When The reaction below demonstrates which characteristic of a base?

Upper C U upper S upper O subscript 4 (a q) plus 2 upper N a upper O upper H (a q) right arrow upper C u (upper O upper H) subscript 2 (s) plus upper N a subscript 2 upper S upper O subscript 4 (a q).does phenolphthalein turn pink?

Answer:

90/38 Sr

Explanation:

Gamma decay involves rearrangements in an atomic nucleus usually following a nuclear change. When a nuclear change occurs, the daughter nuclei is in excited state, a high energy state. The nucleons quickly rearrange themselves and rapidly lower the energy of the daughter nucleus. The corresponding amount of excited state energy is emitted as the daughter nuclei return to ground state as short range, high energy electromagnetic gamma rays.

Hence, the number of nucleons in the nucleus remain the same before and after a gamma decay.

Answer:

11.88

Explanation:

The capacity of container to be able to hold any amount of moles is dependent on size of container.

Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.

It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.

It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Molar mass has units of gram per mole . In case of molecules, where molar mass  in grams present in one mole  of atoms is its atomic mass.

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Answer:

Q< K hence a precipitate will not form.

Explanation:

First convert the concentration to molL-1

For number of moles of K^+ = 0.05×20/1000 = 1×10^-3 moles

If we have 1×10^-3 moles in 20cm3

Then in 1000cm^3 we have 1×10^-3 moles×1000/20= 0.05 M

For ClO4^-= 0.50×80/1000= 0.04 moles

If we have 0.04 moles in 80cm3

Then in 1000cm^3 we have 0.04 moles×1000/80= 0.5 M

Q= [K^+] [ClO4^-]

Q= [0.05] [0.5]

Q= 0.025= 2.5×10-2

Q< K hence a precipitate will not form.

Answer:2.50 gN2

Explanation:

Pair A is:

2.50 g N2

Pair B is:

21.5 g N2

Pair C is:

0.081 CO2

Answer:

Explanation:

Hi there,

To get started, recall the Heat-specific heat capacity equation of a substance:

Q=mCΔT

Our final temperature is 47.8 °C, since this is the way it is worded in the response (from temperature X to temperature Y)

Quite simply, we can go ahead and plug in mass, specific heat capacity, and change in temperature as all the units match up!

[tex]Q= (525 g)(0.21 \ cal\ / g*C^{o} )(47.8C^{o}-13.0 C^{o})=3836.7 \ cal[/tex]

Study well and persevere.

thanks,

Final answer:

3823.8 calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C, calculated using the formula Q = mcΔT with a specific heat of 0.21 cal/g°C.

Explanation:

To find how many calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C, we use the formula for heat transfer Q = mcΔT, where 'm' is the mass of the substance, 'c' is the specific heat of the substance, and ΔT is the change in temperature.

First, we calculate the change in temperature (ΔT): ΔT = final temperature - initial temperature = 47.8°C - 13.0°C = 34.8°C. Using the provided values: m (mass) = 525g c (specific heat of Aluminum) = 0.21 cal/g°C ΔT (change in temperature) = 34.8°C

We plug these values into the equation to find the heat (Q): Q = m·c·ΔT Q = (525g)(0.21 cal/g°C)(34.8°C) Q = 3823.8 calories.

Therefore, 3823.8 calories of heat are required to raise the temperature of 525g of Aluminum from 13.0°C to 47.8°C.

Answer:

The average rate of formation of I₂(g) is 6.00x10⁻⁵Ms⁻¹

Explanation:

The question is:

The average rate of formation of I2 over the same time period is______M s-1.

Based in the reaction:

2 HI(g) ⇄ H₂(g) + I₂(g)

If 2 moles of HI(g) disappears in a rate of 1.20x10⁻⁴Ms⁻¹, 1 mole of  I₂(g) will appears at:

1 mole I₂(g) × (1.20x10⁻⁴Ms⁻¹ / 2 mol) = 6.00x10⁻⁵Ms⁻¹ I₂(g)

This is an incomplete question, here is a complete question.

The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?

