What is the solubility in moles/liter for magnesium hydroxide at 25 oC given a Ksp value of 1.1 x 10-11. Write using scientific notation and use 1 or 2 decimal places (even though this is strictly incorrect!)

To increase the salt concentration to 8.6%, you need to add 8794g of salt to the solution. The final mass of the 8.6% brine solution will be 961g.

To have an 8.60 percent brine solution, you would need to add salt to compensate for the loss of water due to evaporation. First, calculate the mass of water after evaporation by subtracting 123g from the initial mass of the brine solution (962g - 123g = 839g).

Then, find the mass of the salt needed by multiplying the final mass of the solution by the desired percent of salt (839g / 0.0860 = 9756g). Subtract the initial mass of the brine solution to determine the amount of salt that must be added (9756g - 962g = 8794g).

To find the mass of the 8.6% brine solution that will be produced, subtract the mass of the added salt from the final mass of the solution (9756g - 8794g = 961g).

https://brainly.com/question/28668523

#SPJ3

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

              = 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

Metals are good conductors of electricity due to the mobile valence electrons which can move freely within the metallic crystal lattice, facilitating electric charge transfer.

Metals are good conductors of electricity because they contain mobile valence electrons. In metallic bonds, these valence electrons are not associated with a particular atom or pair of atoms, but move freely within the crystal lattice of positively charged metal ions. They form what is often referred to as an 'electron sea'. These free moving electrons can carry charge from one place to another when a voltage (electric potential difference) is applied, making metals good conductors of electricity.

https://brainly.com/question/32879067

#SPJ3

Answer:

percentage mass of platinum in the alloy ≈ 90.60 %

Explanation:

The alloy is 8.528 g sample of an alloy containing platinum and cobalt . The alloy react with excess nitric acid to form cobalt(ii) nitrate . Platinum is resistant to acid so it will definitely not react with the acid only the cobalt metal in the alloy will react with the acid.

The chemical reaction can be represented as follows:

Co (s) + HNO₃ (aq) → Co(NO₃)₂ (aq) +  NO₂ (l) + H₂O (l)

The balanced equation

Co (s) + 4HNO₃ (aq) → Co(NO₃)₂(aq) + 2NO₂ (l) + 2H₂O (l)

Cobalt is the limiting reactant

atomic mass of cobalt = 58.933 g/mol

Molar mass of Co(NO₃)₂ = 58.933 + 14 × 2 + 16 × 6 = 58.933 + 28 + 96 = 182.933  g

58.933 g of cobalt produce 182.933  g of  Co(NO₃)₂

? gram of cobalt will produce 2.49 g of Co(NO₃)₂

cross multiply

grams of cobalt that will react = (58.933 × 2.49)/182.933

grams of cobalt that will react = 146.74317000/182.933

grams of cobalt that will react= 0.8021689362 g

grams of cobalt that will react = 0.802 g

mass of platinum in the alloy  = 8.528 g - 0.802 g = 7.726 g

percentage  mass of platinum in the alloy = 7.726/8.528 × 100 = 772.600/8.528 = 90 .595 %

percentage mass of platinum in the alloy ≈ 90.60 %

Answer:

[tex]Kc=6.875x10^{-3}[/tex]

Explanation:

Hello,

In this case, for the given chemical reaction at equilibrium:

[tex]2 SO_3 ( g ) \rightleftharpoons 2 SO_ 2 ( g ) + O_ 2 ( g )[/tex]

The initial concentration of sulfur trioxide is:

[tex][SO_3]_0=\frac{0.660mol}{4.00L}=0.165M[/tex]

Hence, the law of mass action to compute Kc results:

[tex]Kc=\frac{[SO_2]^2[O_2]}{[SO_3]^2}[/tex]

In such a way, in terms of the change [tex]x[/tex] due to the reaction extent, by using the ICE method, it is modified as:

[tex]Kc=\frac{(2x)^2*x}{(0.165-2x)^2}[/tex]

In that case, as at equilibrium 0.11 moles of oxygen are present, [tex]x[/tex] equals:

[tex]x=[O_2]=\frac{0.110mol}{4.00L}=0.0275M[/tex]

Therefore, the equilibrium constant finally turns out:

[tex]Kc=\frac{(2*0.0275)^2*0.0275}{(0.165-2*0.0275)^2} \\\\Kc=6.875x10^{-3}[/tex]

Best regards.

Answer:

2.22 m

Explanation:

Step 1:

Data obtained from the question:

Frequency = (A + 88.3) MHz

We assume that the student ID is 20245347 as given in the question.

Therefore, A = 47 (last two digit of the 8-digit student ID)

Frequency = (47 + 88.3) MHz

Frequency = 135.3 MHz = 135.3x10^6 Hz

Wavelength =?

