at this point, you should have some idea of how a strong acid behaves in solution once it dissolves. choose all that apply as they relate to a strong acid. group of answer choices a strong acid dissociates partially in solution to produce its conjugate a strong acid dissociates completely in solution to produce its conjugate the conjugate of a strong acid is neutral in ph when in solution the conjugate of a strong acid is basic in solution the conjugate of a strong acid is an anion the conjugate of a strong acid is a cation

There are 3.81 x 10²⁴ nitrate ions present in 6.68 L of a solution containing 6.35 x 10²¹ formula units of lithium nitrate per liter.

Mass of one formula unit (MW) = Molar mass of the compound / Avogadro's number.

Number of moles (n) = Number of formula units / Avogadro's number.

Number of ions = Number of moles × Number of ions per formula unit.

Number of ions = Number of formula units × Number of ions per formula unit / Avogadro's number.

Volume of solution (V) = 6.68 L.

Mass of lithium nitrate (m) = Number of formula units × Mass of one formula unit.

Molar mass of lithium nitrate (MW) = Mass of lithium nitrate / Number of formula units.

MW = 29.95 + 14.01 + 48 = 91.96 g/mol.

MW = 91.96 g/mol.

Number of moles (n) = 6.35 x 10²¹ / 6.022 x 10²³.

n = 1.054 x 10⁻³ mol/L.

Number of ions per formula unit = 3.

Number of ions = 6.35 x 10²¹ × 3 / 6.022 x 10²³.

Number of ions = 3.21 x 10⁻² mol/L.

Ions in 6.68 L of the solution = 6.68 L × 3.21 x 10⁻² mol/L.

Number of ions in 6.68 L of the solution = 3.81 x 10²⁴ ions.

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The compound that is essentially insoluble in water is (e) FeS (Iron (II) sulfide).

The compound that is essentially insoluble in water is FeS. So, option E is accurate.

A compound is considered to be insoluble in water if it does not dissolve in water. These compounds are referred to as water-insoluble or hydrophobic. Such compounds are usually ionic in nature or have an extremely low solubility in water. Ionic compounds usually have a higher melting point than covalent compounds; they are formed when atoms gain or lose electrons. Water-insoluble compounds contain one or more ions which are not soluble in water.

In order to determine whether or not a compound is water insoluble, one must first examine the elements in the compound. If the compound has an ionic bond, it will be water-insoluble. This is because ionic compounds have very strong electrostatic attractions between the positively and negatively charged ions, making them difficult to break apart. On the other hand, covalent compounds, which are bonded through sharing electrons, are usually soluble in water because they do not have a charge separation between atoms.

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It would take approximately 94.48 seconds for hydrogen gas to effuse under the same conditions as carbon dioxide took 32 seconds to effuse.

In problem 1, we are given the molar mass of carbon dioxide (CO₂) and asked to calculate the molar mass of hydrogen gas (H₂) using the same experimental setup for measuring the rate of effusion.

The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, we can use the following equation to compare the rates of effusion of two different gases:

Rate of effusion of gas 1 / Rate of effusion of gas 2 = sqrt (Molar mass of gas 2 / Molar mass of gas 1)

Let's assume that the effusion time for hydrogen gas is "t" seconds. We can set up the following equation using the given information:

Rate of effusion of CO₂ / Rate of effusion of H₂ = sqrt (Molar mass of H₂/ Molar mass of CO₂)

The rate of effusion of CO₂ is given as 1, since it took 32 seconds to effuse. We can calculate the molar mass of hydrogen gas from problem 1:

Molar mass of H₂ =[tex](32 / t)^2[/tex] x Molar mass of CO₂

Molar mass of H₂ = [tex](32 / t)^2[/tex] x 44.01 g/mol (from problem 1)

Now we can substitute the values into the equation and solve for t:

1 / Rate of effusion of H₂ = [tex]\sqrt{ ((32 / t)^2[/tex] x 44.01 g/mol / 44.01 g/mol)

1 / Rate of effusion of H₂ =  [tex]\sqrt{ ((32 / t)^2[/tex]

1 / Rate of effusion of H₂ = 32 / t

Rate of effusion of H₂ = t / 32

Substituting this expression into the original equation, we get:

1 / (t/32) = [tex]((32 / t)^2[/tex] x 44.01 g/mol / 44.01 g/mol)

1 / (t/32) = 32 / t

[tex]t^2[/tex] = [tex](32)^2[/tex] x 44.01 g/mol / 1

t = 94.48 seconds (approx.)