[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]

Answer : The value of [tex]K_p[/tex] is, [tex]5.79E^{-7}[/tex]

Explanation :

The relation between [tex]K_c[/tex] and [tex]K_p[/tex] is:

[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]

where,

[tex]K_p[/tex] = equilibrium constant at constant pressure  = [tex]3.3\times 10^{-5}[/tex]

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 700 K

[tex]\Delta n[/tex] = change in number of gaseous moles = Product moles - Reactant moles = (2+1) - (2) = 3 - 1 = 1 mol

[tex]K_c[/tex] = equilibrium constant

Now put all the given values in the above expression, we get:

[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]

[tex]3.3\times 10^{-5}=K_c(0.0821\times 700)^{1}[/tex]

[tex]K_c=5.7\times 10^{-7}=5.7E^{-7}[/tex]

Therefore, the value of [tex]K_p[/tex] is, [tex]5.7E^{-7}[/tex]

The value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]

The given parameters;

The change in the number of gaseous moles is calculated as follows;

[tex]\Delta n = product \ - reactant\\\\\Delta n = (2+1) - 2 = 1 \ mole[/tex]

The value of the equilibrium constant (Kc) is calculated as follows;

[tex]K_p = K_c \times (RT)^{\Delta n}\\\\3.3 \times 10^{-5} = K_c \times (0.0821 \times 700)^1\\\\3.3 \times 10^{-5} = 57.47K_c\\\\K_c = \frac{3.3 \times 10^{-5} }{57.47} \\\\K_c = 5.74 \times 10^{-7} \[/tex]

Thus, the value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]

"Your question is not complete, it seems to be missing the following information;"

The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?

[tex]2SO_3(g) \ ---> \ 2SO_2(g) \ + \ O_2[/tex]

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Answer:

Explanation:

During a phase change, the temperature remains constant because the energy supplied via heat is used differently. While not changing states, the heat (energy) goes into changing the kinetic energy of every particle in the body that receives it.

Answer:

1.12 g of oxygen.

Explanation:

First step is to determine the decomposition reaction:

Ammonium perchlorate → NH₄ClO₄

2NH₄ClO₄ (s) →  N₂(g) + Cl₂(g) + 2O₂(g) + 4H₂O(g)

We define the moles of the reactant:

4.1 g . /117.45 g/mol = 0.0349 moles

As ratio is 2:2, If I use 0.0349 moles of perchlorate I would produce the same amount of moles of oxygen.

0.0349 mol of O₂ are produced by the reaction

We convert to mass:  0.0349 mol . 32 g /1mol = 1.12 g

Answer:

absorbed

Explanation:

Answer:

4.77 is the pH of the given buffer .

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})[/tex]

We are given:

[tex]K_a[/tex] = Dissociation constant of propanoic acid = [tex]1.3\times 10^{-5}[/tex]

[tex][CH_3CH_2COONa]=0.254 M[/tex]

[tex][CH_3CH_2COOH]=0.329 M[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})[/tex]

pH = 4.77

4.77 is the pH of the given buffer .

Final answer:

To determine the pH of a buffer with given concentrations of propanoate and propanoic acid, apply the Henderson-Hasselbalch equation using the given Ka value for propanoic acid and the concentrations of the acid and its conjugate base.

Explanation:

To calculate the pH of a buffer consisting of 0.254 M sodium propanoate (CH3CH2COO-Na+) and 0.329 M propanoic acid (CH3CH2COOH), we use the Henderson-Hasselbalch equation:

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

Where:

First, we calculate the pKa:

pKa = -log(Ka) = -log(1.3 x 10-5)

Then plug the values into the Henderson-Hasselbalch equation:

pH = pKa + log(0.254/0.329)

After calculating the above expression, we find the pH of the buffer solution.

Answer:

278 balloons can be aired.

Explanation:

We apply the Ideal Gases Law to determine the total available volume for each baloon of 0.75L. So we need to divide the volume we obtain by the volume of each baloon.

We determine the moles of CO₂

375 g. 1 mol / 44g = 8.52 moles

V = (n . R . T) / P → (8.52 mol . 0.082 . 298K) / 1 atm = 208.3 L

That's the total volume from the tank, so we can inflate (208.3 / 0.75) = at least 278 balloons

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

First, we will convert 375 g of CO₂ to moles using its molar mass (44.01 g/mol).

[tex]375 g \times \frac{1mol}{44.01g} = 8.52 mol[/tex]

1 mole of CO₂ at 1.0 atm and 298 K occupies 24.4 L. The volume occupied by 8.52 moles of CO₂ is:

[tex]8.52 mol \times \frac{24.4L}{mol} = 208 L[/tex]

8.52 moles of CO₂ at 1.0 atm and 298 K occupies 208 L. The number of 0.75 L-balloons that can be filled is:

[tex]208 L \times \frac{1balloon}{0.75L} \approx 277 balloon[/tex]

A tank that contains 375 g of CO₂ at 1.0 atm and 298 K can be used to fill 277 0.75 L-balloons.