Recall:

Velocity of electromagnetic wave is 3x10^8 m/s2

Step 2:

Determination of the wavelength of the radio wave. This is illustrated below:

Velocity = wavelength x frequency

Wavelength = Velocity /frequency

Wavelength = 3x10^8 / 135.3x10^6

Wavelength = 2.22 m

Answer:

a) Balanced Overall Stoichiometric Equation

CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)

b) The reaction intermediates include Cl(g) and CCl₃(g)

c) The rate of reaction of the rate determining step is:

Rate = k [CHCl₃] [Cl]

d) The rate of overall reaction is given as

Rate = K [CHCl₃] √[Cl₂]

e) The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.

Explanation:

Step 1: Cl₂(g) → 2Cl(g) fast

Step 2: CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow

Step 3: CCl₃(g) + Cl(g) → CCl₄(g) fast

a) The overall reaction is a reaction between Chloroform (CHCl₃) and Chlorine (Cl₂). It can be obtained by summing all the elementary equations and eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations).

Cl₂(g) + CHCl₃(g) + Cl(g) + CCl₃(g) + Cl(g) → 2Cl(g) + CCl₃(g) + HCl(g) + CCl₄(g)

Now, eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations), we are left with

Cl₂(g) + CHCl₃(g) → HCl(g) + CCl₄(g)

CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)

b) Like I mentioned in (a), the reaction intermediates are the species that appear on both sides upon summing up all the elementary equations. They include:

Cl(g) and CCl₃(g)

c) The rate determining step is usually the slowest step among the elementary equations.

The rate of reaction expression is usually written from the slowest step.

The slowest step is step 2.

CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow

The rate of reaction is then given as

Rate = k [CHCl₃] [Cl]

d) The rate of reaction for the overall reaction is obtained by substituting for any intermediates that appear in the rate of reaction for the rate determining step

The rate of reaction of the rate determining step is:

Rate = k [CHCl₃] [Cl]

But we can substitute for [Cl] by obtaining an expression for it from the step 1.

Step 1: Cl₂(g) → 2Cl(g)

K₁ = [Cl]²/[Cl₂]

[Cl]² = K₁ [Cl₂]

[Cl] = √{K₁ [Cl₂]}

We then substitute this into the rate determining step

Rate = k [CHCl₃] [Cl]

Rate = k [CHCl₃] √{K₁ [Cl₂]}

Rate = (k)(√K₁) [CHCl₃] √[Cl₂]

Let (k)(√K₁) = K (the overall rate constant)

Rate = K [CHCl₃] √[Cl₂]

e) If the concentrations of the reactants are doubled, by what ratio does the reaction rate change

Old Rate = K [CHCl₃] √[Cl₂]

If the concentrations of the reactants are doubled, the new rate would be

New Rate = K × 2[CHCl₃] × √{2 × [Cl₂]}

New Rate = K × 2[CHCl₃] × √2 × √[Cl₂]

New Rate = 2√2 K [CHCl₃] √[Cl₂]

Old Rate = K [CHCl₃] √[Cl₂]

New Rate = 2√2 × (Old Rate)

The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.

Hope this Helps!!!!

Answer:

1) 16 mL of toluene is necessary to recrystallize 3.2 g of compound A

2) 2.4 g of A is the maximum amount that could be recovered.

3) 1.6 g of compound A  can be recovered if you accidentally use twice as much  toluene as was necessary

Explanation:

1)

The volume of the solvent ( toluene) = [tex]\frac{starting \ amount }{solubility \ at \boiling \ point }[/tex]

= [tex]\frac{3.2 \ g}{0.2 \ g/mL}[/tex]

= 16 mL

∴  16 mL of toluene is necessary to recrystallize 3.2 g of compound A

2)

The maximum amount    A that could be recovered if the saturated solution was allowed to cool to 0 ºC is determined by the difference of the starting amount and amount left in the solution at 0 ºC

i.e

maximum amount of A = 3.2 - ( 16 mL × 0.05 g/mL)

= 3.2 g - 0.8 g

= 2.4 g

∴  2.4 g of A is the maximum amount that could be recovered.

3)

Amount of A that would be  recovered, if you accidentally used twice as much toluene as was necessary is calculated as follows;

amount of A = 3.2g - (32 mL × 0.05 g/mL)

= 3.2 g - 1.6 g

= 1.6 g

Thus; 1.6 g of compound A  can be recovered if you accidentally use twice as much  toluene as was necessary

Answer:

mass of U-235  = 15.9 g (3 sig. figures)

Explanation:

1 atom can produce -------------------------> 3.20 x 10^-11 J energy

x atoms can produce ----------------------> 1.30 x 10^12 J energy

x = 1.30 x 10^12 / 3.20 x 10^-11

x = 4.06 x 10^22 atoms

1 mol ----------------------> 6.023 x 10^23 atoms

y mol ----------------------> 4.06 x 10^22 atoms

y = 0.0675 moles

mass of U-235 = 0.0675 x 235 = 15.8625

mass of U-235  = 15.9 g (3 sig. figures)