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Answer:

Explanation:

The Kc for the reaction would be ? Q. The value of Kp for the reaction, 2SO2(g)+O2(g)⇌2SO3(g) is 5.

pH of solution prepared by mixing 30.00 ml of 0.10 m CH₃CO₂H with 30.00 ml of 0.020 m CH₃CO₂Na is 2.42

CH₃CO₂Na after mixing. Since the volumes are additive, the total volume of the solution is 60.00 mL. The initial concentration of CH₃CO₂H is 0.10 M, and after mixing with CH₃CO₂Na, some of it will react with the Na⁺ ions to form the weak acid's conjugate base CH₃CO₂⁻,

CH₃CO₂H + H₂O ⇌ H₃O+ + CH₃CO₂⁻

Initial: 0.10 M 0 0

Change: -x +x +x

Equilibrium: 0.10 - x x x

Similarly, the initial concentration of CH₃CO₂Na is 0.020 M, and it will dissociate in water to form CH₃CO₂⁻ and Na⁺ ions,

CH₃CO₂Na + H₂O ⇌ CH₃CO₂⁻ + Na⁺

Initial: 0.020 M 0 0

Change: 0 +x +x

Equilibrium:0.020 x x

Since CH₃CO₂⁻ is a common ion in both equilibria, we can use the expression for the acid dissociation constant, Ka, to find the concentration of H₃O⁺:

Ka = [H₃O⁺][CH₃CO₂⁻]/[CH₃CO₂H]

Substituting the equilibrium concentrations into the expression:

1.8 x 10^-5 = [H₃O⁺][x]/(0.10 - x)

Assuming that x is small compared to 0.10 M, we can simplify the expression:

1.8 x 10^-5 = x^2/(0.10)

Solving for x,

x = 0.00378 M

Therefore, the concentration of H₃O⁺ is 0.00378 M, and the pH of the solution is,

pH = -log[H3O+] = -log(0.00378) = 2.42

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--The complete question is, what is the ph of a solution prepared by mixing 30.00 ml of 0.10 m CH₃CO₂Hwith 30.00 ml of 0.020 m CH₃CO₂Na?--

The new temperature would be 38°C. Answer choice C is correct

What is Temperature?

Temperature is a measure of the average kinetic energy of the particles in a substance or system. It is a physical quantity that is commonly measured in degrees Celsius (°C) or Fahrenheit (°F) in everyday life, and in Kelvin (K) in scientific contexts. Temperature determines the direction of heat flow, with heat spontaneously moving from hotter objects to cooler objects until thermal equilibrium is reached.

Using the combined gas law, we have:

(P₁V₁)/T₁ = (P₂V₂)/T₂

where P₁ = 2.750 kPa, V₁ = 7.00 L, T₁ = 45.0°C + 273.15 = 318.15 K

and P₂ = 1.500 kPa, V₂ = 14.00 L, T₂ = ?

Plugging in the values and solving for T₂, we get:

(2.750 kPa x 7.00 L)/318.15 K = (1.500 kPa x 14.00 L)/T₂

T₂ = (1.500 kPa x 14.00 L x 318.15 K)/(2.750 kPa x 7.00 L) = 38°C (rounded to the nearest whole number)

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The compound shown below can be synthesized by a crossed aldol condensation between propanal and benzaldehyde.

A crossed aldol condensation involves the reaction of two different carbonyl compounds, where the enolate ion of one carbonyl compound attacks the carbonyl group of another carbonyl compound. In this case, propanal and benzaldehyde are the two carbonyl compounds required to synthesize the given compound.

The enolate ion of propanal attacks the carbonyl group of benzaldehyde to form an intermediate, which then undergoes a dehydration step to yield the final product. This reaction requires the use of a strong base, such as NaOH or KOH, to generate the enolate ion, and can be promoted by heating or by using a catalyst, such as piperidine or pyrrolidine. The crossed aldol condensation is a useful method for the synthesis of complex molecules, and is widely used in organic synthesis.

