Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which are separated by a distance of 7.5 nm

In the given chemical equation, zinc is undergoing reduction as it gains electrons.

Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.

A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .

The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.

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Answer:

A. Cl, Se, Br

B. I, Xe, Ba

Explanation:

The elements arranged in their increasing atomic size using periodic table positions are:

a) Cl < Br < Se

b) Xe < I < Ba.

To rank elements by increasing atomic size, we need to refer to their positions on the periodic table. Atomic size generally increases as we move down a group and decreases as we move across a period from left to right.

(a) Se, Br, Cl

These elements are all in Group 16 (Se), 17 (Br), and 17 (Cl), respectively. Since size increases down a group and decreases across a period:

(b) I, Xe, Ba

These elements are in Group 17 (I), 18 (Xe), and 2 (Ba), respectively. Comparing their positions:

Explanation:

A solution is said to saturated when it cannot dissolve any extra solute in it. The extra solute put remains undissolved.

A solution is said to unsaturated, when the concentration of solute is less as compared to solubility of the solution it is said to be unsaturated.

A solution is said to be super saturated when it contains more of the solute than the solvent  can dissolve under normal conditions is called super saturated.

We can determine whether the solution is unsaturated, saturated, or supersaturated by knowing the amount of solute in the solution.

A solution is said to be saturated when it cannot dissolve any extra solute in it, a solution is said to be unsaturated solution, when the concentration of solute is less as compared to solubility of the solution and the solution is able to dissolve more solute.

Whereas, a solution is said to be super saturated when it contains more of the solute than the solvent can dissolve under normal conditions.

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Answer:

45, 34, 36

Explanation:

The atomic number of Selenium is 34 and the atomic number is 79 also the atom has gained two electron denoted by superscript -2

number of neutrons = mass number - atomic number = 79 - 34 = 45

number of proton = atomic number = 34

number of electron = 34 + 2 = 36. In an atom the number of proton is always equal to number of electron if the atom is neutral but this Se atom has gain two so the number of electron will exceed the number of proton by 2.

The Se ion has 34 protons, 45 neutrons and 36 electrons.

The mass number (A) is given by the sum of the protons and neutrons:

A = protons + neutrons = 79

From the Periodic Table, we can see that the chemical element Selenium (Se) has an atomic number (Z) of 34, which is equal to the number of protons of a chemical element:

Z = protons = 34

Thus, we calculate the number of neutrons as the difference between A and Z:

neutrons = A - Z = 79 - 34 = 45

In a neutral atom (without electric charge), the number of electrons is equal to the number of protons. Since Se ion has 34 protons and a charge of -2, it has 34 electrons to be neutral and then it gained 2 electrons, so the number of electrons is equal to:

electrons = protons + 2 = 34 + 2 = 36

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Answer:

Explanation: i)  Mass per capacity of the tant = (0.15015/100100) kg/L = 0.0000015 kg/L

Amount salt of concentrated salt left = ( 0.03003 - 0.0000015) kg/L = 0.0300285 kg/L

∴  mass of salt in the tank = 0.0300285 kg/L X 77 L/min = 2.31 kg

ii) Capacity of tank at 0.01001 kg/L: 2.31 kg/0.01001 kg/L = 230.77 L

∴ time taken for the concentration of the salt = 230.77/(77 L/min) = 3 minutes.

Answer:

surface tension.

Explanation:

Answer:

For A: The value of [tex]K_p[/tex] for the given equation is [tex]1.64\times 10^{-4}[/tex]

For B: The value of [tex]K_p[/tex] for the given equation is 50.2

For C: The value of [tex]K_c[/tex] for the given equation is [tex]5.312\times 10^{-39}[/tex]

For D: The value of [tex]K_c[/tex] for the given equation is [tex]4.60\times 10^{-3}[/tex]

Explanation:

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

[tex]K_p=K_c(RT)^{\Delta ng}[/tex]         ..........(1)

where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure

[tex]K_c[/tex] = equilibrium constant in terms of concentration

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}

[/tex]

T = temperature

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}[/tex]

The given chemical equation follows:

[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]

We are given:

[tex]K_c=0.50\\T=400^oC=[400+273]K=673K\\\Delta n_g=2-4=-2[/tex]

Putting values in equation 1, we get:

[tex]K_p=0.50\times (0.0821\times 673)^{-2}\\\\K_p=1.64\times 10^{-4}[/tex]

Hence, the value of [tex]K_p[/tex] for the given equation is [tex]1.64\times 10^{-4}[/tex]

The given chemical equation follows:

[tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]

We are given:

[tex]K_c=50.2\\T=448^oC=[448+273]K=721K\\\Delta n_g=2-2=0[/tex]

Putting values in equation 1, we get:

[tex]K_p=50.2\times (0.0821\times 721)^{0}\\\\K_p=50.2[/tex]

Hence, the value of [tex]K_p[/tex] for the given equation is 50.2

The given chemical equation follows:

[tex]Na_2SO_4.10H_2O(s)\rightleftharpoons Na_2SO_4(s)+10H_2O(g)[/tex]

We are given:

[tex]K_p=4.08\times 10^{-25}\\T=25^oC=[25+273]K=298K\\\Delta n_g=10-0=10[/tex]

Putting values in equation 1, we get:

[tex]4.08\times 10^{-25}=K_c\times (0.0821\times 298)^{10}\\\\K_c=5.312\times 10^{-39}[/tex]

Hence, the value of [tex]K_c[/tex] for the given equation is [tex]5.312\times 10^{-39}[/tex]

The given chemical equation follows:

[tex]H_2O(l)\rightleftharpoons H_2O(g)[/tex]

We are given:

[tex]K_p=0.122\\T=50^oC=[50+273]K=323K\\\Delta n_g=1-0=1[/tex]

Putting values in equation 1, we get:

[tex]0.122=K_c\times (0.0821\times 323)^{1}\\\\K_c=4.60\times 10^{-3}[/tex]

Hence, the value of [tex]K_c[/tex] for the given equation is [tex]4.60\times 10^{-3}[/tex]

Kp and Kc are the equilibrium constant. Kp for A. [tex]1.64 \times 10^{-4}[/tex], for B. 50.2 and Kc for C is [tex]5.312 \times 10^{-39}[/tex] and D. [tex]4.60\times 10^{-3}.[/tex]

Kp is the equilibrium constant given relative to the partial pressure whereas, Kc is given relative to the concentration. The relation between Kp and Kc can be shown as:

[tex]\rm K_{p} = K_{c} (RT)^{\Delta\;ng}[/tex]

For reaction A the balanced reaction is shown as:

[tex]\rm N_{2}(g)+3H_{2}(g) \leftrightharpoons 2NH_{3}(g)[/tex]

The value of Kp is:

[tex]\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\&= 0.50\times (0.0821\times 673)^{-2}\\\\&= 1.64\times 10^{-4}\end{aligned}[/tex]

Thus, the Kp for A is [tex]1.64\times 10^{-4}.[/tex]

For reaction B the balanced reaction is shown as:

[tex]\rm H_{2} + I_{2} \rightleftharpoons 2HI[/tex]

The value of Kp is:

[tex]\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\&= 50.2\times (0.0821\times 721)^{0}\\\\&= 50.2 \end{aligned}[/tex]

Thus, the Kp for B is 50.2.

For reaction C the balanced reaction is shown as:

[tex]\rm Na_{2}SO_{4} .10H_{2}O(s) \rightleftharpoons Na_{2}SO_{4}(s)+10H_{2}O(g)[/tex]

The value of Kc is:

[tex]\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\4.08\times 10^{-25} &= \rm K_{c} \times (0.0821\times 298)^{10}\\\\\rm K_{c} &= 5.312\times 10^{-39} \end{aligned}[/tex]

Thus, the Kc for C is [tex]5.312\times 10^{-39}.[/tex]

For reaction D the balanced reaction is shown as:

[tex]\rm H_{2}O(l) \rightleftharpoons H_{2}O[/tex]

The value of Kc is:

[tex]\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\0.122 &= \rm K_{c} \times (0.0821\times 323)^{1}\\\\\rm K_{c} &= 4.60\times 10^{-3} \end{aligned}[/tex]

Thus, the Kc for D is [tex]4.60\times 10^{-3}.[/tex]

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Answer :  The condensed ground-state electron configuration for each is:

(a) [tex][Ar]4s^23d^{2}[/tex]

(b) [tex][Ne]3s^23p^5[/tex]

(c) [tex][Ar]4s^23d^{3}[/tex]

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Noble-Gas notation : It is defined as the representation of electron configuration of an element by using the noble gas directly before the element on the periodic table.

(a) The given element is, Ti (Titanium)

As we know that the titanium element belongs to group 4 and the atomic number is, 22

The ground-state electron configuration of Ti is:

[tex]1s^22s^22p^63s^23p^64s^23d^{2}[/tex]

So, the condensed ground-state electron configuration of Ti in noble gas notation will be:

[tex][Ar]4s^23d^{2}[/tex]

There are 4 number of valence electrons shown in orbital diagram.

(b) The given element is, Cl (Chlorine)

As we know that the chlorine element belongs to group 17 and the atomic number is, 17

The ground-state electron configuration of Cl is:

[tex]1s^22s^22p^63s^23p^5[/tex]

So, the condensed ground-state electron configuration of Cl in noble gas notation will be:

[tex][Ne]3s^23p^5[/tex]

There are 7 number of valence electrons shown in orbital diagram.