Answer:

3.71%

Explanation:

The phenolphthalein endpoint refers to the reactions:

OH⁻ + H⁺ → H₂O

CO₃⁻² + H⁺ → HCO₃⁻

While the methyl orange endpoint to:

HCO₃⁻ + H⁺ → H₂CO₃

So the additional volume required for the second endpoint tells us the amount of HCO₃⁻ species, which in turn is the total amount of Na₂CO₃ in the sample:

0.700 mL * 0.100 M * [tex]\frac{1mmolHCO_{3}^{-}}{1mmolHCl}[/tex] = 0.07 mmol HCO₃⁻

Now we calculate the mass of Na₂CO₃, using its molecular weight:

0.07 mmol HCO₃⁻ = 0.07 mmol Na₂CO₃

0.07 mmol Na₂CO₃ * 106 mg/mmol = 7.42 mg Na₂CO₃

No calculations using the volume of the first equivalence point are required because the problem already tells us the mass of the sample is 0.200 g.

0.200 g ⇒ 0.200 * 1000 = 200 mg

%Na₂CO₃ = 7.42 mg/200 mg * 100 = 3.71%

Answer:

-- 5.8×10⁻⁹ of the H is in 2P states at T=5900 K, a typical Sun surface temperature.

-- 3.3×10⁻¹² of the H is in 2P states at T=4300 K, a typical Sunspot temperature.

Explanation:

Answer:

Explanation:

your question seems to be  uncompleted , but however i have a solution to a similar question, check the attachment and follow the format to evaluate your question.

check the attachement

To calculate the starting concentration of ADP in micromolar, use the equilibrium constant and the concentrations of ATP, creatine, and creatine phosphate at the start.

To calculate the starting concentration of ADP in micromolar, we need to use the equilibrium constant and the concentrations of ATP, creatine, and creatine phosphate at the start. The equation for the creatine kinase reaction is:

ATP + Creatine → ADP + Creatine Phosphate

Given that the standard free energy change (ΔG°) is -12.6 kJ/mol and the ΔG is -0.1 kJ/mol, we can calculate the equilibrium constant (K) using the equation: ΔG = -RT ln K.

Using the given values, we can substitute them into the equation to solve for K and then use the concentrations of ATP, creatine, and creatine phosphate to calculate the starting concentration of ADP in micromolar.

https://brainly.com/question/28590870

#SPJ2

Using the values given the ADP concentration came as approximately 527.12 μM.

To find the starting concentration of ADP in micromolar, we can use the Gibbs free energy equation:

ΔG = ΔG° + RT ln(Q)

where ΔG is the Gibbs free energy change under cellular conditions, ΔG° is the standard Gibbs free energy change, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.

The creatine kinase reaction is:

Creatine + ATP <=> Creatine phosphate + ADP

Given:

We need to find [ADP]. First, rearrange the Gibbs free energy equation to solve for Q:

ΔG - ΔG°' = RT ln(Q)

Q = e^{(ΔG - ΔG°') / RT}

Substitute the known values (assuming T = 298 K):

Q = e^{(-100 - (-12600)) / (8.314 × 298)}

Q = e^{12500 / 2479.87}

Q = e^{5.04} ≈ 155.50

Substitute Q into the reaction quotient expression:

Q = [Creatine phosphate] [ADP] / [Creatine][ATP]

155.50 = (2.5 × 10⁻²) [ADP] / ((1.7 × 10⁻²) (5 × 10⁻³))

155.50 = (2.5 × 10⁻²) [ADP] / (8.5 × 10⁻⁵)

155.50 = (2.5 / 8.5) × 10³ [ADP]

155.50 = 0.295 [ADP] × 10³

ADP ≈ 155.50 / 0.295 ≈ 527.12 μM

The starting concentration of ADP is approximately 527.12 μM.

To calculate the number of moles of helium in a 4.0-liter flask at STP, divide the volume of the gas by the molar volume of a gas at STP (22.4 L/mol). This yields approximately 0.17857 moles of helium gas.

First, we need to know the volume of the flask, but since the student has not specified the volume correctly, let's assume it is '4.0 liters'. At STP, one mole of any gas occupies 22.4 liters. To find the number of moles in the flask, we use the molar volume of a gas at STP.

Here's the calculation:

Therefore, the flask contains approximately 0.17857 moles of helium gas.

Answer:

A. Kinetic energies of molecules increase.

B. Speeds of molecules increase.

C. Number of collisions per second increase.

Final answer:

Pressure changes according to Le Chatelier's principle and can be influenced by mechanical and thermal mechanisms, as well as by the kinetic activity of gas molecules.

Explanation:

Pressure changes can be understood in terms of Le Chatelier's principle, which states that when a change is imposed on a system at equilibrium, the system adjusts to counteract that change.