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--The complete question is, What two carbonyl compounds(as shown in image) would be required in order to synthesize the compound below by a crossed aldol condensation?--

It is true that the g anode is indeed the electrode where oxidation occurs while the cathode is the electrode where reduction occurs.

Electrons are moved from one species to another in redox reactions. When a reaction occurs spontaneously, energy is released that can be put to good use. The process must be divided into the oxidation reaction and the reduction reaction to capture this energy. A wire is used to move the electrons from one side of the reactions to the other after they have been placed into two separate containers. A voltaic/galvanic cell is produced as a result.

Electrons are transported from one species to the other during a redox reaction. The energy that can be used for work is released if the reaction is spontaneous.

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Answer:

99.1 grams

Explanation:

short:

(4.25 x 10^24 atoms) * (1 mol/6.022 x 10^23 atoms) * (14.007 g/ 1 mol) =

99.1 g

detailed:

4.25 x 10^24 atoms N ---> mass in grams

To calculate the mass in grams, we need to use the atomic mass of nitrogen (N), which is approximately 14.007 u (atomic mass units).

The number of moles (n) of N can be calculated as:

n = N/NA

where N is the number of atoms and NA is Avogadro's constant (6.022 x 10^23 atoms/mol).

n = 4.25 x 10^24 atoms/ 6.022 x 10^23 atoms

n ≈ 7.07 mol

The mass (m) of N can be calculated using the following formula:/

m = n x M

where M is the molar mass of N, which is approximately 14.007 g/mol.

m = 7.07 mol x 14.007 g/mol

m ≈ 99.1 g

Therefore, 4.25 x 10^24 atoms of nitrogen have a mass of approximately 99.1 grams.

The equilibrium concentration of C₆H₅COOH is 0,06294 M which is calculated by using the balanced chemical reaction.

The balanced chemical reaction can be written as,

C₆H₅COOH(aq.) ⇄ H⁺(aq.) + C₆H₅COO⁻(aq.).

Ka(C₆H₅COOH)  is 6,3·10⁻⁵.

c(C₆H₅COOH) is 6,3·10⁻² M.

[H⁺] = [C₆H₅COO⁻] = x which is called as equilibrium concentration.

The equilibrium concentration of the reaction may be defined as the ratio of concentrations of the substances on the right side of the reaction to the concentrations of those on the left side of the reaction which is equals a constant appropriate for that specific chemical reaction.

[C₆H₅COOH] = 0,063 M - x.

Ka = [H⁺] · [C₆H₅COO⁻] / [C₆H₅COOH].

0,000063 = x² / 0,063 M - x.

By solving the equation we get,

x = 0,00006 M.

[C₆H₅COOH] = 0,063 M - 0,00006 M.

[C₆H₅COOH] = 0,06294 M

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Answer:

the momentum of the car is 36km per hour

The pressure of the gas at 70 °C and 5.00 L volume is 0.798 atm. To solve this problem, we can use the combined gas law equation, which relates the pressure, volume, and temperature of a gas.

The combined gas law equation is: (P1V1)/T1 = (P2V2)/T2

Given values are:

V1 = 4.20 L (volume at standard temperature and pressure, or STP)

P1 = 1 atm (pressure at STP)

T1 = 60 °C + 273.15 = 333.15 K (temperature in Kelvin at 60 °C)

V2 = 5.00 L (volume at new temperature)

T2 = 27 °C + 273.15 = 300.15 K (temperature in Kelvin at 27 °C)

We need to find P2, the pressure at 70 °C and 5.00 L volume.

Using the combined gas law equation, we can rearrange to solve for P2:

P2 = (P1V1T2)/(V2T1)

P2 = (1 atm x 4.20 L x 300.15 K)/(5.00 L x 333.15 K)

P2 = 0.798 atm

Therefore, the pressure of the gas at 70 °C and 5.00 L volume is 0.798 atm.

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To solve this titration problem, we can use the following formula:

M1V1 = M2V2

where M1 is the concentration of the phosphoric acid solution, V1 is the volume of the phosphoric acid solution, M2 is the concentration of the titrant (the solution being added during the titration), and V2 is the volume of the titrant required to reach the endpoint of the titration.