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Answer:

pH after the addition of 10 ml NaOH = 4.81

pH after the addition of 20.1 ml NaOH = 8.76

pH after the addition of 25 ml NaOH = 8.78

Explanation:

(1)

Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles  = 2 x 10⁻³ moles,

Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles

                          CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O

Initial conc.            2 x 10⁻³                 1 x 10⁻³           0            

Equilibrium             1 x 10⁻³                   0                  1 x 10⁻³

Final volume = 20 + 10 = 30 ml = 0.03 lit

So final concentration of Acid = [tex]\frac{0.001}{0.03} = 0.03mol/lit[/tex]

Final concentration of conjugate base [CH₃CH₂CH₂COONa][tex]=\frac{0.001}{0.03} = 0.03 mol/lit[/tex]

Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .

Using Henderson Hasselbalch equation to find the pH

[tex]pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81[/tex]

During titration of butanoic acid with NaOH, we can calculate the pH at various points using the Henderson-Hasselbalch equation for buffer scenarios. After 10.00mL of NaOH, the pH will be 4.74. After 20.10 mL, the pH will be 8.27, and after 25.00 mL, the pH will be 12.30.

This involves calculating the pH at various stages during a titration procedure. Here the titration involves a weak acid, butanoic acid, with a strong base, NaOH. We can simplify the reaction as follows: CH₃CH₂CH₂COOH + OH- --> CH₃CH₂CH₂COO- + H₂O.

(a) After 10.00 mL of NaOH is added, the system isn't at equivalence. Here, the reaction hasn't fully completed and a buffer solution is present. Using the Henderson-Hasselbalch equation, we can find the pH: pH = pKa + log([base]/[acid]). After calculating, we can find pH = 4.74.

(b) After 20.10 mL NaOH is added, the system reaches past equivalence. The pH can be determined by finding pOH using the remaining OH- concentration and then subtracting from 14. After the calculation, the pH = 8.27.

(c) For 25.00 mL of NaOH, the system is beyond equivalence and extra OH- ions increase pH. The pH calculation is like previous step and the result will be pH = 12.30.

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Answer:

The gaps are filled using the following;

1. removes

2. from

3. less

4. Protonated

5. Ortho, para

Explanation:

Electron density is removed from nitrogen in acetyl group which is why it is less basic

The complete sentences is now written as;

The acetyl group removes some of the electron density from the nitrogen, making it much less basic; the nitrogen of this amide is protonated under the reaction conditions. The N retains enough electron density to share with the benzene ring, so the NHCOCH3 group is still an activating ortho, para director, though weaker than NH2.

Answer:

B) 655, 485, 433, 409 nm

Explanation:

655, 485, 433, 409 nm

The greater the energy change involved in a transition, the shorter the wavelength.

Transitions to n = 2 from n = 3, 4, 5 and 6 give lines of wavelength 655, 485, 433 and 409 nm. Other answer choices represent the Lyman. Brackett, and Pfund series.

Answer:

Explanation:

check the attachment for the propose neutral  structure for each compound that is  consistent with the data.

The structures are:

A: 1-phenylethyl lactate

B:lactic acid

C: 1-phenylethanol

D: 1-phenylethanol

E: Ester formed from 1-phenylethanol lactic acid chloride

To solve this problem, we will follow the information given in the question step by step to deduce the structures of compounds A, B, C, D, and E.

1. Saponification Reaction: The ester A reacts with NaOH to produce an alcohol and a carboxylate salt. The fact that 5.30 mL of 1.00 M HCl is required to titrate the unused NaOH tells us that the rest of the NaOH was used up in the saponification reaction. We can use the moles of NaOH reacted to find the moles of ester A that reacted.

First, calculate the moles of NaOH used in the saponification:

Since the reaction is 1:1, the moles of ester A are also 9.70 mmol.

2. Molecular Formula of A: Since we have 2.00 g of A a[tex]\[ \text{Moles of NaOH used} = (15.00 \text{ mL} - 5.30 \text{ mL}) \times \frac{1.00 \text{ mmol}}{\text{mL}} \] \[ \text{Moles of NaOH used} = 9.70 \text{ mmol} \][/tex]nd we know the moles of A, we can calculate the molar mass of A:

[tex]\[ \text{Molar mass of A} = \frac{\text{mass of A}}{\text{moles of A}} \] \[ \text{Molar mass of A} = \frac{2.00 \text{ g}}{9.70 \text{ mmol}} \] \[ \text{Molar mass of A} \approx 206.19 \text{ g/mol} \][/tex]

Given that A contains only C, H, and O, we can use the molar mass to propose a molecular formula. A common approach is to start with the simplest ratio and increase it until the calculated molar mass is close to the experimental value. For example, if we start with C₃H₆O₂ (molar mass 88.09 g/mol), we can see that it's not enough. We can keep increasing the number of carbons until we reach a molar mass close to 206.19 g/mol. Let's assume the formula is C₁₂H₂₄O₄ (molar mass 248.34 g/mol), which is a common ester formula that would give us a molar mass close to the experimental value.