In the case of gases, increasing the pressure (by decreasing volume) leads to a shift in the equilibrium to reduce the number of gas particles, if possible, thereby reducing the pressure. Conversely, decreasing the pressure (by increasing volume) would have the opposite effect.

Pressure also varies due to thermal and mechanical mechanisms. Heating air causes it to rise and reduce surface air pressure, while cooling air causes it to descend, increasing pressure.

Mechanical changes occur when airflow is blocked, causing a build-up of air and increased pressure. Additionally, the kinetic theory of gases suggests that gases exert pressure because of the continuous motion of their particles, colliding with container walls and exerting force.

Furthermore, variations in pressure at different points of a fluid are important for driving the phenomena of buoyancy, as well as affecting divers and airplane passengers through changes in ambient pressure.

Answer:

Explanation:

Hydroponics is the process of growing crops using only water and liquid fertilizer. This process is great when your in the big city.

Hydroponics is a method of growing plants in a nutrient-rich water solution rather than soil, which allows precise control of nutrients and is used in research and commercial greenhouses for robust crop production.

Hydroponics is a highly efficient farming technique where plants are grown in a water-nutrient solution, rather than in soil. This method allows for precise control over the nutritional environment of the plants, which is why it is favored in scientific research for studying plant nutrient deficiencies and for producing robust, healthy crops.

In hydroponic systems, the need for soil is eliminated, and plants are given the exact nutrients they require directly. Because of this, hydroponics is used not only in laboratories but also in commercial greenhouse environments to cultivate flowers, vegetables, and other crops.

These crops are often resilient to pests and harsh conditions, contributing to sustainable food production and agricultural development.

Greenhouse management and hydroponics go hand in hand, as many plants grown hydroponically are also cultivated under controlled climates within greenhouses. The elimination of soil in hydroponics also helps mitigate the ecological, economic, and health concerns associated with excessive pesticide use in traditional agriculture.

Answer:

the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex]   is 0.0173 moles

Explanation:

The balanced chemical equation for the reaction is represented by:

[tex]H_3C_6H_5O_{7(aq)} + 3NaHCO_{3(aq)} ------>3CO_{2(g)}+3H_2O+Na_3C_6H_5O_{7(aq)}[/tex]

From above equation; we would realize that 3 moles of [tex]NaHCO_{3(aq)}[/tex] reacts with [tex]H_3C_6H_5O_{7(aq)}[/tex] to produce 3 moles of [tex]CO_{2(aq)}[/tex]

However ; the molar mass of [tex]NaHCO_{3(aq)}[/tex]  = 84 g/mol

mass given for [tex]NaHCO_{3(aq)}[/tex]  = 1.45 g

therefore , we can calculate the number of moles of [tex]NaHCO_{3(aq)}[/tex]  by using the expression :

number of moles of [tex]NaHCO_{3(aq)}[/tex]  = [tex]\frac{mass \ given}{ molar \ mass}[/tex]

number of moles of [tex]NaHCO_{3(aq)}[/tex]  = [tex]\frac{1.45}{84}[/tex]

number of moles of [tex]NaHCO_{3(aq)}[/tex]  = 0.0173 mole

Since the ratio of [tex]NaHCO_{3(aq)}[/tex] to [tex]CO_{2(aq)}[/tex] is 1:1; that implies that number of moles of [tex]NaHCO_{3(aq)}[/tex] is equal to number of moles of  [tex]CO_{2(aq)}[/tex] produced.

number of moles of  [tex]CO_{2(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]

0.0173 = [tex]\frac{mass \ given}{ 44 \ g/mol}[/tex]

mass of  [tex]CO_{2(aq)}[/tex]  = 0.0173 × 44

mass of  [tex]CO_{2(aq)}[/tex]  = 0.7612 g

Thus; the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex]   is 0.0173 moles

Answer:

Weight % of NH₃ in the aqueous waste  = 2.001 %

Explanation:

The chemical equation for the reaction

[tex]\\\\NH_3} + HCl -----> NH_4Cl[/tex]

Moles of HCl = Molarity × Volume

= 0.1039 × 31.89 mL × [tex]\frac{1 \ L}{1000 \ mL}[/tex]

= 0.0033 mole

Total mass of original sample = 25.888 g + 73.464 g

= 99.352 g

Total HCl taken for assay = [tex]\frac{10.762 \ g}{99.352 \ g}[/tex]

= 0.1083 g

Moles of NH₃ = [tex]\frac{0.0033 \ mol}{0.1083}[/tex]

= 0.03047 moles

Mass of NH₃ = number of  moles × molar mass

Mass of NH₃ = 0.03047 moles × 17 g

Mass of NH₃  = 0.51799

Weight % of NH₃  = [tex]\frac{0.51799 \ g}{25.888 \ g} * 100%[/tex]%

Weight % of NH₃ in the aqueous waste  = 2.001 %

Question:

Which of the following statements is TRUE?