(c) The given element is, V (Vanadium)

As we know that the vanadium element belongs to group 5 and the atomic number is, 23

The ground-state electron configuration of Ti is:

[tex]1s^22s^22p^63s^23p^64s^23d^{3}[/tex]

So, the condensed ground-state electron configuration of V in noble gas notation will be:

[tex][Ar]4s^23d^{3}[/tex]

There are 5 number of valence electrons shown in orbital diagram.

To draw valence shell electron configurations and orbital diagrams for Ti, Cl, and V, you place electrons in orbitals according to the electron configuration up to the penultimate shell. Titanium's valence shell configuration is [Ar]3d2 4s2, Chlorine's is [Ne]3s2 3p5, and Vanadium's is [Ar]3d3 4s2.

In high school chemistry, understanding electron configurations is fundamental. Here's how you can predict valence shell electron configurations and draw orbital diagrams for titanium (Ti), chlorine (Cl), and vanadium (V).

Titanium has an atomic number of 22. Its electron configuration up to the penultimate shell is [Ar]3d2 4s2. To show the valence electron configuration:

Chlorine has an atomic number of 17. Its electron configuration up to the penultimate shell is [Ne]3s2 3p5. For the valence electrons:

Vanadium has an atomic number of 23. Its electron configuration up to the penultimate shell is [Ar]3d3 4s2. For its valence shell:

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Final answer:

The reaction enthalpy (ΔHrxn) for the combustion of stearic acid is -10500 kJ/mol. When 206 grams of stearic acid are combusted, 7602 kJ of heat is released.

Explanation:

Calculating ΔHrxn for the Combustion of Stearic Acid

The combustion reaction for stearic acid (C₁₈H₃₆O₂) can be written as follows:

C₁₈H₃₆O₂(s) + 26O₂(g) → 18CO₂(g) + 18H₂O(l)

To calculate the reaction enthalpy (ΔHrxn), we use the sum of the enthalpies of formation (ΔHf) of the products minus the sum of ΔHf of reactants:

ΔHrxn = [18(ΔHf of CO₂) + 18(ΔHf of H₂O)] - [ΔHf of stearic acid + 26(ΔHf of O₂)]

Since ΔHf for elemental oxygen (O₂) is zero, we simplify the equation to:

ΔHrxn = [18(-394 kJ/mol) + 18(-242 kJ/mol)] - (-948 kJ/mol)

ΔHrxn = (-7092 kJ/mol + -4356 kJ/mol) - (-948 kJ/mol)

ΔHrxn = -11448 kJ/mol + 948 kJ/mol

ΔHrxn = -10500 kJ/mol

To find the heat (q) released when 206 g of stearic acid reacts, we first convert the mass of stearic acid to moles using its molar mass (284.48 g/mol):

Moles of stearic acid = 206 g / 284.48 g/mol = 0.724 mol

Then, multiply the moles by the ΔHrxn:

q = 0.724 mol * -10500 kJ/mol

q = -7602 kJ

Therefore, 7602 kJ of heat is released in the combustion of 206 g of stearic acid.

The question is incomplete, here is the complete question:

Calculate the pH at of a 0.10 M solution of anilinium chloride [tex](C_6H_5NH_3Cl)[/tex] . Note that aniline [tex](C6H5NH2)[/tex] is a weak base with a [tex]pK_b[/tex] of 4.87. Round your answer to 1 decimal place.

Answer: The pH of the solution is 5.1

Explanation:

Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).

To calculate the pH of the solution, we use the equation:

[tex]pH=7-\frac{1}{2}[pK_b+\log C][/tex]

where,

[tex]pK_b[/tex] = negative logarithm of weak base which is aniline = 4.87

C = concentration of the salt = 0.10 M

Putting values in above equation, we get:

[tex]pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1[/tex]

Hence, the pH of the solution is 5.1

The pH of the solution is 5.1.

Anilinium chloride refers to the salt that should be created by the combination of a weak base (aniline) and a strong acid (HCl).

here the following equation should be used.

ph = 7-1/2(pkb+ logc)

here pkb = negative logarithm of the weak base i.e. aniline = 4.87

And, C = concentration of the salt = 0.10 M

Now the ph should be

= 7-1/2(4.87 + log(0.10))

= 5.1

Hence, The pH of the solution is 5.1.

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Answer:

1. Gender role

2. Sex

Explanation:

Answer:

Explanation:

The detailed and step by step calculation is as shown in the attachment.

