Calculate The Theoretical Percentage Of Water For The Following Hydrates. (a) Manganese(II) Monohydrate, (2024)

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8 cm³

mass of gold cube = 20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

[tex]n_i=8[/tex]

[tex]\lambda=3745\ nm[/tex]

The amount of energy change during the transition is given by :

[tex]\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]

And

[tex]\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}][/tex]

Plugging all the values we get :

[tex]\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5[/tex]

So, the final level of the electron is 5.

Answer:

343 m/s

Explanation:

Velocity Formula (Wave): v = fλ

v - velocity

f - frequency

λ - wavelength

We are given f = 23.0 and λ = 14.9. Simply plug it into the formula:

v = 23.0(14.9)

v = 342.7

v ≈ 343

Answer:

343 m/s

Explanation:

There is a formula to calculate the velocity of a wave given the frequency and wavelength.

Velocity = Frequency × Wavelength

v = f λ

v = 23 × 14.9

v = 342.7 ≈ 343

Answer:

pH = 7.0

Explanation:

When HCl reacts with NaOH, H₂O and NaCl are produced, thus:

HCl + NaOH → H₂O + NaCl

At equivalence point, all HCl reacts with NaOH. The only you will have is water.

Equilbrium of water is:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)

K = 1x10⁻¹⁴ = [H⁺] [OH⁻]

As H⁺ = OH⁻ because both are produced from the same water-

1x10⁻¹⁴ = [H⁺]²

1x10⁻⁷M = [H⁺]

As pH = -log= [H⁺]

pH = 7.0

-The pH at equivalence point in the titration of a strong acid with a strong base is always 7.0-

Answer:

E) n=4 l=2 ml=1 ms= 1/2

Explanation:

Arsenic is a member of group 15 in the periodic table. Its electronic configuration is;

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3. Its condensed electronic configuration can be written as [Ar]4s2 3d10 4p3. This electronic configuration shown here can now enable us to consider each option given in the question in order to meaningfully arrive at a logical answer.

If we look at option E, the data given for that electron is; n=4 l=2 ml=1 ms= 1/2. This refers to an electron in a 4d orbital. In the ground state configuration of arsenic shown above, there is no 4d orbital, hence option E must be the correct answer.

Answer:

1. Chlorophyll A has a yellow-green pigment and absorbs at a wavelength of 430-662 nm. Chlorophyll B has a blue-green pigment and absorbs at a wavelength of 453-642 nm. Carotene has an orange pigment and absorbs at a wavelength of 460-550 nm. Xanthophyll has a yellow pigment and absorbs at a wavelength of 700 nm. Anthocyanin has a purple pigment and absorbs at a wavelength of 660 nm.

2. Plant and planktons that survive by photosynthesis do so with the help of pigments. These pigments help them trap light ( e.g sunlight). When this happens a reflection occurs which gives rise to the pigment colour observed which is usually not absorbed. A typical example is seen in chlorophyll which is green pigmented hence it absorbs all other wavelengths of light the except green.

3. The various ways in which the rate of photosynthesis can be measured include: Measuring the amount of carbon(IV)oxide uptake as well as measuring the amount of oxygen produced.

The reason for the colour change in deciduous trees in the fall is because the chlorophyll in the leaves of the plants is broken down leading to the loss of the green colouration. Also, the presence of anthocyanin can also be responsible for the yellow pigmentation.

5. Due to the evolution of species, and the fact that life originated from microbes, microbes have evolved from the single membrane to more complex systems with specialized organs like the mitochondria and chloroplasts. Early microbes were photosynthetic in nature e.g phytoplanktons which have evolved specialized systems controlled by the DNA to ensure their survival.

Explanation:

1. Chlorophyll A has a yellow-green pigment and absorbs at a wavelength of 430-662 nm. Chlorophyll B has a blue-green pigment and absorbs at a wavelength of 453-642 nm. Carotene has an orange pigment and absorbs at a wavelength of 460-550 nm. Xanthophyll has a yellow pigment and absorbs at a wavelength of 700 nm. Anthocyanin has a purple pigment and absorbs at a wavelength of 660 nm.

