pH} = 14 - \text{pOH} \][/tex]

[tex]\[ \text{pH} = 14 - 7.6576 \][/tex]

[tex]\[ \text{pH} = 6.3424 \][/tex]

Intermolecular forces.

The branch of science that relates with the interaction of matter and electromagnetic radiation is spectroscopy. Thus, option C is correct.

Electromagnetic radiations are given as the beam of protons with the presence of the perpendicular magnetic and electrical waves travelling at the speed of light.

The interaction of the electromagnetic radiations with the matter results in the propagation of the valence electrons of matter to the higher energy level and creating transitions.

These interactions between the electromagnetic radiations and matter are studies with spectroscopy. Thus, option C is correct.

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PH = 7.58

KOH will interact with HClO to provide KClO. you may have an answer that contains unreacted HClO and KOH. this is often a solution.

HClO + KOH → KClO + H2O

HClO reacts with KOH in a very 1: 1 molar ratio

Mole of acid in 50 ml of an answer of 0.150 m = 50/1000 * 0,150 = 0.0075 mole of acid

Mole of KOH in 30 ml of 150 M solution = 30/1000 * 0,150 = 0.0045 mole of KOH

This reacts to provide 0.0045 moles of KClO and there are 0.0030 moles of HClO unreacted

The volume of the answer = 50 ml + 30 ml = 80 ml = 0.080 l

Deposition of acid in solution = 0.0030 / 0.080 = 0.0375M

KClO deposit in solution = 0.0045 / 0.080 = 0.0562 m

pKa HClO = -log (4.0 * 10 ^ -8) = 7.40

PH = pKa + log ([KClO] / [HClO])

PH = 7.40 + notes (0.0562 / 0.0375)

PH = 7.40 + 1.50 notes

PH = 7.40+ 0.18

PH = 7.58

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Details

Grade: College

Subject: Chemistry

Keyword: mole, acid, HClO

Final answer:

In this titration of HClO with KOH, the pH at different volumes of added base is calculated. At 0.00 mL, the pH requires the ionization constant of HClO. At 15.0 mL, the pH is 0.903. At 25.0 mL, the pH is 7. At 30.0 mL, the pH is 14.

Explanation:

In this titration, we are diluting a 0.150 M solution of HClO with a 0.150 M solution of KOH. The goal is to calculate the pH at different volumes of added base.

In summary, the pH at different volumes of added base in the titration of HClO with KOH is as follows: 0.00 mL: pH = ? (Requires ionization constant of HClO); 15.0 mL: pH = **0.903**; 25.0 mL: pH = **7**; 30.0 mL: pH = **14**.

Answer:

Explanation:

1) Word equation (given)

2) Kind of reaction

This is a single replacement reaction.

Alkali metals is the group of metals in the group 1 of the periodic table. These are the most active metals, meaning that they react violently with water releasing hydrogen gas.

         Metal + Water → hydroxide + gaseous hydrogen

            X + H₂O → XOH(aq) + H₂ (g) ↑

3) Chemical equation

4) Balance

There is the same number of each atom on both sides: 2 K in the left and in the right, 4 H in the left and in the right, and 2 oxygen in the left and in the right. So, the equation is balanced.

The reaction of potassium metal with water produces potassium hydroxide and hydrogen gas, represented by the balanced chemical equation [tex]2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)[/tex]

The balanced chemical equation for the reaction between potassium metal (K) and liquid water (H2O) is:

[tex]2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2(g)[/tex]

In this reaction, potassium metal reacts vigorously with water to produce potassium hydroxide (KOH), which is an alkali, and hydrogen gas ([tex]H_2[/tex]). This is an exothermic reaction and heat is also released. The products include the aqueous potassium hydroxide and gaseous hydrogen. The equation is balanced with two atoms of potassium and two molecules of water on the reactant side, and two formula units of potassium hydroxide and one molecule of hydrogen gas on the product side.

i believe B is the answer

Answer: Alkene

Explanation: Apex

Answer : The standard enthalpy of the reaction is 20.2 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equilibrium reaction follows:

[tex]CS_2(g)+2H_2O(l)\rightleftharpoons CO_2(g)+2H_2S(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[n_{(CO_2)}\times \Delta H^o_f_{(CO_2)}+n_{(H_2S)}\times \Delta H^o_f_{(H_2S)}]-[n_{(H_2O)}\times \Delta H^o_f_{(H_2O)}+n_{(CS_2)}\times \Delta H^o_f_{(CS_2)}][/tex]

We are given:

[tex]\Delta H^o_f_{(CS_2(g))}=116.7kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2S(g))}=-20.60kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(1mol\times -393.5kJ/mol)+(2mol\times -20.60kJ/mol)]-[(1mol\times 116.7kJ/mol)+(2mol\times -285.8kJ/mol)]=20.2kJ[/tex]

Therefore, the standard enthalpy of the reaction is 20.2 kJ

The standard enthalpy of reaction is 20.2  kJ/mol.