Charge of the electron: [tex]-2.3\cdot 10^{-19}C[/tex]

Explanation:

In Millikan experiment, it was discovered that the electric charge on the oil drops is discrete - and its value is always an integer multiple of a certain charge [tex]e[/tex], called fundamental charge (the charge of the electron). This is because an oil drop always contains an integer number of electrons, so the charge must be a multiple of [tex]e[/tex].

This means that we can write the charge on an oil drop as

[tex]Q=Ne[/tex]

For the drop recorded in this experiment, we have:

[tex]Q_A = N_A e = -6.9\cdot 10^{-19}C[/tex]

[tex]Q_B = N_B e = -9.2\cdot 10^{-19}C[/tex]

[tex]Q_C = N_C e = -11.5\cdot 10^{-19}C[/tex]

[tex]Q_D = N_D e = -4.6\cdot 10^{-19}C[/tex]

By dividing drop A by drop D, we get

[tex]\frac{Q_A}{Q_D}=\frac{3}{2}[/tex]

Also by dividing deop B by drop D we get

[tex]\frac{Q_B}{Q_D}=\frac{4}{2}[/tex]

And also, by dividing drop C by drop D we get

[tex]\frac{Q_C}{Q_D}=\frac{5}{2}[/tex]

This means that the charges on drop A, B, C and D are in the ratio

3 : 4 : 5 : 2

And therefore, the fundamental charge must be half of the charge on drop D:

[tex]e=\frac{Q_D}{2}=-2.3\cdot 10^{-19}C[/tex]

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Answer:

Only sketch B has the water molecules in the right form/state that the temperature presented predicts!

Explanation:

N.B - With the initial assumption that all the processes or water states exist at normal conditions of atmospheric pressure and temperature!

In the image attached to this solution, sketch A is at -23°C, sketch B is at 237°C and sketch C is at 60°C.

But for water, it's boiling point is 100°C; meaning that the this is the temperature where water molecules change form from fairly free to move around, almost incompressible liquid state to the gaseous state in which the water molecules (now called steam) are totally free to move around.

Its melting point is 0°C; that is, this is the temperature where the water molecules change form from the orderly solid form (called ice) where motion is totally restricted to only vibrations into the more free liquid state.

This explanation indicates that water molecules at temperatures below 0°C exist in the orderly solid form.

Water molecules at temperatures between 0°C and 100°C exist as the fairly free liquid and at temperatures higher than 100°C, the water molecules exist in the free to move about gaseous state.

In the sketches attached to this solution, sketch A evidently shows the water molecules in the fairly free to move about form (that is, liquid form), but matches this state with a temperature of -23°C which corresponds more to the solid, orderly state of water molecules shown in sketch C. Hence, that is a mismatch.

Sketch B shows water molecules in the very freeing state of gaseous form and rightly matches that form with a temperature way above the boiling point of water, 237°C. Thereby indicating a correct match between temperature and the sketch.

Sketch C however shows water molecules in their very organized solid form but mismatches this form to 60°C which corresponds more to the liquid state sketch in sketch A.

Only sketch B has the water molecules in the right form/state that the temperature presented predicts!

Hope this helps!!!

Answer:

(a) AL

(b) Sc

(c)Al

Explanation:

Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.

The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.

By looking at electron configuration we can figure out which electron will need more energy.

1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹

Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

Li₃ ⇒ 1s², 2s¹

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

B₅ ⇒ 1s², 2s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

Answer:

A. Al

B. Sc

C. Al

Explanation:

The third ionisation energy is the energy required to an extra electron from a +2 ion or the energy required to remove the third electron from an element.

Lithium - 1s2 2s1

Sodium - 1s2 2s2 2p6 3s1

Magnesium - 1s2 2s2 2p6 3s2

Aluminium - 1s2 2s2 2p6 3s2 3p1

Potassium - 1s2 2s2 2p6 3s2 3p6 4s1

Calcium- 1s2 2s2 2p6 3s2 3p6 4s2

Boron - 1s2 2s2 2p1

Scandium - 1s2 2s2 2p6 3s2 3p6 3d1 4s2

Removing 2 electrons,

Li2+- 1s1

Na2+ - 1s2 2s2 2p5

Mg2+ - 1s2 2s2 2p6

Al2+ - 1s2 2s2 2p6 3s1

K2+ - 1s2 2s2 2p6 3s2 3p5

Ca2+ - 1s2 2s2 2p6 3s2 3p4

Boron - 1s2 2s1

Scandium - 1s2 2s2 2p6 3s2 3p6 4s1

So comparing,

A. Na, Mg, Al

The third electron is lost from a p- orbital and the energy level of p- is less than s- orbital but 3s is way less than the 2p so the lowest third ionisation energy is Al

B. K, Ca, Sc

The third electrons are lost from the 3p orbital in K and Ca but in 4s in Sc and if you remember, 4s has a lesser energy level than 3p orbital. So, Sc has the lowest third ionisation energy.