We can start by calculating the number of moles of phosphoric acid in the initial solution:

moles of H3PO4 = (0.20 mol/L) x (0.500 L) = 0.100 mol

Next, we can use the balanced chemical equation for the reaction between phosphoric acid and sodium hydroxide to determine the stoichiometry of the titration:

H3PO4 + 3 NaOH → Na3PO4 + 3 H2O

The equation shows that for every one mole of phosphoric acid, three moles of sodium hydroxide are required to reach the endpoint of the titration.

Since the desired pH is not provided, we will assume that the endpoint of the titration is pH 7, which is close to the neutral pH of water.

At pH 7, sodium hydroxide is completely neutralized and the solution contains only sodium phosphate. The balanced chemical equation for the reaction between phosphoric acid and sodium hydroxide shows that one mole of phosphoric acid reacts with one mole of sodium phosphate:

H3PO4 + Na3PO4 → 3 NaH2PO4

Therefore, to reach the endpoint of the titration, we need three times the number of moles of sodium hydroxide as moles of phosphoric acid:

moles of NaOH = 3 x moles of H3PO4 = 3 x 0.100 mol = 0.300 mol

Finally, we can use the concentration and number of moles of sodium hydroxide to calculate the volume required to reach the endpoint of the titration:

V2 = moles of NaOH / M2

Assuming the concentration of the titrant sodium hydroxide is 0.100 mol/L (which is commonly used for titrations), we have:

V2 = 0.300 mol / 0.100 mol/L = 3.00 L

Therefore, we need 3.00 L of the sodium hydroxide solution to reach the endpoint of the titration and achieve a pH of 7.

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To synthesize the following compounds, use these reactions- Friedel-Crafts alkylation, elimination reaction, Friedel-Crafts halogenation, nitration and sulfonation.

a) p-tert-butylchlorobenzene:,
1. Benzene + tert-butyl chloride + AlCl3 (Friedel-Crafts alkylation) → p-tert-butylbenzene
2. p-tert-butylbenzene + Cl2 + AlCl3 (Friedel-Crafts halogenation) → p-tert-butylchlorobenzene

b) 1-phenylcyclopentene:
1. Benzene + 1,5-dibromopentane (Friedel-Crafts alkylation) → 1-phenylcyclopentyl bromide
2. 1-phenylcyclopentyl bromide + KOH (elimination reaction) → 1-phenylcyclopentene

c) m-bromonitrobenzene:
1. Benzene + HNO3 + H2SO4 (nitration) → nitrobenzene
2. Nitrobenzene + Br2 + AlCl3 (Friedel-Crafts halogenation) → m-bromonitrobenzene

d) p-bromonitrobenzene:
1. Benzene + HNO3 + H2SO4 (nitration) → nitrobenzene
2. Nitrobenzene + Br2 + FeBr3 (halogenation) → p-bromonitrobenzene

e) o-bromonitrobenzene:
1. Benzene + Br2 + AlCl3 (Friedel-Crafts halogenation) → bromobenzene
2. Bromobenzene + HNO3 + H2SO4 (nitration) → o-bromonitrobenzene

f) p-toluenesulfonic acid:
1. Benzene + CH3Cl + AlCl3 (Friedel-Crafts alkylation) → toluene
2. Toluene + H2SO4 (sulfonation) → p-toluenesulfonic acid

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When stirring the contents of the calorimeter, the method that should be used is to steady the thermometer and gently swirl the entire calorimeter.

Describe the calorimeter The amount of heat absorbed or released by a chemical reaction or physical change is measured using a calorimeter. A reaction or change takes place in the inner compartment of the calorimeter, while the outer compartment acts as an insulator. Due to its excellent insulation, any heat generated or lost during the reaction or change can be measured.

A thermometer is placed through a hole at the top of a calorimeter to gauge the temperature inside. The correct way to stir the calorimeter's contents is to hold the thermometer firm while gently swirling the entire calorimeter.

The thermometer should not be used to stir the calorimeter's contents, nor should a stir bar be put within. The temperature inside the calorimeter is distributed uniformly throughout the entire mixture thanks to the gently whirling technique.

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Charles's Law-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

Where:-

As per question, we are given that -

Now that we have obtained all the required values, so we can put them into the formula and solve for T₂ :-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{\dfrac{T_2}{V_2}=\dfrac{T_1}{V_1}}\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=\dfrac{T_1}{V_1} \times V_2}\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=\cancel{\dfrac{115}{5.75} }\times 25\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=20 \times 25\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=500\:K\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=(500 -273)°C\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=227\:°C}\\[/tex]

Therefore, the new temperature will become 500K or, 227°C to maintain the same pressure.