2. Plant and planktons that survive by photosynthesis do so with the help of pigments. These pigments help them trap light ( e.g sunlight). When this happens a reflection occurs which gives rise to the pigment colour observed which is usually not absorbed. A typical example is seen in chlorophyll which is green pigmented hence it absorbs all other wavelengths of light the except green.

3. The various ways in which the rate of photosynthesis can be measured include: Measuring the amount of carbon(IV)oxide uptake as well as measuring the amount of oxygen produced.

The reason for the colour change in deciduous trees in the fall is because the chlorophyll in the leaves of the plants is broken down leading to the loss of the green colouration. Also, the presence of anthocyanin can also be responsible for the yellow pigmentation.

5. Due to the evolution of species, and the fact that life originated from microbes, microbes have evolved from the single membrane to more complex systems with specialized organs like the mitochondria and chloroplasts. Early microbes were photosynthetic in nature e.g phytoplanktons which have evolved specialized systems controlled by the DNA to ensure their survival.

Answer:

[tex]m_{solution}=470g[/tex]

Explanation:

Hello,

In this case, a solution is formed when a solute is completely dissolved in a solvent, thus, for this situation, the sugar is the solute and the water the solvent but in addition to them we find spices which are also considered in the total mass of the solution. In such a way, for computing the total mass we must add the mass of three constituents (115 g sugar, 350 g water and 5 g spices) as shown below:

[tex]m_{solution}=115g+350g+5g\\\\m_{solution}=470g[/tex]

Best regards.

Answer:

The freezing point of the solution is -1.4°C

Explanation:

Freezing point decreases by the addition of a solute to the original solvent, freezing point depression formula is:

ΔT = kf×m×i

Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.

Molality of the solution is:

-moles CaBr₂ (Molar mass:

189.9g ₓ (1mol / 199.89g) = 0.95 moles

Molality is:

0.95 moles CaBr₂ / 3.75kg water = 0.253m

Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:

CaBr₂(s) → Ca²⁺ + 2Br⁻

3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.

Replacing:

ΔT = kf×m×i

ΔT = 1.86°C/m×0.253m×3

ΔT = 1.4°C

The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,

The freezing point of the solution is -1.4°C

Answer:

THE FREEZING POINT IS -1.41 °C

Explanation:

Using the formula of change in freezing point:

ΔTf = i Kf m

i = 3 (1 Ca, 2 Br)

i is the number of the individual elements in the molecules

Kf of water = 1.86 °C/m

mass of CaBr2 = 189.9 g

Calculate the Molar mass of CaBr2:

Molar mass = ( 40 + 80*2) = 200 g/mol

Calculatee the molarity:

molarity = 189.9 g * 1 mole / 200 g/mol / 3.75 kg of water

molarity = 0.2532 M

So therefore, the change in freezing point is:

ΔTf = 1 Kf * M

ΔTf = 3 * 1.86 * 0.2532

ΔTf = 1.41 °C

The freezing point = old freezing point - change in freezing point

The freezing point = 0 - 1.41 °C = - 1.41 °C

The freezing point therefore is -1.41 °C

The question is incomplete, the complete question is;

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO How many grams of NO and of H20 form? Enter your answers numerically separated by a comma. 4NH3(g) +502(g)------->4NO(g)+6H2O(g)

In a certain experiment, 1.10 g of NH3 reacts with 2.02 g of O2. How many grams of the excess reactant remain after the limiting reactant is completely consumed? Express your answer in grams to three significant figures.

Answer:

Mass of excess ammonia 0.034 g of ammonia

Mass of water formed= 1.37g

Mass of NO formed = 1.50g

Explanation:

The limiting reactant is the reactant that yields the least number of moles of product.