C. Li, Al, B

Al has the lowest third ionisation energy because Li loses its from 1s which is closest to the nucleus and B from 2s which is also close to the nucleus.

Given the pKa values of serine's α‑carboxyl and α‑amino groups, the carboxyl group is fully ionized and carries a -1 charge while the amino group carries a +1 charge at a pH of 8.80. Therefore, the total net charge on serine at this pH is zero.

Your question asks to calculate the average net charge on the amino acid serine at a pH of 8.80, given the pKa values for the α‑carboxyl group (2.21) and the α‑amino group (9.15). This is related to the concept of acid dissociation constants (pKa) and buffer solutions in chemistry.

At a pH of 8.80, the pH is higher than the pKa of the α‑carboxyl group but lower than the pKa of the amino group. For the α‑carboxyl group whose pKa is 2.21, the pH is significantly higher. This means, it is fully ionized and carries a -1 charge.

On the other hand, the α‑amino group has a pKa of 9.15, which is higher than the pH of 8.80. This means it is predominantly in its protonated form and carries a +1 charge.

Therefore, the total charge on serine at this pH is the sum of the charges of the α‑carboxyl group and the α‑amino group, which is -1 + 1 = 0. Hence, the average net charge of serine in a solution with pH 8.80 is zero.

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Answer:

A

Explanation:

A volcanic eruption affects the Earth's atmosphere mainly through the release of ash and poisonous gases. These materials can impact global climate, air travel, human health, and regional weather patterns.

A volcanic eruption impacts the Earth's atmosphere in a number of ways, largely through the release of volcanic gases and ash. The most significant aspect is option A, eruptions shooting out ash and poisonous gases into the air.

These gases, including sulfur dioxide, can contribute to the formation of aerosols in the higher layers of the atmosphere, potentially affecting global climate patterns. For example, a large volcanic eruption can lead to cooler temperatures worldwide for a few years.

Additionally, the ash particles ejected during a volcanic eruption can have a range of effects, from impacting air travel to affecting human health and altering regional weather patterns.

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Answer:

mole fraction methanol = 0.76

mole fraction ethanol = 0.24

Explanation:

Raoult´s law  gives us the partial vapor pressure of a  component in solution as the product of the mole fraction of the component and the value of its pure pressure:

PA  = X(A) x Pº(A)

where PA is the partial vapor pressure of component A, X(A) is the mole fraction of A, and  Pº(A) its pure vapor pressure.

From reference literature the pure pressures of methanol, and ethanol are at 25 ºC :

PºCH₃OH = 16.96 kPa

PºC₂H₅OH =  7.87 kPa

Given that we already have the mole fractions, we can calculate the partial vapor pressures as follows:

PCH₃OH = 0.600 x 16.96 kPa = 10.18 kPa

PC₂H₅OH = 0.400 x 7.87 kPa = 3.15 kPa

Now the total pressure in the gas phase is:

Ptotal = PCH₃OH + PC₂H₅OH  = 10.18 kPa + 3.15 kPa = 13.33 kPa

and the mole fractions in the vapor will be given by:

X CH₃OH  = PCH₃OH / Ptotal = 10.18 kPa/ 13.33 kPa = 0.76

X C₂H₅OH = 1 - 0.76 = 0.24

Answer: The theoretical yield and percent yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] is 3.93 g and 30.53 % respectively

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of [tex]CoCl_2.6H_2O[/tex] = 4.00 g

Molar mass of [tex]CoCl_2.6H_2O[/tex] = 238 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol[/tex]

The chemical equation for the reaction of [tex]CoCl_2.6H_2O[/tex] to form  [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] follows:

[tex]CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of [tex]CoCl_2.6H_2O[/tex] produces 1 mole of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex]

So, 0.0168 moles of [tex]CoCl_2.6H_2O[/tex] will produce = [tex]\frac{1}{1}\times 0.0168=0.0168mol[/tex] of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex]

Now, calculating the mass of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] from equation 1, we get:

Molar mass of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 234 g/mol

Moles of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 0.0168 moles

Putting values in equation 1, we get:

[tex]0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g[/tex]

To calculate the percentage yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex], we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 1.20 g

Theoretical yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] = 3.93 g

Putting values in above equation, we get:

[tex]\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%[/tex]

Hence, the theoretical yield and percent yield of [tex][Co(NH_3)_4(H_2O)_2]Cl_2[/tex] is 3.93 g and 30.53 % respectively

Answer:

ion - dipole interactions

Explanation:

Ion - dipole interactions -

It refers to the interaction between the ion and a dipole , ( any species which is capable to get produce slight positive and slight negative charge ) , is known as ion - dipole interactions .