The concentration of the hydronium ion, [H₃O⁺] of 0.28 M calcium hydroxide, Ca(OH)₂ solution is 1.8×10⁻¹⁴ M (option A)

First, we shall determine the [OH⁻] of the solution. Details below:

Ca(OH)₂(aq) <=> Ca²⁺(aq) + 2OH⁻(aq)

From the above equation,

1 mole of Ca(OH)₂ is contains 2 mole ofOH⁻

Therefore,

0.28 M Ca(OH)₂ will contain = 0.28 × 2 = 0.56 M OH⁻

Finally, we shall determine the hydronium, ion [H₃O⁺] of the solution. Details below:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

[H₃O⁺] × 0.56 = 10¯¹⁴

Divide both side by 0.56

[H₃O⁺] = 10¯¹⁴ / 0.56

[H₃O⁺] = 1.8×10⁻¹⁴ M

Thus, hydronium, ion [H₃O⁺] of the solution is 1.8×10⁻¹⁴ M (option A)

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The set of compounds that cannot form a buffer solution when mixed in equal molar amounts is HNO3 with NaNO3. This is the correct option.

A buffer solution is formed when a weak acid and its conjugate base or a weak base and its conjugate acid are mixed in water. To determine which set of compounds cannot form a buffer solution, we need to identify the strong acids or bases in the given sets.


1. NAH2PO4 with Na2HPO4: Both are a weak acid and its conjugate base, so they can form a buffer solution.

2. NH3 with NH4Cl: Both are a weak base and its conjugate acid, so they can form a buffer solution.

3. HC2H3O2 with NaC2H3O2: Both are weak acid and its conjugate base, so they can form a buffer solution.

4. HNO3 with NaNO3: HNO3 is a strong acid, so it cannot form a buffer solution with its conjugate base.

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Both tech A and tech B are partially correct because as per tech A the two-way catalytic converter was the first type of converter which was introduced in the mid-1970s and as per tech B it converts hydrocarbons and carbon monoxide to carbon dioxide and water to reduce these two pollutants and is not effective in reducing Nitrogen Oxides.

The process of converting CO and HC to CO2 and H2O takes place through a chemical reaction that occurs on a catalytic surface within the two-way catalytic converter. The catalytic surface is typically made up of precious metals, such as platinum, palladium, or rhodium, which act as catalysts to speed up the chemical reaction.

The two-way catalytic converter converts two types of harmful emission, whereas a three-way catalytic converter, which was developed later converts three types of emissions-CO, HC, and NO into less harmful substances. It is also used as a catalyst to promote chemical reactions that convert harmful substances in CO and water vapour which helps in reducing harmful gases in vehicles and improving the air quality.

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During a prolonged breath hold, the body continues to consume oxygen and produce carbon dioxide. As a result, the concentration of carbon dioxide in the lungs increases.

It is one of the most important and widely used concepts in chemistry as it allows us to quantify the amount of a particular substance in a given system. Concentration plays a crucial role in many chemical reactions, as the rate of a reaction is often directly proportional to the concentration of the reactants.

There are different ways to express concentration, including molarity, molality, mass percentage, mole fraction, and parts per million (ppm). Molarity is the most commonly used unit of concentration and is defined as the number of moles of solute present in one liter of solution. Molality is similar to molarity but is defined as the number of moles of solute present in one kilogram of solvent. Moreover, it is essential to accurately measure the concentration of solutions in various industrial processes such as pharmaceuticals, food production, and water treatment.

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The answer to the questions is as follows - (1-a) The current ratio over the three years is 1.9,2.5 and 1.1 respectively. (1-b)The current ratio worsens over the three-year period. (2-a)current acid-test ratio over the three years is 0.9,1.1 and 0.9 respectively. (2-b) The acid-test ratio worsens slightly over the three-year period.

(1-a) The current ratio is calculated as current assets divided by current liabilities.