We can calculate the standard enthalpy of reaction from the standard heat of formation using the formula;

ΔHreaction = ∑ΔH∘f products - ΔH∘f reactants

The equation of the reaction is; CS2(g)+2H2O(l)→CO2(g)+2H2S(g)

ΔH∘f CS2 = 116.7 KJ/mol

ΔH∘f  H2O(l) = −285.8 KJ/mol

ΔH∘f CO2 = −393.5 kJ/mol

ΔH∘f H2S= −20.60 kJ/mol

Substituting values, we have;

ΔHreaction = ∑[(−393.5) + 2 ×(−20.60) - (116.7) + 2 ×(−285.8)] kJ/mol

ΔHreaction = 20.2  kJ/mol

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HNO2 IS WEAK ACID IF U LOOK AT Ka VALUES U FIND ONE LISTED FOR HNO2

I believe the answer is A. However, I would double check the formula.

The isomer of the given structure of the Butane is Iso-Butane. Therefore, option (B) is correct.

Structural isomers are those isomers with the same molecular formulas but the atoms are completely arranged in a different manner. These are the molecules having the same molecular formula with different connectivities among the atoms.

For example, the structure of alkane (C₄H₁₀) is representing structural isomers with different structures. As there is an increase in the number of carbon atoms, the structural isomers of the alkane molecule increase.

The phenomenon of this structural isomer is known as structural isomerism. Structural isomerism is also known as constitutional isomerism as per the IUPAC.

The molecular formula of the butane is C₄H₁₀ where the linear chain of the four carbon atoms of butane is n-butane isomer. While the branched chain of the butane is an iso-butane isomer in which one carbon is tertiary.

Therefore, the structure in option (B) is isobutane, which is an isomer of butane.

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 CO3^2- and O3

B.  It can damage molecules in skin cells such as DNA. Prolonged exposure can actually cause skin cancer.

Ultraviolet light from the Sun primarily damages molecules in skin cells, particularly the DNA, which can lead to skin aging, skin cancer, and wrinkle formation. However, it also aids in Vitamin D production in the skin that has several health benefits. Apart from this, UV light is also used effectively for disinfection purposes.

Ultraviolet light from the Sun primarily damages molecules in skin cells, particularly the DNA. This happens through the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine, and cytosine, on the same strand of DNA. The extent of the damage is often believed to be proportional to the amount of ultraviolet radiation received, which is known as the linear hypothesis.

All types of UV radiation can damage collagen fibres, causing an acceleration of skin aging and wrinkle formation. Overexposure to the Sun when young has been linked to the development of skin cancer like melanoma in later life. However, UV-B radiation from sunlight does have some beneficial effects too, such as Vitamin D production in the skin, reducing risks of certain types of cancer and osteoporosis.

Lastly, UV light can sometimes be used effectively for disinfection. It forms thymine dimers in the DNA of microbes, leading to mutations that can kill the microorganisms. These properties of UV rays are used in water purification systems and germicidal lamps.

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The spoon will get warmer as it takes in the heat of the water

Competitive inhibitor

No you can’t do that, that’s idiotic

Answer:

Part A: 10.7.

Part B: 2) the solution is basic.

Part C: 5.9.

Part D: 1) the solution is acidic.

Part E: 8.1.

Part F: 2) the solution is basic.

Explanation:

Part A [OH−]= 4.5×10−4 M . Express your answer using two significant figures.  

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[OH⁻] = 4.5×10⁻⁴ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻]  = 10⁻¹⁴/(4.5×10⁻⁴  M) = 2.22 × 10⁻¹¹ M.

∵ pH = - log[H₃O⁺]

∴ pH = - log(2.22 × 10⁻¹¹) = 10.65 ≅ 10.7.

Part B 1) the solution is acidic 2) the solution is basic 3) the solution is neutral  

The solution is basic, because the pH is higher than 7.

Part C [OH⁻]= 7.7×10⁻⁹ M . Express your answer using two significant figures.  

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[OH⁻] = 7.7×10⁻⁹ M.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻]  = 10⁻¹⁴/(7.7×10⁻⁹ M) = 1.29 × 10⁻⁶ M.

∵ pH = - log[H₃O⁺]

∴ pH = - log(1.29 × 10⁻⁶) = 5.88 ≅ 5.9.

Part D 1) the solution is acidic 2) the solution is basic 3) the solution is neutral  

The solution is acidic, because the pH is lower than 7.

Part E A solution in which [OH−] is 140 times greater than [H+]. Express your answer using two significant figures.  

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∵  [OH⁻] = 140[H₃O⁺]

∴ [H₃O⁺](140[H₃O⁺]) = 10⁻¹⁴.

∴ 140 [H₃O⁺]² = 10⁻¹⁴.

[H₃O⁺]² = 10⁻¹⁴/ 140 = 7.14 x 10⁻¹⁷.