For NH3, molar mass of ammonia = 17g mol-1

Number of moles of ammonia reacted= 1.10g/17 gmol-1 = 0.065 moles of ammonia

According to the reaction equation;

4 moles of ammonia yields 4 moles of NO

Hence 0.065 moles of ammonia will yield 0.065 ×4/4 = 0.065 moles of NO

For oxygen, molar mass of oxygen gas = 32gmol-1

Number of moles of oxygen gas= 2.02g/32gmol-1 = 0.063 moles of oxygen

From the reaction equation;

5 moles of oxygen gas yields 4 moles of NO

0.063 moles of oxygen will yield 0.063 ×4 /5 = 0.050 moles of NO

Hence oxygen is the limiting reactant and ammonia is the excess reactant.

Amount of excess ammonia = Amount of ammonia - amount of oxygen

Amount of excess ammonia= 0.065-0.063= 2×10^-3 moles

Mass of excess ammonia = 2×10^-3 moles × 17 gmol-1 = 0.034 g of ammonia

Mass of NO formed is obtained from the limiting reactant. Since molar mass of is 30gmol-1. Then mass of NO formed = 0.050 moles of NO × 30gmol-1 = 1.50 g of NO

For water;

5 moles of oxygen yields 6 moles of water

Hence 0.063 moles of oxygen yields 0.063 × 6/5 = 0.076 moles of water

Molar mass of water = 18gmol-1

Hence mass of water = 0.076 moles × 18gmol-1 = 1.37g of water

Answer:

The vapor pressure of carbon tetrachloride, CCl4, is 0.354 atm,

and the vapor pressure of chloroform, CHCl3, is 0.526 atm at

316 K. A solution is prepared from equal masses of these two

compounds at this temperature. Calculate the mole fraction of

the chloroform in the vapor above the solution. If the vapor

above the original solution is condensed and isolated into a

separate flask, what would the vapor pressure of chloroform be

above this new solution?

ANSWER

0.346atmatm

Chapter 13

Solutions



Chemistry Structure and Properties

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Video Transcript

problem. 1 13 asks what the, um the mole fraction of chloroform is if equal masses of carbon tetrachloride and core former mixed together. And then it asks you to condense the vapor above such a solution and figure out what the pressure of core form is above that new solution. So the problem gives you that the vapor pressure of pure carbon tetrachloride is 0.354 atmospheres, and the vapor pressure of pure chloroform is 0.5 to 6 atmospheres. Both of these at 316

Answer:

Kc = [CH₄] / [H₂]²

Kp = [CH₄] / [H₂]² * (0.082*T)^-1

Explanation:

Equilibrium constant, Kc, is defined as the ratio of the concentrations of the products over the reactants. Also, each concentration of product of reactant is powered to its coefficient.

Pure solids and liquids are not taken into account in an equilibrium

Thus, for the reaction:

C(s)+ 2H₂(g) ⇌ CH₄(g)

Equilibrium constant is:

Kc = [CH₄] / [H₂]²

Now, using the formula:

Kp = Kc* (RT)^Δn

Where R is gas constant (0.082atmL/molK), T is the temperature of the reaction and Δn is difference in coefficients of gas products - coefficients of gas reactants (1 - 2= -1)

Replacing:

Kp = [CH₄] / [H₂]² * (0.082*T)^-1

Answer:

Option C. Electrons are shared between two atoms

Explanation:

Covalent bonding is a type of bonding which exist between two non metals.

In this bonding, electrons are shared between the two atoms involved in order to attain a stable octet configuration.

This can be seen when hydrogen atom combine with chlorine atom to form hydrogen chloride as shown below:

H + Cl —> HCl

Hydrogen has 1 electron in it's outmost shell and it requires 1 electron to attain a stable configuration.

Chlorine has 7 electrons in it's outmost shell and requires 1 electron to attain a stable configuration.

During bonding, both hydrogen and chlorine will contribute 1 electron each to form bond, thereby attaining a stable configuration. The bond formed in this case is called covalent bond as both atoms involved shared electron to attain a stable configuration.

C. Electrons are shared between two atoms.

What is Covalent Bonding?

A covalent bond is fashioned among non-metals which have comparable electronegativities. Neither atom is "strong" sufficient to draw electrons from the other.It is formed when pairs of electrons are shared by atoms.Atoms will covalently bond with different atoms that allows you to benefit extra stability, that is gained through forming a complete electron shell. By sharing their outer most (valence) electrons, atoms can replenish their outer electron shell and gain stability.