3. Structure of A: Since A is not oxidized by K₂Cr₂O₇ and does not decolorize Br₂ in CH₂Cl₂, it suggests that A does not have any carbon-carbon double bonds or carbon-carbon triple bonds. Therefore, A is a saturated ester. Given that A is optically active, it must contain a chiral center. A simple saturated ester with a chiral center could be butanoic acid ethyl ester with a methyl group replaced by a hydroxyl group on the alpha carbon, making it chiral.

4. Structure of B and C: B is the acid product, and C is the alcohol product from the saponification of A. Since C is oxidized to acetophenone by K2Cr2O7, C must be a secondary alcohol with a benzene ring attached to the carbon adjacent to the hydroxyl group. Therefore, C is 1-phenylethanol. B, being the acid counterpart to C, must be benzoic acid.

5. Structure of D: D is formed by the reduction of acetophenone with NaBH4. This reaction would convert the ketone to a secondary alcohol. Therefore, D is 1-phenylethyl alcohol, which is the same as C.

6. Structure of E: E is formed when D reacts with the acid chloride derived from B. Since A is reformed, E must be the product of the reaction between benzoyl chloride (from B) and 1-phenylethyl alcohol (D). E is therefore a benzoate ester with a phenylethyl group.

Based on the above analysis, the proposed structures are:

A: 1-phenylethyl lactate

B:lactic acid

C: 1-phenylethanol

D: 1-phenylethanol

E: Ester formed from 1-phenylethanol lactic acid chloride

This proposal is consistent with the given data and the reactivity of the compounds involved.

The reaction will consume 3 moles of iron(III) oxide and will produce 3 moles of aluminum oxide and 12 moles of iron, following the mole ratio from the balanced chemical equation.

Starting from the balanced chemical equation, we can see that the ratio of "Fe2O3" (iron(III) oxide) to "Al" (aluminum) is 1:2. This ratio allows us to calculate the amount of moles that will react or produce. We can apply these ratios in the following way:

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Answer:

Explanation:

pKa = 3.86

-log Ka = pKa

- log Ka = 3.86

Ka = 1 / ( 10^(3.86)) = 1.38 × 10⁻⁴

lactic → (H⁺) + lactate

but Ka ( equilibrium constant ) = ( H⁺) (lactate) / ( lactic acid)

when HCl dissociate, it forms

HCl → H ⁺ + Cl⁻

0.08 M of HCl will yield 0.08 M H⁺, lactate ion = 0.1 M

1.38 × 10⁻⁴ = (0.08) (lactate) / 0.1

(1.38 × 10⁻⁴  × 0.1) / 0.08 = lactate

lactate = 1.73 × 10⁻⁴

The Electronic configuration, of an element is 1s² 2s² 2p¹. The number of valance electrons present in the element is 3.

Electronic configuration refers to the arrangement of electrons within the orbitals of an atom or ion. It describes how the electrons are distributed among the various energy levels, sublevels, and orbitals within an atom.

The electronic configuration is often represented using a shorthand notation known as the Aufbau principle or the electron configuration notation. In this notation, the principal energy levels (denoted by the quantum number n) are represented by numbers (1, 2, 3, etc.), and the sublevels (s, p, d, f) are represented by letters. The superscript numbers indicate the number of electrons occupying each sublevel.

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Final answer:

The element with the electron configuration 1s2 2s2 2p1 has a total of 3 valence electrons, from the occupied 2s and 2p subshells in the second energy level.

Explanation:

The Electronic configuration, of an element is 1s² 2s² 2p¹. The number of valance electrons present in the element is 3.

Electronic configuration refers to the arrangement of electrons within the orbitals of an atom or ion. It describes how the electrons are distributed among the various energy levels, sublevels, and orbitals within an atom.

The electronic configuration is often represented using a shorthand notation known as the Aufbau principle or the electron configuration notation. In this notation, the principal energy levels (denoted by the quantum number n) are represented by numbers (1, 2, 3, etc.), and the sublevels (s, p, d, f) are represented by letters. The superscript numbers indicate the number of electrons occupying each sublevel.

Answer:

11.3 g of H₂O will be produced.

Explanation:

The combustion is:

2C₈H₁₈ +  25O₂→  16CO₂  +  18H₂O

First of all, we determine the moles of the reactants in order to find out the limiting reactant.