∴ [H₃O⁺] = √(7.14 x 10⁻¹⁷) = 8.45 x 10⁻⁹ M.

∵ pH = - log[H₃O⁺]

∴ pH = - log(8.45 x 10⁻⁹) = 8.07 ≅ 8.1.

Part F 1) the solution is acidic 2) the solution is basic 3) the solution is neutral

The solution is basic, because the pH is higher than 7.

The [H+] for solutions with given [OH−] are calculated using the ion product of water, with solutions having [H+] less than 1.0 × 10-7 being basic and those with more being acidic. The solutions for Part A and E were found to be basic while the solution for Part C was found to be acidic.

The [H+] of a solution can be determined from its [OH−] by using the ion product of water, which is 1.0 × 10-14 at 25°C. This can be represented by the equation [H+][OH−] = 1.0 × 10-14.

For Part A, you needed to calculate [H+] for a solution with [OH−] = 4.5 × 10-4M. [H+] = 1.0 × 10-14 / [OH−] = 1.0 × 10-14 / 4.5 × 10-4= 2.2 × 10-11 M. Since [H+] is less than 1.0 × 10-7, the solution is basic.

For Part C, [OH−] is 7.7 × 10-9M. [H+] = 1.0 × 10-14 / 7.7 × 10-9= 1.3 × 10-6M. Because [H+] is higher than 1.0 × 10-7, the solution is acidic.

In Part E, the solution's [OH−] is 140 times that of [H+]. Therefore, if we let [H+] = x, we get [OH−] = 140x. Since [H+][OH−] = 1.0 × 10-14, we get x(140x) = 1.0 × 10-14. Solving for x, we get [H+] = x = 5.37 × 10-14M. Since [H+] is less than 1.0 × 10-7, the solution is basic.

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A. Zinc (Zn)

Like all the other metals, it has a cloud of an electron because of the free electrons in the valence shell. These electrons are easier to move.

When some electric force is applied, the electrons start moving towards the positive end due to the attraction. This causes the flow of electrons, and, ultimately, the generation of electricity. Zinc one has 27%

PV=nRT  

P = 85.0 x 10^3 Pa  

T = 273 + 20 = 293 K  

n = 2.0 moles  

R = 8.314 m3. pa / mol .K  

V = 2.0 x 293 x 8.314 / 85.0x10^3 =  0.195623529 m^3

pOH = 3.08

NX3 + H2O <----> NHX3+ + OH-  

Kb = 4.0 x 10^-6

Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).

c(NH₄⁺) = c(OH⁻) = x.

x² = Kb · c(NH₃)

x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = c(OH⁻) = √(7.0 × 10⁻⁷)

    = 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

        =- log ( 8.367 × 10⁻⁴)

        = 3.08

Final answer:

To calculate the pOH of a 0.175 M aqueous solution of NX₃ with Kb of 4.0×10⁻⁶, we simplify the dissociation equation and find the OH- concentration to be approximately 1.67×10⁻³ M, resulting in a pOH of about 2.78.

Explanation:

To find the pOH of a 0.175 M aqueous solution of NX₃, given that Kb for NX3 is 4.0×10⁻⁶, we start by setting up the base dissociation reaction NX₃ + H₂O → NX₃H+ + OH−. The equilibrium expression associated with this dissociation is Kb = [NX3H+][OH−]/[NX3]. We can assume the concentrations of NX₃H+ and OH− to be equal (x) and much smaller than the initial concentration of NX₃, thus [NX₃] approximately equals 0.175 M. With these assumptions, Kb can be rewritten as 4.0×10⁻⁶ = x₂ / (0.175 − x).

Assuming x is much smaller than 0.175, we simplify to 4.0×10⁻⁶ ≈ x2 / 0.175. Solving for x gives us x ≈ √(4.0×10⁻⁶ ∗ 0.175) ≈ 1.67×10⁻³ M, which is the concentration of OH−. The pOH is calculated using the formula pOH = −log[OH−], resulting in pOH ≈ 2.78.

Answer:

C) 8.9x10-19M

Explanation:

Whe disolving a solid in water, this solid splits in anions and cations, in our example for CuS solid:

CuS (s) -------> Cu+2 (ac)  +   S-2 (ac)

Whe this solution is saturated, and balance is reached, this balance can be represented with Ksp formula (solubility constant), as following:

Ksp= [Cu+2][S-2], and for tgis ecuation, you only consider aqueous ions, but not the solid salt

When balance is reached, you may suppose that "X" moles of CuC gives you "X" moles of Cu+2 and "X" moles of S-2 (according to the stoicheometry of the above reaction):

XCuS (s) ------> XCu+2(ac)  +  XS-2(ac) , for which:

[Cu+2]=X and [S-2]=X so this gives us: Ksp=8x10-37=X.X=X²

X= √8x10-37 and finally X=[Cu+2]=8.9x10-19M (expressed in molar units)