For example:

In H₂ molecule; there is a covalent bond formation between two hydrogen atoms as the electron from each hydrogen atom is shared leading to the formation of hydrogen molecule.

Learn more:

https://brainly.com/question/3447218

Hey there!:

Molar mass NaCl = 58.44 g/mol

Number of moles = mass of solute / molar mas

Number of moles = 0.900 / 58.44

Number of moles = 0.0154 moles of NaCl

Volume in liters of solution :

100.0 mL / 1000 => 0.1 L

Therefore:

Molarity = number of moles / volume in liters

Molarity = 0.0154 / 0.1

Molarity = 0.154 M

Hope this helps!

Answer:

Be∠ Li∠ Ba∠ Na∠ Cs

Explanation:

Beryllium does not react with water. It is the only alkaline earth metal that does not react with water because of its small size and high ionization energy. Beryllium differs considerably from other members of group two, its compounds when anhydrous show a considerable degree of covalent character.

As the atomic number of the group two elements increases, their ionization energies decreases and their electrode potentials become more negative hence their reactivity increases down the group. This implies that barium will have a very negative electrode potential comparable to that of the alkali metals, hence it reacts considerably with water.

The reactivity of alkali metals with water increases down the group. Lithium reacts quietly with water, sodium and potassium react with water with increasing vigour while rubidium and cesium react with water with exceptional violence.

This little explanation, is the reason behind the option chosen as the answer.

Answer:

The density would be larger than the correct value.

Explanation:

First off, the realtionship between denisty and volume is given in the equation below;

Density = Mass / Volume

From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.

Assuming all thing's being normal;

Mass = 2g

Volume = 10ml

Density = 2 / 10 = 0.2 g/ml

Second case scenario;

'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"

Lets have a value of 8ml for our volume. Mass remains constant.

Density = 2 / 8 = 0.25 g/ml

The density would be larger than the correct value.

Answer: The density would be larger than the correct value.

First off, the relationship between density and volume is given by:

Density = Mass / Volume

From this equation, Density is inversely proportional to volume. This means as the volume increases, the density decreases and as the volume decreases the density increases.

Assuming all thing's being normal;

Mass = 2g

Volume = 10ml

Density = [tex]\frac{2}{10}=0.2[/tex] g/ml

Second case scenario;

'your pipet was incorrectly calibrated so that it transferred less than 10.00 mL"

Lets have a value of 8ml for our volume. Mass remains constant.

Density = [tex]\frac{2}{8}= 0.25[/tex] g/ml

The density would be larger than the correct value.

Learn more: https://brainly.com/question/11107994

Answer:

i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances

Explanation:

Answer:

None

Explanation:

Answer on Edge 2022

Answer:

Metallic bonding

Explanation:

Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.

The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.

In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.

Answer:

See explanation

Explanation:

In this case, we have to check two variables:

1) The leaving group

2) The carbon bonded to the leaving group.

Let's check one by one:

2-chloro-3-methylbutane

In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a secondary substrate.

1-phenylpropan-1-ol

In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a secondary substrate.

(E)-pent-3-en-2-yl 4-methylbenzenesulfonate

In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an allylic substrate.

3a-bromooctahydro-1H-indene

In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a tertiary substrate.

1-iodo-3-methylbutane

In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a primary substrate.

See figure 1

I hope it helps!

Answer:

[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

Explanation:

In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).

[tex]Q=m*Cp*ΔT[/tex]

The next step is to identify what values we have:

[tex]Q~=~1870~J[/tex]

[tex]m~=~85.01~g[/tex]

[tex]ΔT~=~45.2~^{\circ}C[/tex]

[tex]Cp~=~X[/tex]

Now, we can plug the values and solve for "Cp":

[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]

[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]

[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]

The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]

I hope it helps!

Answer:

e. Na_I

Explanation:

a. Na–O

b. Na–F

c. Na–Cl

d. Na–Br

e. Na–I

Bond polarity can be calculated, examining the Pauling scale electronegativity values of the two atoms. The difference between these values will determine the predominant type of bond between the respective atoms.