Water is a polar compound , and due to more electronegative oxygen atom , it can have slight negative charge and correspondingly , hydrogen atom can attain slight positive charge , and thereby generates a dipole .

Now , from the question,

The salt when dissolved in water , breaks down to ions , cations and anion , and these ions interacts with the polar water molecules , giving rise to the ion - dipole interactions .

Answer:

35.5 % NaCl

Explanation:

Total mass of mixture = 30.126 g

Mass of sand = 15.976 g

Mass of iron = 3.455 g

Mass of NaCl = ? → Total mass - 15.976 g - 3.455 g = 10.695 g

% mass of NaCl in the sample → (Mass of NaCl / Total mass) . 100

(10.695 g / 30.126 g) . 100 = 35.5 %

Answer:

Explanation:

Two constitutional isomers of C₂H₇P

1 ) CH₃-CH₂ - PH₂ ( Ethyl phosphane )

2 ) CH₃ - PH - CH₃ ( Dimethyl phosohane )

The chemical formulae of constitutional isomers are the same, but their connectivities differ. Constitutional isomers include ethanol and dimethyl ether, as well as n-butane and isobutane.

Although structural (constitutional) isomers share the same chemical formula, their atoms are bonded in different ways. Stereoisomers share the same atomic configurations and chemical formulae. Only the spatial arrangement of the groups within the molecule separates them from one another.

Compounds with the same chemical formula but different properties are known as isomers. Constitutional isomers are isomers that have different atom connections.

Two constitutional isomers of C₂H₇P

1 ) CH₃-CH₂ - PH₂ ( Ethyl phosphane )

2 ) CH₃ - PH - CH₃ ( Dimethyl phosphane )

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Answer: The major product of the reaction between Hydrobromic Acid and 2-bromo-1-butene in the presence of ether and acid is 2,2-dibromobutane.

Explanation:

The mechanism of the reaction is supported by the Markovnikov's rule which explains that in the addition reaction of alkenes by hydrogen-halogen compounds, the incoming halogen substituent goes to the more substituted Carbon. It can also be stated that incoming hydrogen atom goes to the Carbon with more Hydrogen atoms.

The only case when the reverse of Markovnikov's rule takes place is when Hydrogen peroxide is present and the addition reagent is HBr.

This case is not like that and it simply follows the Markovnikov's rule.

I'll add an attachment of the reaction to this now.

Answer:

On the reaction the product is 2-2-dibromobutane.

Explanation:

2-Bromo-1 butene is given as in the figure. On the breaking of the double bond 2 local electrophilic and nucleophilic radicals will be formed in the 2-Bromo-1-butene and H-Br respectively.

Due to the Markovnikov Rule the nucleophilic radical of the attacking compound bonds with the carbon atom with least number of H atoms so the product formed will be 2-2-dibromobutane as indicated in the figure.

Answer : The pH will be, 3.2

Explanation :

As we known that the value of solubility constant of ferric hydroxide at [tex]25^oC[/tex] is, [tex]2.79\times 10^{-39}[/tex]

Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L

The given solubility of iron convert from mg/L to mol/L.

[tex]1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L[/tex]

The chemical reaction will be:

[tex]Fe(OH)_3\rightarrow Fe^{3+}+3OH^-[/tex]

The expression of solubility constant will be:

[tex]K_{sp}=[Fe^{3+}]\times [3OH^-]^3[/tex]

Now put all the given values in this expression, we get the concentration of hydroxide ion.

[tex]2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3[/tex]

[tex][OH^-]=1.5\times 10^{-11}M[/tex]

Now we have to calculate the pOH.

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

[tex]pOH=-\log (1.5\times 10^{-11})[/tex]

[tex]pOH=10.8[/tex]

Now we have to calculate the pH.

[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2[/tex]

Therefore, the pH will be, 3.2

To precipitate all but 0.3 mg/L of iron, the pH must be raised to a certain value. This can be calculated by using the solubility product constant (Ksp) and the balanced equation for the precipitation reaction. By solving for pOH and then pH, we can determine the pH at which the majority of the iron will precipitate.