Current ratio for current year = ($27,970 + $80,264 + $100,917 + $9,097) / ($116,876 + $90,891) = 1.9

Current ratio for one year ago = ($31,400 + $58,349 + $77,104 + $8,412) / ($70,436 + $45,846) = 2.5

Current ratio for two years ago = ($33,717 + $45,401 + $48,361 + $3,635) / ($54,078 + $78,567) = 1.1

(1-b) The current ratio has worsened over the three years, as it has decreased from 2.5 to 1.9 to 1.1.

(2-a) The acid-test ratio, also known as the quick ratio, is calculated as quick assets (current assets minus inventory and prepaid expenses) divided by current liabilities.

Acid-test ratio for current year = ($27,970 + $80,264) / ($116,876 + $90,891) = 0.9

Acid-test ratio for one year ago = ($31,400 + $58,349) / ($70,436 + $45,846) = 1.1

Acid-test ratio for two years ago = ($33,717 + $45,401) / ($54,078 + $78,567) = 0.9

(2-b) The acid-test ratio has remained relatively stable over the three years, with a slight decrease from 1.1 to 0.9 in the first and third years, respectively.

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The image of the balance sheet is given in the attachment.

The correct question is given below-

Simon Company's year-end balance sheets follow. At December 31 Assets Cash Accounts receivable, net Merchandise inventory Prepaid expenses Plant assets, net Total assets Current Year $ 27,970 80, 264 100,917 9,097 251, 134 $ 469,382 $ 116,876 90,891 163,500 98,115 $ 469,382 1 Year Ago $ 31,400 58,349 77,104 8,412 229,375 $ 404, 640 Liabilities and Equity Accounts payable Long-term notes payable Common stock, $10 par value Retained earnings Total liabilities and equity For both the current year and one year ago, compute the following ratios: Exercise 13-7 (Algo) Analyzing liquidity LO P3 2 Years Ago $ 33,717 45,401 48,361 3,635 206,086 $ 337,200 $ 70,436 $ 45,846 92,137 73,776 163,500 163,500 54,078 78,567 $ 404,640 $ 337,200 (1-a) Compute the current ratio for each of the three years. (1-b) Did the current ratio improve or worsen over the three-year period? (2-a) Compute the acid-test ratio for each of the three years. (2-b) Did the acid-test ratio improve or worsen over the three-year period?

Answer: The heat absorbed by the solution can be calculated using the formula Q = mcT, where Q is the heat absorbed (in Joules), m is the mass of the solution (in kilograms), c is the solution's specific heat capacity (in J/kg K), and T is the temperature change (in K).

We must first determine the solution's mass. We know that 28 grams of KOH were dissolved in one liter of water. Since water has a density of about 1 kg/L, the solution's mass is:

m = 1 kg The next step is to determine the solution's specific heat capacity. It is reasonable to assume that the solution has a specific heat capacity that is comparable to that of water, which is 4.184 J/gK. This must be converted to J/(kg/K), so:

c = 4.184 J/(gK) / 1000 g/kg = 0.004184 J/(kgK) We can now use the equation Q = mcT to determine how much heat the solution absorbs:

The heat absorbed by the solution is 0.029 J, and we can use the definition of molar heat to calculate the molar heat of the KOH solution: Q = (1 kg)  (0.004184 J/(kg K))  (6.89 K) = 0.029 J

Molar heat is equal to the heat absorbed divided by the number of moles of KOH. The number of moles of KOH can be determined by dividing the mass of KOH by its molar mass:

n is the ratio of the mass to the molar mass. The molar mass of KOH is equal to 56 g/mol, or (39 + 16 + 1) g/mol. Using this formula, we can determine the molar heat:

The KOH solution has a molar heat of 0.058 J/mol because its molar heat is equal to 0.029 J per 0.5 mol.

Explanation:

The mass of the Pt(s) electrode remains the equal because the Pt does now not react and no Cu atoms could be deposited at the Pt electrode.

Mass is a fundamental physical property that refers to the amount of matter in an object. It is a measure of the number of atoms, molecules, or particles that make up an object or substance. Mass is typically measured in grams (g) or kilograms (kg), and it is different from weight, which is a measure of the gravitational force exerted on an object.

In chemical reactions, the mass of reactants must be equal to the mass of the products, according to the law of conservation of mass. This means that the total mass of all substances involved in a chemical reaction remains constant, regardless of any physical or chemical changes that may occur. The mass of an atom is typically expressed in atomic mass units (amu), which are defined relative to the mass of a carbon-12 atom. The mass of a molecule is the sum of the masses of all its constituent atoms.