Therefore, Na-I would be least polar bond. (The difference in the electronegativies of the two atom is least)

Answer:

pH = 9.58

Explanation:

First of all, we need to determine the molarity of the solution.

We determine the molar mass of morphine:

12g/m . 17 + 1 g/m . 19 + 14 g/m + 16 g/m . 3 = 285.34 g/m

molar mass g/m, is the same as mg/mm

25 mg . 1 mmol / 285.34 mg = 0.0876 mmoles / 100 mL = 8.76×10⁻⁴ M

In diltuted solution, we must consider water.

Mass balance for morphine = [Morphine] + [Protonated Morphine]

8.76×10⁻⁴ M = [Morphine] + [Protonated Morphine]

As Kb is too small, I can skipped, the [Protonated Morphine]

8.76×10⁻⁴ M = [Morphine]

In the charge balance I will have:

[OH⁻] = [H⁺ morphine] + [H⁺]

Let's go to the Kb expression

Morphine + H₂O ⇄ MorphineH⁺ + OH⁻ Kb

Kb = [MorphineH⁺] [OH⁻] / [Morphine]

Kb = [MorphineH⁺] [OH⁻] / 8.76×10⁻⁴ M

So now, we need to clear [MorphineH⁺] to replace it in the charge balance

Kb . 8.76×10⁻⁴ M / [OH⁻] = [MorphineH⁺]

Now, the only unknown value is the [OH⁻]

[OH⁻] = Kb . 8.76×10⁻⁴ M / [OH⁻] + Kw/[OH⁻]

Remember that Kw = [H⁺] . [OH⁻]

[H⁺] = Kw/[OH⁻]

[OH⁻]² = 1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴

[OH⁻] = √(1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴)

[OH⁻] = 3.76×10⁻⁵ → - log [OH⁻] = pOH = 4.42

pH = 14 - pOH → 14 - 4.42 = 9.58

Answer:

[tex]=C_3H_4O_3[/tex]

Explanation:

When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,

Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C

Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

Moles H = 4.58/3.41 = 1.34

Moles O = 3.41/3.41 = 1

Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

Moles H = 1.34x3 = 4

Moles O = 1x3 = 3

Empirical Formula [tex]=C_3H_4O_3[/tex]

Answer:

Polarizing power refers to an atoms ability to pull an electron toward it, polarizing the atom the electron comes from. Since cations are positive, they are able to attract electrons toward themselves. Anions are negative and so do not attract more electrons.

Therefore, Be2+ has a higher polarizing power because it has a higher quantitiy of protons, hence a higher polarizing power.

Answer:

[H₂A] = 5.0409x10⁻⁷M

[HA⁻] = 0.001951M

[A²⁻] = 0.234

11.29 = pH

Explanation:

When Na₂A = A²⁻is in equilibrium with water, the reactions that occurs are:

A²⁻(aq) + H₂O(l) ⇄ HA⁻(aq) + OH⁻(aq)

By definition of Ka of reaction, you can find Kb (the inverse basic reaction), thus:

Kb1 = KwₓKa2 =

Where Kw is K of equilibrium of water, 1x10⁻¹⁴

1x10⁻¹⁴/ 6.19x10⁻⁹ =

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

And HA⁻ will be in equilibrium as follows:

HA⁻(aq) + H₂O(l) ⇄ H₂A(aq) + OH⁻(aq)

Kb2 = KwₓKa1 = 1x10⁻¹⁴/ 7.68x10⁻⁵ =

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

Thus, the Na₂A has 2 equilibriums, for the first reaction, concentrations wil be:

[HA⁻] = X

[OH⁻] = X

[A²⁻] = 0.236M - X

X as how much will react, Reaction coordinate

Replacing in Kb1:

1.6155x10⁻⁶ = [HA⁻] [OH⁻] / [A²⁻]

1.6155x10⁻⁶ = [X] [X] / [0.236-X]

3.8126x10⁻⁶ - 1.6155x10⁻⁶X = X²

3.8126x10⁻⁶ - 1.6155x10⁻⁶X - X² = 0

Solving for X:

X = -0.00195 → False solution. There is no negative concentrations

X = 0.001952.