8 g / 114g/mol = 0.0701 moles of octane

37g / 32 g/mol = 1.15 moles of oxygen

The limiting reagent is the octane. Let's see it by this rule of three:

25 moles of oxygen react to 2 moles of octane so  

1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.

We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:

2 moles of octane produce 18 moles of water

Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.

We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.

Answer:

1.64g

Explanation:

The reaction scheme is given as;

2-bromocyclohexanol  --> 1,2-epoxycyclohexane + HBr

From the reaction above,

1 mol of 2-bromocyclohexanol produces 1 mol of 1,2-epoxycyclohexane

3.0 grams of trans-2-bromocyclohexanol.

Molar mass = 179.05 g/mol

Number of moles = mass / molar mass = 3 / 179.05 = 0.016755 mol

This means  0.016755 mol of 1,2-epoxycyclohexane would be produced.

Molar mass =  98.143 g/mol

Theoretical  yield = Number of moles * Molar mass

Theoretical  yield =  0.016755 * 98.143 ≈ 1.64g

Answer:

vhgbvbhdf

Explanation:

(a) It is given that the gas is ideal. Formula for change in entropy of the gas according to the first law of thermodynamics is as follows.

       [tex]\Delta U = dQ - dW[/tex]

For isothermal process, [tex]\Delta U = 0[/tex] at constant temperature.

So,   [tex]\Delta U = dQ - dW[/tex]

      [tex]0 = dQ - dW[/tex]

or,       dQ = dW = 730 J

Now, according to the second law of thermodynamics the entropy change is as follows.

           [tex]\Delta S = \frac{dQ}{dT}[/tex]

                        = [tex]\frac{730 J}{293.15 K}[/tex]

                        = 2.490 J/K

Therefore, the entropy change of the gas is 2.490 J/K.

(b)  In the given process, at constant temperature the gas will be compressed slowly because then kinetic energy of the gas molecules will also be constant. The volume decreases so that the movement of molecules increases as a result, entropy of molecules will also increase.

This means that [tex]\Delta S = 2.490 J/K > 0[/tex]

An ideal gas is a hypothetical gas in which there are no intermolecular attractions between the molecules of a gas. The collision between the molecules is perfectly elastic.

The answers are as follows:

(a) The ideal gas is given, in which the change of entropy of the gas can be calculated by the first law of thermodynamics.

[tex]\Delta[/tex] U = dQ - dW

In an isothermal process, the \Delta U = 0 at constant temperature, such that:

[tex]\Delta[/tex] U = dq - dW

0 = dQ - dW

dQ = dW = 730 J (given)

Now, based on the second law of thermodynamics, the entropy change will be:

[tex]\Delta S = \dfrac {\text {dQ}}{\text {dW}}\\\\\Delta S = \dfrac {730 \text J}{293.15 \text K}[/tex]

[tex]\Delta[/tex]S = 2.490 J/K

Thus, the entropy change for the gas is 2.490 J/K.

(b) At constant temperature, the gas will be compressed slowly because the kinetic energy of the molecules will be constant. The decrease in the volume will be due to the increase in the movement of the molecules, which will cause an increase in the entropy. This means [tex]\Delta[/tex] S will be greater than  2.490 J/K.

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Answer:

0.1 mol O2

Explanation:

1 mol of any gas at STP = 22.4 L

2.24L/22.4L = 0.1

1mol * 0.1 = 0.1 mol

you have 0.1 mol of O2

The number of moles of [tex]O_2[/tex]  is required.

The number of moles of [tex]O_2[/tex] is 0.2 moles.

V = Volume = 2.24 L

[tex]\rho[/tex] = Density of [tex]O_2[/tex] at STP = 1.429 g/L

M = Molar mass of [tex]O_2[/tex] = 16 g/mol

Density is given by

[tex]\rho=\dfrac{m}{V}\\\Rightarrow m=\rho V\\\Rightarrow m=1.429\times 2.24\\\Rightarrow m=3.20096\ \text{g}[/tex]

Molar mass is given by

[tex]M=\dfrac{m}{n}\\\Rightarrow n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{3.20096}{16}\\\Rightarrow n=0.2\ \text{moles}[/tex]

The number of moles of [tex]O_2[/tex] is 0.2 moles.

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Answer: Option (b) is the correct answer.

Explanation:

When the student opens up the bottle lid then due to the difference in atmospheric pressure and pressure inside the bottle will lead the carbon dioxide to form more number of bubbles.

Also, when the bottle opens up then tiny bubbles around the neck of bottle will tend to float towards the open surface. Hiss sound appears because on opening the lid of bottle excess gas tends to come out of the bottle causing the sound.

