Consider the generic acid, HA, and how it interacts with water:HA (aq) + H2O (l) ⇌ H3O+ (aq) + A-Consider the following statements and answer true or false. The stronger A- is as a weak base, the stronger HA will be as a weak acid.a. True b. False Having a greater number of electron withdrawing groups is typically the best way to stabilize a conjugate base.a. True b. False If HA is a strong acid, A- will be a relatively strong weak conjugate base.a. True b. False An acid with a Ka value of 1.2 x 10-4 is a weaker acid than an acid with a Ka value of 1.5 x 10-8.a. True b. False

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 boiling point = 89.69 °C

Explanation:

Step 1: Data given

Mass of naphthalane = 7.0 grams

Mass of benzene = 14.4 grams

The Kbp of the solvent = 2.53 K/m

The normal boiling point is 80.1°C

Naphthalene, C10H8 , is a non-electrolyte, which means that the van't Hoff factor for this solution will be 1

Step 2: Calculate moles naphthalene

Moles naphthalene = mass / molar mass

Moles naphthalene = 7.0 grams / 128.17 g/mol

Moles naphthalene = 0.0546 moles

Step 3: Calculate molality

Molality = moles naphthalene / mass water

Molality = 0.0546 moles / 0.0144 kg

Molality = 3.79 molal

Step 4:

ΔT = i*Kb*m

ΔT = 1*2.53 K/m * 3.79 molal

ΔT = 9.59 °C

The boiling point = 80.1 °C + 9.59 °C  = 89.69 °C

The boiling point of a solution made by dissolving 7 g of naphthalene in 14.4 g of benzene, with a Kbp of 2.53 K/m, is 89.68 degrees C.

To calculate the boiling point of a solution made by dissolving naphthalene in benzene, we can use the boiling point elevation formula: \(\Delta T = i \cdot K_{bp} \cdot m\), where \(\Delta T\) is the boiling point elevation, \(i\) is the van't Hoff factor (which is 1 for non-electrolytes like naphthalene), \(K_{bp}\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution.

The molality (\(m\)) is calculated using the formula: \(m = \frac{moles\ of\ solute}{kilograms\ of\ solvent}\). Naphthalene's molar mass is 128.17 g/mol. Thus, the moles of naphthalene are \(\frac{7\ g}{128.17\ g/mol} = 0.0546\ moles\). The mass of benzene is 14.4 g, which is 0.0144 kg. So, the molality is \(\frac{0.0546\ moles}{0.0144\ kg} = 3.79\ m\).

Now, we can find the boiling point elevation: \(\Delta T = 1 \cdot 2.53\ K/m \cdot 3.79\ m = 9.58\ K\). Convert K to \(\degree C\) by using the normal boiling point of benzene (80.1 \(\degree C\)) plus the boiling point elevation: \(80.1 \(\degree C\) + 9.58 \(\degree C\) = 89.68 \(\degree C\)\).

The boiling point of this solution is 89.68 degrees C.

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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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.

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:(a) 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)

(b) 5.55*10^37photons

Explanation:

(a) Here we have to write a balanced thermochemical equation for formation of 1.00 mol of glucose.

In this question it has been given that Chlorophyll absorbs light in the 600 to 700 nm region,

1st we will write a chemical equation for biochemical process of photosynthesis is that CO2 and H2O form glucose (C6H12O6) and O2.

6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)

The heat change off the reaction can be calculated as,

={(1 mol)(6 mol) }- {(6 mol) [H2O]}

=[1 1273.3 kJ + 6(0)] - [6 (-39.5 kJ) + 6 (-285.840 kJ)]

= 2802.74 or 2802.7 kJ

Thus the balanced equation can be written as,

6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g) = 2802.7 kJ for 1.00 mol of glucose.

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:

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:

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: 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 of the solution is, 0.932

Explanation :

First we have to calculate the moles of HCl and NaOH.

[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.200mole/L\times 0.200L=0.040mole[/tex]

[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.050mole/L\times 0.100L=0.0050mole[/tex]

The balanced chemical reaction will be,

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

From the balanced reaction we conclude that,

As, 1 mole of NaOH neutralizes by 1 mole of HCl

So, 0.0050 mole of NaOH neutralizes by 0.0050 mole of HCl

Thus, the number of neutralized moles = 0.0050 mole

Remaining moles of HCl = 0.040 - 0.0050 = 0.035 moles

Total volume of solution = 200.0 mL + 100.0 mL = 300.0 mL = 0.300 L

Now we have to calculate the concentration of HCl(acid).

[tex]Concentration=\frac{Moles}{Volume}=\frac{0.035mol}{0.300L}=0.117M[/tex]

As we know that, 1 mole of HCl dissociates to give 1 mole of hydrogen ion and 1 mole of chloride ion.

So, concentration of [tex]H^+[/tex] = 0.117 M

Now we have to calculate the pH of solution.

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

[tex]pH=-\log (0.117)[/tex]

[tex]pH=0.932[/tex]

Thus, the pH of the solution is, 0.932

"The pH after the addition of 100.0 mL of 0.050 M NaOH to 200.0 mL of 0.200 M HCl is approximately 2.68.