To calculate the pH at which all but 0.3 mg/L of iron will precipitate, we can use the solubility product constant (Ksp) of the iron precipitate. The balanced equation for the precipitation reaction is:

Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)

The Ksp expression for this reaction is:

Ksp = [Fe3+][OH-]^3

Assuming that Fe(OH)3 is the only significant source of Fe3+ ions, we can set up an equilibrium expression:

[Fe3+] = Ksp / [OH-]^3

Given that the concentration of Fe3+ ions we want to achieve is 0.3 mg/L, we can substitute this value and solve for the hydroxide ion concentration:

[OH-]^3 = Ksp / [Fe3+]

Since Fe(OH)3 is an amphoteric hydroxide and can behave as both an acid and a base, we can assume that the hydroxide ion concentration is equal to the concentration of the hydroxide ion from water:

[OH-] = 10^(-pOH)

Plugging in the values, we get:

(10^(-pOH))^3 = Ksp / [Fe3+]

Simplifying, we obtain:

10^(-3pOH) = Ksp / [Fe3+]

Taking the logarithm of both sides:

-3pOH = log(Ksp / [Fe3+])

Rearranging the equation and solving for pOH:

pOH = -log(Ksp / [Fe3+]) / 3

To find the pH, we can use the relationship:

pH + pOH = 14

Substituting the pOH value we obtained above:

pH = 14 - pOH

Now we can substitute the given values and calculate the pH at which the majority of iron will precipitate.

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Answer:

Zn + 2HCl —> ZnCl2 + H2

Explanation:

This is a displacement reaction, in which Zn displaces H2 from acid. The equation for the reaction is given below:

Zn + 2HCl —> ZnCl2 + H2

Final answer:

The answer to the question is Na3PO4, as it provides the same anion as Na2HPO4 without a conjugate acid or base, making it unable to form a buffer solution with Na2HPO4.

Explanation:

A buffer solution is formed from a weak acid and its conjugate base, or a weak base and its conjugate acid. Therefore, a 0.10 M solution of Na2HPO4 (which is sodium hydrogen phosphate) could be made into a buffer solution with another compound that either provides its conjugate acid or its conjugate base. Na2HPO4 can act as both a weak acid (donating H+) and a weak base (accepting H+).

A buffer solution with Na2HPO4 could be made using the following combinations:

The one that cannot be used to form a buffer with Na2HPO4 is Na3PO4, because Na3PO4 is the fully deprotonated form and provides the same anion as Na2HPO4 without an accompanying conjugate acid or base. Therefore, the correct answer is Na3PO4.

Answer : The pH of the solution is, 5.22

Explanation :

First we have to calculate the moles of acetic acid and sodium acetate.

[tex]\text{Moles of acetic acid}=\text{Concentration of acetic acid}\times \text{Volume of acetic acid}=0.01M\times 0.025L=0.00025mol[/tex]

and,

[tex]\text{Moles of sodium acetate}=\text{Concentration of sodium acetate}\times \text{Volume of sodium acetate}=0.03M\times 0.025L=0.00075mol[/tex]

Now we have to calculate the concentration of acetic acid and sodium acetate in 1 L of solution.

[tex]\text{Concentration of acetic acid}=\frac{\text{Moles of acetic acid}}{\text{Volume of solution}}=\frac{0.00025mol}{1L}=0.00025M[/tex]

and,

[tex]\text{Concentration of sodium acetate}=\frac{\text{Moles of sodium acetate}}{\text{Volume of solution}}=\frac{0.00075mol}{1L}=0.00075M[/tex]

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[\text{Sodium acetate}]}{[\text{Acetic acid}]}[/tex]

[tex]pK_a[/tex]  of acetic acid = 4.74

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

[tex]pH=4.74+\log (\frac{0.00075}{0.00025})[/tex]

[tex]pH=5.22[/tex]

Thus, the pH of the solution is, 5.22

To find the pH of the solution, calculate the concentration of the acetic acid and sodium acetate. Convert the volume of acetic acid and sodium acetate to liters. Use the Henderson-Hasselbalch equation to find the pH of the buffer solution.

To determine the pH of the solution, we need to calculate the concentration of the acetic acid and its conjugate base, sodium acetate. First, convert the volume of acetic acid and sodium acetate to liters. The concentration of acetic acid is 0.01 M (10 mM), and the concentration of sodium acetate is 0.03 M (30 mM). Next, calculate the moles of acetic acid and sodium acetate using the equation moles = concentration x volume. The moles of acetic acid are 0.01 moles, and the moles of sodium acetate are 0.03 moles. The solution contains a weak acid (acetic acid) and its conjugate base (sodium acetate), which makes it a buffer solution. To find the pH of a buffer solution, we need to use the Henderson-Hasselbalch equation: pH = pKa + log ([base]/[acid]). The pKa of acetic acid is 4.74. Substitute the values into the equation: pH = 4.74 + log (0.03/0.01) = 4.74 + log(3) = 4.74 + 0.48 = 5.22. Therefore, the pH of the 1-L solution is 5.22.