Replacing, concentrations for the first equilibrium are:

[HA⁻] = 0.001952[OH⁻] = 0.001952[A²⁻] = 0.234

Now, in the second equilibrium:

[HA⁻] = 0.001952 - X

[OH⁻] = X

[H₂A] = X

Replacing in Kb1:

1.3021x10⁻¹⁰ = [H₂A] [OH⁻] / [HA⁻]

1.3021x10⁻¹⁰ = [X] [X] / [0.001952 - X]

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X = X²

2.5417x10⁻¹³ - 1.3021x10⁻¹⁰X - X² = 0

Solving for X:

X = -5.04x10⁻⁷ → False solution. There is no negative concentrations

X = 5.0409x10⁻⁷

Replacing, concentrations for the second equilibrium are:

[HA⁻] = 0.001951M[OH⁻] = 5.0409x10⁻⁷M[H₂A] = 5.0409x10⁻⁷M

Thus, you have concentrations of H2A, HA−, and A2−

Now, for pH, you need [OH⁻] concentration that is:

[OH⁻] = 0.0019525

pOH = -log[OH⁻] = 2.709

As 14 = pH+ pOH

14-2.709=pH

11.29 = pH

Answer:

Explanation:

If an antacid has sodium hydrogen carbonate/Calcium carbonate, it reacts with HCl as shown

NaHCO₃+ HCl → NaCl + CO₂+ H₂O

Antacid acid salt gas water

CaCO₃+ 2HCl → CaCl₂+ CO₂+ H₂O

Antacid acid salt gas water

The formation of gas CO₂ is shown by brisk effervescence when the antacid (sodium hydrogen carbonate/calcium carbonate) reacts with HCl (acid). So CO₂ is the additional product formed and its formation is supported by observation of brisk effervescence as HCl is added to the antacid.

Answer:

0.0480 M

Explanation:

The reaction is ...

H₂SO₄ + 2NaOH ⇒ Na₂SO₄ +2H₂O

That is, 2 moles of NaOH react with each mole of H₂SO₄. Then the molarity of the H₂SO₄ is ...

moles/liter = (0.186 M/2)(12.9 mL)/(25.0 mL) ≈ 0.0480 M

you can use stoichiometry which is taking the number in from of each substance and using it as a molar ratio(ones that don't have a number actually have a 1 in front

so 2AgNO3 + 1 Zn -> 2 Ag + Zn (NO3)2

The ratio is of moles of Zn to moles of Silver is 1:2 so for every 1 mole of Zn there's 2 moles of Ag

1.3 × 2 = 2.6 moles of Ag are made

Answer:

1st Equivalent Point

59.5 mL KOH 8.34 pH

2nd Equivalent Point

119 mL KOH 11.25 pH

Explanation:

These are the two neutralization reactions:

H₂CO₃ + OH⁻ ⇄ HCO₃⁻ + H₂O

HCO₃⁻ + OH⁻ ⇄ CO₃⁻² + H₂O

For the first equivalence point we need to know that

mili mol of base = mili mol of acid

0.0378 M . volume (mL) = 17.3 mL . 0.130M

Volume (mL) = (17.3 mL . 0.130M) / 0.0378M = 59.5 mL

Global reaction is:

H₂CO₃ + 2OH⁻ ⇄ CO₃⁻² + 2H₂O

0.0378 M . volume (mL) = 2 . 17.3mL . 0.130M

Volume(mL) = 119 mL

The first equivalence point has only the HCO₃⁻ as predominant, so the pH will be the media, of the two pKa from the H₂CO₃

(pKa₁ + pKa₂) / 2 = (6.35 + 10.32) / 2 = 8.34 → pH₁

In the second equivalence point the predominant is the CO₃⁻² which is a base. First of all, we determine the [CO₃⁻²]