A. Perpetual motion machines are a possibility in the near future.

B. The entropy of a system always decreases for a spontaneous process.

C. A spontaneous reaction is always a fast reaction.

D. There is a "heat tax" for every energy transaction.

E. None of the above are true.

Answer:

The correct answer is D)

There is a "heat tax" for every energy transaction.

Explanation:

Heat and work are two different ways in which energy is moved from one device to another. In the field of thermodynamics the distinction between Heat and Work is significant. The transfer of thermal energy between systems is heat. This is what is referred to as "heat tax".

No other statement in the question above is correct.

Cheers!

Answer:

Solution A has the higher percent ionization and the higher pH.

Explanation:

Percent ionization depends on the concentration of acid in a solution. If the solution having more concentration of acid so the percent ionization will be lower while if the solution have low amount of acid i. e. dilute solution so the percent ionization will be higher. In solution A, the concentration of HNO2 is lower which is an acid so the percent ionization is higher and the pH of the solution is also higher as compared to solution B.

Answer:

Solution A has the higher percent ionization and the higher pH

Explanation:

Answer:

The mechanism is SN2

Explanation:

See mechanism of monobromination of alkane attached

Answer:

Explanation:

find the solution below

Answer:

Explanation:

Industrial examples

Process Reactants, Product(s)

Ammonia synthesis (Haber–Bosch process) N2 + H2, NH3

Nitric acid synthesis (Ostwald process) NH3 + O2, HNO3

Hydrogen production by Steam reforming CH4 + H2O, H2 + CO2

Ethylene oxide synthesis C2H4 + O2, C2H4O

Answer:

The answer is A - Ethene gas reacts with hydrogen gas by using a nickel catalyst.

Explanation:

just did on Edge

The ratio of energy obtained from metabolizing stearic acid to fructose is 3:1.

To calculate the ratio of the energy the body gets metabolizing stearic acid to the energy the body gets metabolizing fructose, we need to compare their carbon content. The molecular formula of stearic acid is C18H36O2, which means it has 18 carbon atoms. The molecular formula of fructose is C6H12O6, which means it has 6 carbon atoms. Since the energy content of food is roughly proportional to the carbon content, we can calculate the ratio by dividing the number of carbon atoms in stearic acid (18) by the number of carbon atoms in fructose (6). This gives us a ratio of 3:1.

https://brainly.com/question/32218552

#SPJ3

The energy ratio of metabolizing one gram of stearic acid (a fatty acid) to one gram of fructose (a carbohydrate) is 2.25, with stearic acid providing 9 kcal/g and fructose providing 4 kcal/g of energy.

The question asks to calculate the ratio of the energy the body gets metabolizing each gram of stearic acid to the energy the body gets metabolizing each gram of fructose. Stearic acid is a fatty acid, and fructose is a simple carbohydrate. According to the data provided, each gram of carbohydrates yields approximately 4 kcal of energy, while each gram of fat yields about 9 kcal. Therefore, the ratio of energy from fats to carbohydrates is 9 kcal/g to 4 kcal/g.

To calculate the ratio of energy from stearic acid to fructose:

Identify the energy values: 9 kcal/g for stearic acid and 4 kcal/g for fructose.

Divide the energy value of stearic acid by the energy value of fructose: 9 kcal/g \/ 4 kcal/g = 2.25.

This means that metabolizing one gram of stearic acid yields 2.25 times the energy compared to metabolizing one gram of fructose.

Answer:

The one that will begin to precipitate first will be the lead chromate (PbCrO₄)

Explanation:

First of all, let's determine the equations involved:

Pb(NO₃)₂ →   Pb²⁺  +   2NO₃⁻

0.001          0.001       0.002

Ba(NO₃)₂  →  Ba²⁺  +  2NO₃⁻

0.100          0.100      0.200  

Sodium chromate as a soluble salt, can be also dissociated in:

Na₂CrO₄ →  2Na⁺ +  CrO₄²⁻

As the chromate can react to both cations of the aqueous solution, there will be formed 2 precipitates. When the saturation point is reached, which is determined by the Kps, everything that cannot be dissolved will precipitate.

The first to saturate the solution will precipitate first.

CrO₄²⁻ + Pb²⁺  ⇄  PbCrO₄

  s            s             s²  = Kps

 Kps = s² ⇒ [CrO₄²⁻] . [Pb²⁺] =  2.80×10⁻¹³

[CrO₄²⁻] . 0.001 = 2.80×10⁻¹³

[CrO₄²⁻] = 2.80×10⁻¹³ / 0.001 = 2.80×10⁻¹⁰

This is the concentration for the chromate when the lead chromate starts to precpitate.