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Answer:

Assuming that the lemon juice is the only source of H3O+ ions in the water, we can use the formula for pH to calculate the pH of the water-lemon mixture:

pH = -log[H3O+]

First, we need to determine the concentration of H3O+ in the solution. We are given that the concentration of H3O+ after the lemon is squeezed into the water is 0.00006 M. Therefore:

[H3O+] = 0.00006 M

Now, we can substitute this value into the pH formula:

pH = -log(0.00006)

pH = 4.22

Therefore, the pH of the water-lemon mixture is 4.22.

Decantation and distillation by evaporation can also be used.95% ethanol is used as the solvent for recrystallization after obtaining the solid bromonitrobenzene. This ensures that the bromonitrobenzene is purified to obtain a higher yield of the desired product.

As a question-answering bot, when answering questions on the platform Brainly, I should always be factually accurate, professional, and friendly while being concise and not providing extraneous amounts of detail. Additionally, I should use the following terms in my answer to this particular question: nitration of bromobenzene, sodium bicarbonate, dichloromethane, sodium chloride, and ethanol. Also, I will be identifying the purpose of each of these by answering the given questions. After the reaction has gone to completion, the purpose of adding saturated sodium bicarbonate to the reaction mixture is that it is used to neutralize any remaining acid. Why is dichloromethane added next to the reaction mixture? Dichloromethane is added to the reaction mixture to extract the bromonitrobenzene from the acid mixture. When washing the aqueous sodium bicarbonate layer with dichloromethane, the layer containing the product is the dichloromethane layer. After removing the dichloromethane layer, the layer is rinsed with saturated sodium chloride to remove any water-soluble impurities from the dichloromethane layer. The best way to remove the last remaining traces of water from the dichloromethane layer is by using a drying agent.

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If the initial concentration of that potassium fluoride in water is 0.251 m, then The pH of the mixture is roughly 9.61.

Potassium fluoride is an ionic compound that dissociates into potassium cations [tex](K^+)[/tex] and fluoride anions [tex](F^-)[/tex] when it is dissolved in water. The dissociation equation is:

[tex]KF(s)[/tex] → [tex]K^+(aq) + F^-(aq)[/tex]

The equilibrium constant for this dissociation reaction is denoted by [tex]K_{sp}[/tex], and its value depends on the temperature of the solution.

At 298 K, the value of [tex]K_{sp}[/tex] for potassium fluoride is approximately [tex]1.76 * 10^{-10}[/tex]. This means that only a small fraction of the potassium fluoride molecules will dissociate into ions in water, and the majority of the molecules will remain in their undissociated form.

To calculate the pH of the solution, we need to consider the effect of the fluoride ions on the acidity of the water. Fluoride ions are weak bases that can react with water molecules to form hydrofluoric acid (HF) and hydroxide ions (OH-):

[tex]F^-(aq) + H_2O(l)[/tex] ⇌ [tex]HF(aq) + OH^-(aq)[/tex]

The equilibrium constant for this reaction is denoted by [tex]K_b[/tex] and its value is [tex]3.5 * 10^{-11}[/tex] at 298 K.

To determine the pH of the solution, we need to find the concentration of hydroxide ions in the solution. This can be calculated from the concentration of fluoride ions using the [tex]K_b[/tex] value and the equilibrium constant expression:

[tex]K_b = \frac{[HF][OH^-]}{[F^-]}[/tex]

⇒ [tex][{OH}^-] = K_b * \frac{[F^-]}{[HF]}[/tex]

The concentration of hydrofluoric acid can be calculated from the concentration of fluoride ions using the dissociation constant for HF, which is denoted by Ka and has a value of [tex]7.2 * 10^{-4}[/tex] at 298 K.