Thus, we can conclude that when the student opens the bottle and hears a loud hiss as gas under pressure escapes from the bottle the bubbles will grow, and more may appear.

When a carbonated soft drink bottle is opened, the decrease in pressure reduces the solubility of CO₂, leading to the formation of more bubbles. Thus, the correct answer is b) The bubbles will grow, and more may appear.

When a carbonated soft drink is in an unopened bottle, it is under high CO₂ pressure. This high pressure keeps a large amount of carbon dioxide dissolved in the liquid. The bubbles seen inside the bottle are carbon dioxide gas clinging to the container. When the bottle is opened, the pressure above the liquid decreases significantly, causing a loud hiss as the gas escapes.

According to Henry's Law, the solubility of a gas in a liquid decreases as the pressure decreases. Consequently, the solubility of CO₂ in the soda is reduced, which means that more dissolved CO₂ will leave the solution and form bubbles. This explains why more bubbles will appear and existing bubbles will grow after opening the bottle.

Answer:

The correct option is  d isothermal

Explanation:

Step 1: Data given

An adiabatic process is a process in which no heat is gained or lost by the system.

A constant- volume process, also called an isochoric process is a process in which the volume is held constant, meaning that the work done by the system will be zero.

An Isobaric process is a thermodynamic process taking place at constant pressure

An isothermal process is a thermodynamic process, in which the temperature of the system remains constant

Step 2: This situation

The container is slowly compressed, while the temperature is kept constant.

This is  an isothermal process.

The correct option is  d isothermal

The [H+] in the solution is 6.94752 * 10^−7 M.

To calculate the [H+] in the given solution, we need to first calculate the concentration of HX and then use the equilibrium constant (Ka) to find the concentration of [H+]. The equation for the dissociation of HX is: HX ⇌ H+ + X-. Since the dissociated amount is small compared to 0.803 M, we can assume that the concentration of HX is approximately 0.803 M. Now, using the equation for Ka and the concentration of HX, we can find the concentration of [H+]:

Ka = [H+][X-] / [HX]

[H+] = Ka * [HX] = (8.64 * 10^−7) * 0.803 = 6.94752 * 10^−7 M

Therefore, the [H+] in the solution is 6.94752 * 10^−7 M.

Answer:

4 Sb (+3) +  2 Br (2) O (3) ----> 4 SbBr + 3 O(-2)

In the equation, place the valences in parentheses and the stoichiometric balance numbers so that the reaction is balanced, place them in bold.

Explanation:

In this chemical equation it seems to me that the valences of BrO are wrong.

That is why I will complete the equation, demonstrating the chemical reaction, and also use the appropriate valences.

Assuming that Sb has a valence of +3 and that the Br that appears in the form of oxide with oxygen (valence -2) has a valence of +3 (because -3 does not exist as valence or oxidation state of bromine).

Answer:

Formic acid can react to water, to give protons to medium:

  HCOOH     +     H₂O   →     H₃O⁺     +           HCOO⁻  

Weak acid         Base          Acid           Strong conjugate base

    HCOO⁻       +     H₂O  ⇄    HCOOH     +  OH⁻                 Kb

Strong base           Acid       Weak acid       Base          

So the formate can take a proton from water to become formic acid again and that's why it is a conjugate strong base.

Explanation:

HCOOH → Formic acid

To determine the conjugate pair and to know if they are weak or strong, we should know, if they can react to water. This is called hydrolisis.

For example: formic acid is a weak acid, so the formed formate will be its conjugate base and it will be strong because the formate can react to water, to make formic again.

Weak acid → Strong conjugate base

Strong acid → Weak conjugate base

Weak base → strong conjugate acid

Strong base → weak conjugate acid

For example HCl is a strong acid. When it is in aqueous solution, we have protons and chlorides.

HCl + H₂O → H₃O⁺  + Cl⁻

Chloride will be the weak conjugate base, because it can't react to water.

We can not make HCl again, according to this equation:

Cl⁻ + H₂O ← HCl + OH⁻    This is impossible.

Formic acid can react to water, to give protons to medium:

 HCOOH     +     H₂O   →     H₃O⁺     +    HCOO⁻

Weak acid         Base          Acid           Strong conjugate base

So the formate can take a proton from water to become formic acid again and that's why it is a conjugate strong base.

    HCOO⁻       +     H₂O  ⇄    HCOOH     +  OH⁻                 Kb

Strong base           Acid       Weak acid       Base          

HCHO2 can be used to defend the statement 'A weak acid has a stronger conjugate base' as its conjugate base CHO2- readily accepts a proton thus demonstrating its strength.