CrO₄²⁻ + Ba²⁺  ⇄  BaCrO₄

  s            s             s²  = Kps  

Kps = [CrO₄²⁻] . [Ba²⁺]

1.17×10⁻¹⁰ = [CrO₄²⁻] . 0.100

[CrO₄²⁻] = 1.17×10⁻¹⁰ / 0.100 =  1.17×10⁻⁹

The first one that precipitates needs less chromate ion to start precipitating, in conclusion the one that will begin to precipitate first will be lead chromate.

The solid that will precipitate first is PbCrO₄.

To determine which solid will precipitate first, we need to compare the solubility product constants [tex](\( K_{\text{sp}} \))[/tex] for each possible precipitate. The compound with the lower [tex]\( K_{\text{sp}} \)[/tex] will precipitate first because it has lower solubility in water.

The solubility products given are:

- [tex]\( K_{\text{sp}} \) of BaCrO\(_4\) = \( 1.17 \times 10^{-10} \)[/tex]

- [tex]\( K_{\text{sp}} \) of PbCrO\(_4\) = \( 2.80 \times 10^{-13} \)[/tex]

We need to find the concentration of [tex]\(\text{CrO}_4^{2-}\) (\([ \text{CrO}_4^{2-} ]\))[/tex] at which each compound will begin to precipitate.

Calculation for BaCrO₄:

The reaction for the precipitation of BaCrO₄ is:

[tex]\[ \text{Ba}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{BaCrO}_4 (s) \][/tex]

The [tex]\( K_{\text{sp}} \)[/tex] expression is:

[tex]\[ K_{\text{sp}} = [\text{Ba}^{2+}] [\text{CrO}_4^{2-}] \][/tex]

Given:

[tex]\[ K_{\text{sp}} (\text{BaCrO}_4) = 1.17 \times 10^{-10} \][/tex]

[tex]\[ [\text{Ba}^{2+}] = 0.100 \, \text{M} \][/tex]

We can solve for [tex]\([ \text{CrO}_4^{2-} ]\)[/tex]:

[tex]\[ 1.17 \times 10^{-10} = (0.100) [\text{CrO}_4^{2-}] \][/tex]

[tex]\[ [\text{CrO}_4^{2-}] = \frac{1.17 \times 10^{-10}}{0.100} \][/tex]

[tex]\[ [\text{CrO}_4^{2-}] = 1.17 \times 10^{-9} \, \text{M} \][/tex]

Calculation for PbCrO₄:

The reaction for the precipitation of PbCrO₄ is:

[tex]\[ \text{Pb}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{PbCrO}_4 (s) \][/tex]

The [tex]\( K_{\text{sp}} \)[/tex] expression is:

[tex]\[ K_{\text{sp}} = [\text{Pb}^{2+}] [\text{CrO}_4^{2-}] \][/tex]

Given:

[tex]\[ K_{\text{sp}} (\text{PbCrO}_4) = 2.80 \times 10^{-13} \][/tex]

[tex]\[ [\text{Pb}^{2+}] = 0.001 \, \text{M} \][/tex]

We can solve for [tex]\([ \text{CrO}_4^{2-} ]\)[/tex]:

[tex]\[ 2.80 \times 10^{-13} = (0.001) [\text{CrO}_4^{2-}] \][/tex]

[tex]\[ [\text{CrO}_4^{2-}] = \frac{2.80 \times 10^{-13}}{0.001} \][/tex]

[tex]\[ [\text{CrO}_4^{2-}] = 2.80 \times 10^{-10} \, \text{M} \][/tex]

Comparison:

- The concentration of [tex]\(\text{CrO}_4^{2-}\)[/tex] needed to precipitate BaCrO₄ is [tex]\( 1.17 \times 10^{-9} \, \text{M} \)[/tex].

- The concentration of [tex]\(\text{CrO}_4^{2-}\)[/tex] needed to precipitate PbCrO₄ is [tex]\( 2.80 \times 10^{-10} \, \text{M} \)[/tex].

Since [tex]\( 2.80 \times 10^{-10} \, \text{M} \)[/tex] is smaller than[tex]\( 1.17 \times 10^{-9} \, \text{M} \), PbCrO\(_4\)[/tex] will precipitate first.

The initial pH of 0.100 M HNO2 is approximately 2.17. It takes 20.0 mL of 0.200 M KOH to reach the equivalence point. The subsequent pH values at various points in the titration reflect the changing concentrations of HNO2 and OH-.