[CO₃⁻²] = mili moles of acid / initial volume + volume for the 2nd point

[CO₃⁻²] = 0.130 M . 17.3 mL / 17.3 mL + 119mL = 0.0165 M

CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁

pKb₁ = 14 - pKa₂ → 14 - 10.32 = 3.68

Kb₁ = 10⁻³°⁶⁸ = 2.09 ₓ10⁻⁴

CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁

0.0165 - x x x

Kb₁ = x² / 0.0165 - x

2.09 ₓ10⁻⁴ ( 0.0165 - x ) - x² = 0

This is a quadractic. We can not use, only 0.0165 in the denominator because the Kb is in order of 10⁻⁴)

3.4485×10⁻⁶ - 2.09×10⁻⁴x - x²

a= -1

b= - 2.09×10⁻⁴

c = 3.4485×10⁻⁶

(- b +- √(b² - 4ac) ) / 2

x = 1.75×10⁻³

pOH = - log [OH⁻] =2.75

pH = 14 - pOH → 11.25

The first Equivalent Point of 59.5 mL KOH 8.34 pH. The second Equivalent Point of 119 mL KOH 11.25 pH.

Neutralization reactions:

These are the two neutralization reactions:

H₂CO₃ + OH⁻ ⇄ HCO₃⁻ + H₂O

HCO₃⁻ + OH⁻ ⇄ CO₃⁻² + H₂O

GIVEN:

0.0378 M . volume (mL) = 17.3 mL . 0.130M

Volume (mL) = (17.3 mL . 0.130M) / 0.0378M = 59.5 mL

Overall reaction is:

H₂CO₃ + 2OH⁻ ⇄ CO₃⁻² + 2H₂O

0.0378 M . volume (mL) = 2 . 17.3mL . 0.130M

Volume(mL) = 119 mL

The first equivalence point has only the HCO₃⁻ as predominant, so the pH will be the media, of the two pKa from the H₂CO₃

(pKa₁ + pKa₂) / 2 = (6.35 + 10.32) / 2 = 8.34 → pH₁

In the second equivalence point the predominant is the CO₃⁻² which is a base.

First of all, we determine the [CO₃⁻²]

[CO₃⁻²] = mili moles of acid / initial volume + volume for the 2nd point

[CO₃⁻²] = 0.130 M . 17.3 mL / 17.3 mL + 119mL = 0.0165 M

CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁

pKb₁ = 14 - pKa₂ → 14 - 10.32 = 3.68

Kb₁ = 10⁻³°⁶⁸ = 2.09 ₓ10⁻⁴

CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁

0.0165 - x x x

Kb₁ = x² / 0.0165 - x

2.09 ₓ10⁻⁴ ( 0.0165 - x ) - x² = 0

3.4485×10⁻⁶ - 2.09×10⁻⁴x - x²

a= -1

b= - 2.09×10⁻⁴

c = 3.4485×10⁻⁶

Using the formula for quadratic equation:

(- b +- √(b² - 4ac) ) / 2

x = 1.75×10⁻³

pOH = - log [OH⁻] =2.75

pH = 14 - pOH →

pH= 11.25

Find more information about Neutralization reaction here:

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Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C

(58 g . 0.100) / 0.580 = 10 g of H

48 g of C / 12 g/mol = 4 mol

10 g of H / 1g/mol = 10 moles

The molecular formula of the compound is C4H10.

At STP;

P = 1 atm

T = 273 K

V = 33.6 mL or 0.0336 L

R = 0.082 atmLK-1mol-1

n = ?

Hence;

n = PV/RT

n = 1 atm × 0.0336 L/0.082 atmLK-1mol-1 × 273 K

n = 0.0015 moles

Number of moles = mass/molar mass

Molar mass= Mass/Number of moles

Molar mass = 0.087 g/0.0015 moles

Molar mass = 58 g/mol

Mass of carbon = (58 g × 0.480) / 0.580 = 48 g of C

Mass of hydrogen = (58 g × 0.100) / 0.580 = 10 g of H

Number of moles of carbon = 48 g of C / 12 g/mol = 4 mol

Number of moles of hydrogen = 10 g of H / 1g/mol = 10 moles

Formula of the compound must then be C4H10.

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