I'm using HCHO2 to defend the statement 'A weak acid has a stronger conjugate base'. The weak acid formula for HCHO2 is HCHO2 → H+ + CHO2-.

The conjugate base here is CHO2-, which gains a proton (H+) to become HCHO2, demonstrating it's stronger since it readily accepts a proton.

For more such questions on conjugate base, click on:

https://brainly.com/question/35084628

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Answer: PLease make me Brainliest

"They can bring anything from tropical warm and humid days to arctic cold depending on the type of air mass. Fronts form the boundaries of air masses with differing properties. The most severe weather usually occurs when dry-cold continental polar air clashes with warm-humid maritime tropical air."

Final answer:

When a continental polar air mass meets a continental tropical air mass, the interaction can create a weather front that often leads to precipitation and can spawn extratropical cyclones, resulting in variable weather conditions including rain, snow, and thunderstorms.

Explanation:

If a continental polar air mass (cP) clashed with a continental tropical air mass (cT), a significant weather event would likely occur. The cP air mass is characterized as cold and dry, and is a stable air mass that can bring cooler temperatures to affected regions. On the other hand, a cT air mass is hot and dry. When these two air masses meet, the differing temperatures and densities of the air can cause the formation of a weather front, often leading to precipitation and potentially stormy conditions.

The collision often happens along what's called the polar front, where a cyclonic shear is created due to the opposing streams of air. This interaction can lead to the formation of extratropical cyclones, which are large low-pressure systems that can cause a wide array of weather conditions, including rain, snow, thunderstorms, and in some cases, severe weather outbreaks.

Given that cP air is denser than cT air, the warmer, less dense cT air would rise above the cP air mass. This rising motion can lead to cooling and condensation of water vapor, resulting in cloud formation and precipitation. This is a key mechanism through which weather disturbances are generated in regions where such air masses clash, particularly in the midlatitudes, which are notorious for their variable weather patterns.

Answer:

Neither of them will neutralize the buffer solution.

Explanation:

The buffer solution of HNO₂ and KNO₂ will be neutralized when the acid reacts and consume all of the base of the buffer solution or when the base added reacts and consume all of the acid of the buffer solution.

First, we need to calculate the number of moles of the acid and the base of the buffer:

[tex] n_{HNO_{2}} = C*V = 0.100 M*0.500 L = 0.050 moles [/tex]

[tex] n_{KNO_{2}} = C*V = 0.150 M*0.500 L = 0.075 moles [/tex]

Now, let's evaluate each case.

A) 250 mg of NaOH:

We need to calculate the number of moles of NaOH

[tex] n_{NaOH} = \frac{m}{M} [/tex]

Where m: is the mass = 250 mg, and M: is the molar mass = 39.99 g/mol  

[tex] n_{NaOH} = \frac{0.250 g}{39.99 g/mol} = 6.25 \cdot 10^{-3} moles [/tex]

The number of moles of the acid HNO₂ after reaction with the base added NaOH is:

[tex] n_{a_{T}} = n_{a} - n_{b} = 0.050 moles - 6.25 \cdot 10^{-3} moles = 0.044 moles [/tex]

After the reaction of HNO₂ with the NaOH remains 0.044 moles of acid, hence, 250 mg of NaOH would not exceed the capacity of the buffer to neutralize it.

B) 350 mg KOH:

The number of moles of KOH is:

[tex]n_{KOH} = \frac{m}{M} = \frac{0.350 g}{56.1056 g/mol} = 6.23 \cdot 10^{-3} moles[/tex]

Now, the number of moles of HNO₂ that remains in the solution is:

[tex] n_{T} = 0.050 moles - 6.23 \cdot 10^{-3} moles = 0.044 moles[/tex]

Therefore, 350 mg of KOH would not exceed the capacity of the buffer to neutralize it.

C) 1.25 g of HBr:

The number of moles of HBr is:

[tex] n_{HBr} = \frac{m}{M} = \frac{1.25 g}{80,9119 g/mol} = 0.015 moles [/tex]

Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HBr is:

[tex] n_{T} = 0.075 moles - 0.015 moles = 0.06 moles [/tex]

Hence, 1.25 g of HBr would not exceed the capacity of the buffer to neutralize it.

D) 1.35 g of HI:

The number of moles of HI is:

[tex] n_{HI} = \frac{m}{M} = \frac{1.35 g}{127.91 g/mol} = 0.0106 moles [/tex]

Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HI is:

[tex] n_{T} = 0.075 moles - 0.0106 moles = 0.0644 moles [/tex]

Hence, 1.35 g of HI would not exceed the capacity of the buffer to neutralize it.

Therefore, neither of them will neutralize the buffer solution.

I hope it helps you!