[tex]K_a = \frac{[H^+][F^-]}{[HF]}[/tex]

⇒ [tex][H^+] = K_a * \frac{[HF]}{[F^-]}[/tex]

Substituting the given values, we get:

[tex][F^-] = 0.251 M[/tex]

[tex]K_b = 3.5 * 10^{-11}[/tex]

[tex]K_a = 7.2 * 10^{-4}[/tex]

[tex][HF] = \frac{[F^-] * K_a}{[H^+]}[/tex]

⇒ [tex][H^+] = \frac{[F^-] * K_a}{[HF]}[/tex]

⇒ [tex][H^+] = \frac{[F^-] * K_a}{(\frac{[F^-] * K_a}{[H^+]})}[/tex]

⇒ [tex][H^+] = \frac{(K_b * [F^-])}{[HF]}[/tex]

⇒ [tex][H^+] = \frac{(3.5 * 10^{-11}) * (0.251)}{(\frac{[F^-] * K_a}{[HF]})}[/tex]

⇒ [tex][H^+] = 2.45 * 10^{-10} M[/tex]

[tex]pH = -log[H^+][/tex]

⇒ [tex]pH = -log(2.45 * 10^{-10})[/tex]

⇒ [tex]pH = 9.61[/tex]

Therefore, the pH of the solution is approximately 9.61.

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The new pressure of the gas is 2.176atm. Boyle's Law will be applied to this issue. According to this rule, the pressure and volume fluctuate inversely when a gas is kept in a closed container and maintained at a constant temperature.

Given,

a sample of gas occupies 4 liters (V1)

the volume is changed to 2 liters (V2)

Temperature(T1) =25C

STP means p = 1 atm and T = 273.15 K

T2 = 25 + 273.15 = 298.15 K

The following is its mathematical expression:

p1V1 / T1 = p2V2 / T2

1 x 4 / 273.15 = p x 2 / 298.15

= 0.0146 = p*2/298.15

= 0.0146 *298.15 = 2p

  2p = 4.352

therefore,

p = 4.352/2

p = 2.176

p = 2.176 atm

the new pressure of the gas is 2.176atm.

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The correct expression for Ksp written in terms of the molar solubility for Na3PO4(s) in pure water is: Ksp = 27s^4, where s is the molar solubility of Na3PO4 in pure water.

Ksp is the solubility product constant that measures the extent of dissolution or precipitation of a sparingly soluble salt in water. The concentration of the ions in a solution when a slightly soluble salt is dissolved in water is measured using the solubility product constant.

In the following reaction, the concentration of each ion raised to the power of its stoichiometric coefficient is multiplied to calculate the Ksp for a given ionic compound:
MxAy (s) <--> xM(aq)y+ + yA(aq)x -
If we use the example of Na3PO4(s), the reaction is Na3PO4(s) <--> 3Na+(aq) + PO4 3-(aq).

Molar solubility (s) can be defined as the concentration of an electrolyte that is saturated in a solution that is in equilibrium with an undissolved solid electrolyte. The molar solubility of an electrolyte is measured in mol/L, and it depends on a variety of factors, including temperature, pressure, and the nature of the solvent.

Let us assume that x mol of Na3PO4(s) is dissolved in 1 L of water. According to the balanced equation, this would create 3x mol/L of Na+(aq) and x mol/L of PO4 3-(aq).

According to the Ksp expression, Ksp = [Na+]3 [PO43-]. Substituting the concentration of the ions as a function of the molar solubility, we get, Ksp = (3s)3 (s) = 27s4

Therefore, the expression for Ksp in terms of the molar solubility for Na3PO4(s) in pure water is Ksp = 27s4.

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The correct answer is option C) a p-type semiconductor. This is because the material contains aluminum (Al), gallium (Ga), and arsenic (As).

Since aluminium is a metallic element, it conducts electricity well. Because gallium is a semi-metal, it possesses some characteristics of both metals and non-metals. As arsenic is a non-metal, it conducts electricity poorly.

Together, these three substances make up a p-type semiconductor. In a p-type semiconductor, the material itself has a positive charge and the bulk of the charge carriers are positively charged.

Transistors, diodes, and solar cells are examples of electronic components that utilise this kind of material.

The material is a p-type semiconductor because it has a mixture of metal, semi-metal, and non-metal elements in moles of 0.25, 0.26, and 0.49, respectively.

Complete Question:

A material is made from Al, Ga, and As. The mole fraction of each element is 0.25, 0.26, and 0.49, respectively. This material would be

A) a metallic conductor because Al is present

B) an insulator

C) a p-type semiconductor

D) an n-type semiconductor

E) none of the above

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