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Answer:

[tex]\large \boxed{\text{0.012 mol}}[/tex]  

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

               CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL:                    70.

c/mol·L⁻¹:             0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-p-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

[tex]\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}[/tex]

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

[tex]\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }[/tex]

Answer:

The van't Hoff factor of the solute is 1.51

Explanation:

Step 1: Data given

Molality = 0.710 molal

The aqueous solution freezes at − 2.00°C

Freezing point depression constant of water = 1.86 °C/m

Step 2: Calculate the van't Hoff factor

ΔT = i*Kf * m

⇒ with ΔT = The difference between the feezing point of pure and solution = 2.00°C

⇒ i the van't Hoff factor = TO BE DETERMINED

⇒ Kf = Freezing point depression constant of water = 1.86 °C/m

⇒ m = the molality of the solution = 0.710 molal

2.00 = i * 1.86 * 0.710

i = 1.51

The van't Hoff factor of the solute is 1.51

Final answer:

To find the van't Hoff factor for a solution that freezes at − 2.00°C with a molality of 0.710m, the van't Hoff factor, i, is approximately calculated as 1.52, indicating dissociation into multiple particles.

Explanation:

To calculate the van't Hoff factor, i, for the solute in a 0.710 m aqueous solution that freezes at − 2.00 ℃, we use the formula for freezing point depression, ΔTf = iKfm, where ΔTf is the freezing point depression, Kf is the molal freezing-point depression constant for the solvent (water in this case, with a value of − 1.86°C/m), and m is the molality of the solution. First, understand that the freezing point of pure water is 0°C, and the solution's freezing point is − 2.00°C, so the depression, ΔTf, is 2.00°C. To find the van't Hoff factor, i, rearrange the equation to i = ΔTf / (Kfm). Substituting the values gives us i = 2.00 °C / (− 1.86°C/m × 0.710 m) = 2.00°C / − 1.32°C = − 1.52. However, since i should be a positive value and considering the potential rounding or measurement error, the magnitude is taken yielding i approximately equal to 1.52, reflecting the number of particles the solute dissociates into in solution.

Answer:

B. False

A. True

B. False in

Explanation:

1.

From Bronsted-Lowrys definition of Acids and bases, a strong acid is a substance that gives up a proton (to form a weak conjugate base), while a strong base is one that willingly accepts a proton.

Therefore in the reaction,

HA(aq) + H2O (l) ⇌ H3O+ (aq) + A-

The stronger HA is, the weaker the A- and vice versa; the weaker HA is, the stronger A-. Example, HCl is a strong acid and its conjugate base, Cl- is a weak base.

B. False

2.

Electron drawing groups are molecules and/or atoms that enable the release of a proton from a specie. They cause inductive as well as mesomeric effects. Examples, -NO2, -COOH, -OH etc.

Fluoride ion is the most stable in this series because it's the most electronegative while carbon is the least stable because it's the least electronegative. Because of this, we were able to say that H-F was the most acidic, because it had the most stable conjugate base.

A. True.

3.

pKa1 = -log[Ka1]

= -log[1.2 x 10-4]

= 3.92

pKa2 = -log[Ka2]

= -log[1.5 x 10-8]

= 7.82

Using Bronsted-Lowry definition, the smaller the pKa value the more ease the acid loses its proton, that is, the smaller the pKa value, the stronger the acid.

Therefore, pKa2 > pKa1, so Ka value of 1.2 x 10-4 is a stronger acid than Ka value of 1.5 x 10-8

B. False.

1. B. False

2. A. True

3. B. False

1.

Bronsted Lowry Concept:

According to Bronsted Lowry, an acid is a proton (H⁺) donor, and a base is a proton acceptor.

The given reaction,

[tex]HA(aq) + H_2O (l)[/tex] ⇌  [tex]H_3O^+ (aq) + A^-[/tex]

The stronger HA is, the weaker the A- and vice versa; the weaker HA is, the stronger A-.

Example, HCl is a strong acid and its conjugate base, Cl- is a weak base.

2.

Electron drawing groups are molecules and/or atoms that enable the release of a proton from a species. Examples, -NO₂, -COOH, -OH etc.

Fluoride ion is the most stable in this series because it's the most electronegative while carbon is the least stable because it's the least electronegative. Because of this, we were able to say that H-F was the most acidic, because it had the most stable conjugate base.

3.

[tex]p_{Ka_1} = -log[Ka_1]\\\\p_{Ka_1}= -log[1.2 * 10^{-4}]\\\\p_{Ka_1}= 3.92[/tex]

[tex]p_{Ka_2} = -log[Ka_2]\\\\p_{Ka_2}= -log[1.5 * 10^{-8}]\\\\p_{Ka_2}= 7.82[/tex]

Using Bronsted-Lowry definition, the smaller the pKa value the more ease the acid loses its proton, that is, the smaller the pKa value, the stronger the acid.

Therefore, [tex]p_{Ka_2} > p_{Ka_1}[/tex], so Ka value of[tex]1.2 * 10^{-4}[/tex] is a stronger acid than Ka value of [tex]1.5 * 10^{-8}[/tex]

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