To find the pH, we first need to determine the moles of HCl and NaOH involved in the reaction:

Moles of HCl = volume (L) — concentration (mol/L)

Moles of HCl = 0.200 L — 0.200 mol/L = 0.0400 mol

 Moles of NaOH = volume (L) — concentration (mol/L)

Moles of NaOH = 0.100 L — 0.050 mol/L = 0.0050 mol

 The balanced equation for the reaction between HCl and NaOH is:

[tex]\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]

From the equation, we see that the reaction between HCl and NaOH is 1:1. Therefore, the moles of NaOH added will neutralize an equal number of moles of HCl.

 Moles of HCl remaining after neutralization = moles of HCl initially - moles of NaOH added

Moles of HCl remaining = 0.0400 mol - 0.0050 mol = 0.0350 mol

The new concentration of HCl after the addition of NaOH is:

[tex]\[ \text{Concentration of HCl} = \frac{\text{moles of HCl remaining}}{\text{total volume}} \] \[ \text{Concentration of HCl} = \frac{0.0350 \text{ mol}}{0.200 \text{ L} + 0.100 \text{ L}} = \frac{0.0350 \text{ mol}}{0.300 \text{ L}} \] \[ \text{Concentration of HCl} = 0.1167 \text{ M} \][/tex]

Since HCl is a strong acid, it dissociates completely in water. Therefore, the concentration of HCl is equal to the concentration of H+ ions in the solution.

[tex]\[ \text{pH} = -\log[\text{H}^+] \] \[ \text{pH} = -\log(0.1167) \] \[ \text{pH} \approx 2.68 \][/tex]

Thus, the pH of the solution after the addition of 100.0 mL of 0.050 M NaOH is approximately 2.68."

Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are [tex]6.023 \times 10^{26}[/tex] atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = [tex]\frac{38}{6.023 \times 10^{26}}[/tex]    

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

            = [tex]\frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}[/tex]

            = [tex]37.06 \times 10^{-26}[/tex]

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = [tex]a^{3}[/tex]

                   = [tex](0.503 \times 10^{-9})^{3}[/tex]

                   = [tex]0.127 \times 10^{-27} m^{3}[/tex]

Formula to calculate density of diamond cell is as follows.

               Density = [tex]\frac{mass}{volume}[/tex]

                             = [tex]\frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}[/tex]

                            = 2918.1 [tex]g/m^{3}[/tex]

or,                         = 0.0029 g/cc       (as 1 [tex]m^{3} = 10^{6} cm^{3}[/tex])

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

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: a) The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.

b) The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.

Explanation:

The formula that relates wavelength of emissions to Rydberg's constant and the n₁ values is

(1/λ) = R ((1/(n₁^2)) - (1/(n2^2))

Where λ = wavelength, R = (10.972 × 10^6)/m, n2 = ∞ (since they're emitted out of the atom already)

a) n₁ = ?

λ = 3282 nm = (3.282 × 10^-6)m

(1/(3.282 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2) (since 1/∞ = 0)

n₁^2 = (3.282 × 10^-6) × (10.972 × 10^6) = 36

n₁ = 6.

The value of n₁ that would produce a series of lines in which the highest energy line has a wavelength of 3282 nm is 6.

b) n₁ = ?

λ = 7460 nm = (7.46 × 10^-6)m

(1/(7.46 × 10^-6)) = (10.972 × 10^6) ((1/(n₁^2)) - (1/(n2^2)) for lowest energy line, n2 = n₁ + 1

(n₁^2)((n₁+1)^2))/(2n₁+1) = (7.46 × 10^-6) × (10.972 × 10^6) = 81.85

(n₁^2)((n₁+1)^2))/(2n₁+1) = 81.85

Solving the quadratic eqn,

n₁ = 5.

The value of n₁ that would produce a series of lines in which the lowest energy line has a wavelength of 7460 nm is 5.

QED!

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

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:

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:

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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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:

D) NaO2CC6H4CO2K

Explanation:

Water is a polar solvent and tends to solvate polar molecules. This allows solute molecules to interact with the solvent and that is why the solubility of a molecule in water increases with the increase in its polarity. So, the salt of phthalic acid is more soluble in water than phthalic acid itself. Although the monosodium and monopotassium salts are also more soluble than phthalic acid, the dialkali phthalate salt (NaO2CC6H4CO2K) is the most soluble due to the highest polarity.

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 amount of p-toluidine hydrochloride contained is 2.4 moles.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

Molarity of p-toluidine hydrochloride solution = 0.167 M

Volume of solution = 70. mL

Putting values in equation 1, we get:

[tex]0.167=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{(0.167\times 70}{1000}=2.38mol=2.4mol[/tex]

Hence, the amount of p-toluidine hydrochloride contained is 2.4 moles.

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:

(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.

Answer:

Explanation:

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

Answer:

1. Gender role

2. Sex

Explanation:

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 : 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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