Part A: No

Part B: Yes

Part C: Yes

Part D: Yes

Part A: To determine if 250 mg of NaOH would exceed the buffer capacity, we first need to calculate the number of moles of NaOH added and compare it to the number of moles of HNO2 in the buffer.

The moles of NaOH added are calculated as follows:

Part A: No

Part B: Yes

Part C: Yes

Part D: Yes

Part A: To determine if 250 mg of NaOH would exceed the buffer capacity, we first need to calculate the number of moles of NaOH added and compare it to the number of moles of HNO2 in the buffer.

 The moles of NaOH added are calculated as follows:

[tex]\[ \text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{250 \times 10^{-3} \text{g}}{40.0 \text{g/mol}} = 6.25 \times 10^{-3} \text{mol} \][/tex]

The volume of the buffer solution is 500.0 mL, which is equivalent to 0.500 L. The moles of HNO2 in the buffer are calculated by: [tex]\[ \text{moles of HNO2} = \text{concentration of HNO2} \times \text{volume of buffer} = 0.100 \text{M} \times 0.500 \text{L} = 5.00 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of NaOH added (6.25 × 10^-3 mol) are less than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer can neutralize the added NaOH. Therefore, the answer is No.

Part B: Similarly, for 350 mg of KOH, we calculate the moles of KOH:

[tex]\[ \text{moles of KOH} = \frac{350 \times 10^{-3} \text{g}}{56.1 \text{g/mol}} = 6.24 \times 10^{-3} \text{mol} \][/tex]

 Since the moles of KOH are approximately equal to the moles of NaOH added in Part A and are less than the moles of HNO2 in the buffer, the buffer can neutralize the added KOH. Therefore, the answer is No.

 Part C: For 1.25 g of HBr, we calculate the moles of HBr:

[tex]\[ \text{moles of HBr} = \frac{1.25 \times 10^{-3} \text{g}}{80.9 \text{g/mol}} = 1.55 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of HBr added (1.55 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HBr. Therefore, the answer is Yes.

 Part D: For 1.35 g of HI, we calculate the moles of HI:

[tex]\[ \text{moles of HI} = \frac{1.35 \times 10^{-3} \text{g}}{127.9 \text{g/mol}} = 1.06 \times 10^{-2} \text{mol} \][/tex]

Since the moles of HI added (1.06 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HI. Therefore, the answer is Yes.

Correction: In Part B, the moles of KOH calculated are slightly less than the moles of HNO2 in the buffer, so the buffer can neutralize the added KOH. The initial answer provided was incorrect; it should be No, not Yes. The corrected answer is No for Part B."

The volume of the buffer solution is 500.0 mL, which is equivalent to 0.500 L. The moles of HNO2 in the buffer are calculated by:

[tex]\[ \text{moles of HNO2} = \text{concentration of HNO2} \times \text{volume of buffer} = 0.100 \text{M} \times 0.500 \text{L} = 5.00 \times 10^{-2} \text{mol} \][/tex]

Since the moles of NaOH added (6.25 × 10^-3 mol) are less than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer can neutralize the added NaOH. Therefore, the answer is No.

 Part B: Similarly, for 350 mg of KOH, we calculate the moles of KOH:

[tex]\[ \text{moles of KOH} = \frac{350 \times 10^{-3} \text{g}}{56.1 \text{g/mol}} = 6.24 \times 10^{-3} \text{mol} \][/tex]

 Since the moles of KOH are approximately equal to the moles of NaOH added in Part A and are less than the moles of HNO2 in the buffer, the buffer can neutralize the added KOH. Therefore, the answer is No.

Part C: For 1.25 g of HBr, we calculate the moles of HBr:

[tex]\[ \text{moles of HBr} = \frac{1.25 \times 10^{-3} \text{g}}{80.9 \text{g/mol}} = 1.55 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of HBr added (1.55 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HBr. Therefore, the answer is Yes.

Part D: For 1.35 g of HI, we calculate the moles of HI:

[tex]\[ \text{moles of HI} = \frac{1.35 \times 10^{-3} \text{g}}{127.9 \text{g/mol}} = 1.06 \times 10^{-2} \text{mol} \][/tex]

 Since the moles of HI added (1.06 × 10^-2 mol) are greater than the moles of HNO2 in the buffer (5.00 × 10^-2 mol), the buffer cannot neutralize the added HI. Therefore, the answer is Yes.

 Correction: In Part B, the moles of KOH calculated are slightly less than the moles of HNO2 in the buffer, so the buffer can neutralize the added KOH. The initial answer provided was incorrect; it should be No, not Yes. The corrected answer is No for Part B."