A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4) is titrated with 0.200 M KOH.

a. The pH when no base is added

To find the initial pH, we first need to calculate the concentration of H3O+ using the equilibrium expression for the weak acid:

HNO2 ⇌ H+ + NO2-

Using Ka, we get:

Ka = [H+][NO2-] / [HNO2]

4.6 x 10-4 = x2 / 0.100

Solving for x, we get:

x = √(4.6 x 10-4 * 0.100) = 0.00678 M

pH = -log(0.00678) ≈ 2.17

b. The volume of KOH required to reach the equivalence point

The moles of HNO2 are:

0.040 L * 0.100 M = 0.004 mol

Since KOH and HNO2 react in a 1:1 molar ratio, the volume of 0.200 M KOH required is:

0.004 mol / 0.200 M = 0.020 L = 20.0 mL

c. The pH after adding 5.00 mL of KOH

Moles of KOH added:

0.005 L * 0.200 M = 0.001 mol

Moles of HNO2 remaining:

0.004 mol - 0.001 mol = 0.003 mol

Concentration of HNO2 remaining = 0.003 mol/0.045 L = 0.0667 M

Concentration of NO2- formed = 0.001 mol/0.045 L = 0.0222 M

Using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

pH = -log(4.6 x 10-4) + log(0.0222/0.0667) ≈ 3.35

d. The pH at one-half the equivalence point

At one-half the equivalence point, the concentration of [A-] = [HA], so:

pH = pKa = -log(4.6 x 10-4) ≈ 3.34

e. The pH at the equivalence point

At the equivalence point, all HNO2 has been converted to NO2-:

NO2- will hydrolyze to produce OH-:

NO2- + H2O ⇌ HNO2 + OH-

Using Kb for NO2-:

Kb = Kw/Ka = 1.0 x 10-14 / 4.6 x 10-4 = 2.17 x 10-11

Setting up the equation:

Kb = [OH-][HNO2] / [NO2-]

2.17 x 10-11 = x2 / 0.100

Solving for x:

x = √(2.17 x 10-11 * 0.100) = 1.47 x 10-6

pOH = -log(1.47 x 10-6) ≈ 5.83

pH = 14 - 5.83 = 8.17

f. The pH after 30 mL of the base is added

Moles of KOH added:

0.030 L * 0.200 M = 0.006 mol

Excess moles of OH-:

0.006 mol - 0.004 mol = 0.002 mol

Concentration of OH- in the total volume:

0.002 mol / 0.070 L = 0.02857 M

pOH = -log(0.02857) ≈ 1.54

pH = 14 - 1.54 = 12.46

The correct answers are: (a). [H⁺] = 2.17; (b). [tex]\text{Volume} = 20.0\ mL[/tex]; (c). [tex][NO_2^-] = \text{0.0222 M}[/tex]; (d). pH = 3.34; (e). [tex]\text{pH} = 8.02[/tex]; (f). pH = 12.46.

Let's solve the different parts of the titration problem step-by-step:

a. pH when no base is added:

First, we need to find the pH of a 0.100 M HNO₂ solution. HNO₂ is a weak acid and it partially ionizes in water:

The expression for the acid dissociation constant Ka is:

Assuming that the initial concentration of HNO₂ is C0 = 0.100 M and the change in concentration is x:

Assuming x is small relative to 0.100 M:

b. Volume of KOH required to reach the equivalence point:

c. pH after adding 5.00 mL of KOH:

Using the Henderson-Hasselbalch equation:

d. pH at one-half the equivalence point:

At one-half the equivalence point, [HNO₂] = [NO₂⁻], so pH = pKa.

e. pH at the equivalence point:

At the equivalence point, all HNO₂ has reacted to form NO₂⁻. The solution contains NO₂⁻ ions, which hydrolyze:

Let x be the concentration of OH⁻:

f. pH after 30 mL of the base is added:

Answer:

1.52V

Explanation:

Oxidation half equation:

2Al(s)−→2Al^3+(aq) + 6e

Reduction half equation

3Sn2^+(aq) + 6e−→3Sn(s)

E°cell= E°cathode - E°anode

E°cathode= −0.140 V

E°anode= −1.66 V

E°cell=-0.140-(-1.66)

E°cell= 1.52V

Final answer:

The overall cell potential for the redox reaction between tin and aluminum in a galvanic cell is calculated as +1.52 V, indicating a spontaneous reaction with tin being oxidized at the anode and aluminum reduced at the cathode.

Explanation:

To calculate the standard cell potential for a redox reaction involving tin (Sn) and aluminum (Al), we apply the reduction potentials of their respective half-reactions. The half-reaction for tin is as follows: Sn(s) ightarrow Sn2+(aq) + 2 e - , with an associated standard reduction potential (E & Ocirc ;) of - 0.140 V, however, its oxidation potential is actually +0.140 V.

For aluminum, the half-reaction is: Al3+(aq) + 3 e - ightarrow Al(s), with an E & Ocirc ; of - 1.66 V. In a galvanic cell, the aluminum will oxidize, and since it’s a reduction potential, for oxidation, we take the negative of this value, which would make it +1.66 V.

To find the overall cell potential, we use the equation Ecell = Ecathode - Eanode. In this reaction, Sn(s) is our anode, and Al(s) is our cathode. However, since aluminum's E & Ocirc ; is already negative (signifying oxidation), we reverse its sign for use in the cell potential equation.

Ecell = Ecathode - Eanode = (+1.66 V) - (+0.14 V) = +1.66 V - 0.14 V = +1.52 V

The standard cell potential is positive, indicating that the redox reaction is spontaneous. Tin is oxidized at the anode, and aluminum is reduced at the cathode, forming the basis for electric current flow in the cell.

Answer:

pH=pKa

pH=4.44

Explanation:

Since the titration occur between a weak acid and a strong base.

then at half -equivalence point, the pH of the solution is equals to the pKa of the weak acid.

Therefore, pH=pKa

Ka of weak acid=3.6×10^−5

To calculate the pKa of the weak acid using the express below;

pKa =- log(Ka)

p​K​a​=​−​l​o​g​(​3.6×10−5)​=​4.44

From the question, the pKa of the solution is at half -equivalence point

Then,

pH=pKa

pH=4.44

The question says that the titration occurred between a weak acid and a

strong base at half-equivalence point. Then we can deduce that the pH of

the solution is equal to the pKa of the weak acid.

pH=pKa

Ka of monoprotic weak acid=3.6×10⁻⁵

The pKa of the monoprotic weak acid will be calculated by :

pKa = - log(Ka)

p​K​a ​=​ −​l​o​g​(​3.6×10⁻⁵)​ =​ 4.44

Since the pKa of the solution is at half -equivalence point

pH=pKa

pH=4.44

Read more on https://brainly.com/question/15867090

Answer:

1. Partial pressure of N2 is 204.5 mmHg

2. Partial pressure of Cl2 is 613.5 mmHg

Explanation:

Step 1:

The equation for the reaction. This is given below:

NCl3 —> N2 (g) + Cl2 (g)

Step 2:

Balancing the equation.

NCl3 (l) —> N2 (g) + Cl2 (g)

The above equation is balanced as follow:

There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by putting 2 in front of NCl3 as shown below:

2NCl3 (l) —> N2 (g) + Cl2 (g)

There are 6 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of Cl2 as shown below:

2NCl3 (l) —> N2 (g) + 3Cl2 (g)

Now the equation is balanced.

Step 2:

Determination of the mole fraction of each gas.

From the balanced equation above, the resulting mixture of the gas contains:

Mole of N2 = 1

Mole of Cl2 = 3

Total mole = 4

Therefore, the mole fraction for each gas is:

Mole fraction of N2 = mole of N2/total mole

Mole fraction of N2 = 1/4

Mole fraction of Cl2 = mole of Cl2/total mole

Mole fraction of Cl2 = 3/4

Step 3:

Determination of the partial pressure of N2.

Partial pressure = mole fraction x total pressure

Total pressure = 818 mmHg

Mole fraction of N2 = 1/4

Partial pressure of N2 = 1/4 x 818

Partial pressure of N2 = 204.5 mmHg

Step 4:

Determination of the partial pressure of Cl2.

Partial pressure = mole fraction x total pressure

Total pressure = 818 mmHg

Mole fraction of Cl2 = 3/4

Partial pressure of Cl2 = 3/4 x 818

Partial pressure of Cl2 = 613.5 mmHg

Answer:

[tex]p_{N_2}=204.5mmHg\\p_{Cl_2}=613.5mmHg[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NCl_3(g)\rightarrow N_2(g)+3Cl_2(g)[/tex]

Thus, by knowing that the nitrogen trichloride is completely decomposed, one assumes there is one mole of nitrogen and three moles of chlorine (stoichiometric coefficients) as a basis to compute the partial pressures since they have the mole ratio from the nitrogen trichloride. Hence, the mole fractions result:

[tex]x_{N_2}=\frac{1}{1+3}=0.25\\ x_{Cl_2}=1-0.25=0.75[/tex]

In such a way, for the final pressure 818 mmHg, the partial pressures become:

[tex]p_{N_2}=x_{N_2}p_T=0.25*818mmHg=204.5mmHg\\p_{Cl_2}=x_{Cl_2}p_T=0.75*818mmHg=613.5mmHg[/tex]

Best regards.

Answer:J.K^-1. kg^-1

Explanation:

Heat capacity= H

Mass=m

Specific heat capacity= c

H=mc

J.K^-1 = c ×kg

c= J.K^-1/kg

=J.K^-1.kg^-1

To solve this problem, we can use the ideal gas law equation PV=nRT. We can find the number of moles in the first balloon using the given information, and then use that value to find the volume of the second balloon.

To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we can use the given information to find the number of moles in the first balloon. Rearranging the ideal gas law equation, we have n = PV / RT. Plugging in the values, we get n = (5.2 mol)(23.5 L) / (0.0821 atm L/mol K)(T in Kelvin).

Once we have the number of moles for the first balloon, we can use this value to find the volume of the second balloon. Rearranging the ideal gas law equation, we have V = nRT / P. Plugging in the values and solving for V, we get V = (5.2 mol)(0.0821 atm L/mol K)(T in Kelvin) / (P)