"A student prepares a solution by dissolving 1.66 g of solid KOH in enough water to make 500.0 mL of solution. Calculate the molarity of K+ ions in this solution.

A 35.00 mL sample of this KOH solution is added to a 1000 mL volumetric flask, and water is added to the mark. What is the new molarity of K+ ions in this solution?"

Answer: 1g of formaldehyde

Explanation:

Molar Mass of benzene (C6H6) = (12x6) + (6x1) = 78g/mol

1mole(78g) of benzene contains 6.02x10^23 molecules.

1g of benzene with contain = (6.02x10^23) /78 = 7.72x10^21 molecules

Molar Mass of formaldehyde (CH2O) = 12 +2+16 = 30g/mol

1mole (30g) of formaldehyde contains 6.02x10^23 molecules.

1g of formaldehyde will contain = (6.02x10^23) /30 = 2.01x10^22 molecules

Molar Mass of TNT(C7H5N3O6) = (12x7)+(5x1)+(14x3)+(16x6) = 227g/mol

1mole(227g) of TNT contains 6.02x10^23 molecules.

1g of TNT will contain = (6.02x10^23)/227 = 2.65x10^21 molecules

Molar Mass of naphthalene (C10H8) = (12x10) +8 =128g/mol

1mole(128g) of naphthalene contains 6.02x10^23 molecules.

1g of naphthalene will contain = (6.02x10^23)/128 = 4.7x10^21 molecules

Molar Mass of glucose(C6H12O6) = (12x6) + 12 +(16x6) = 180g/mol

1mole(180g) of glucose contains 6.02x10^23 molecules.

1g of glucose will contain = (6.02x10^23)/180 = 3.34x10^21 molecules

Therefore, 1g of each of the compound contains the following:

Benzene = 7.72x10^21 molecules

Formaldehyde = 2.01x10^22 molecules

TNT = 2.65x10^21 molecules

Naphthalene = 4.7x10^21 molecules

Glucose = 3.34x10^21 molecules

From the above, we see clearly that formaldehyde has the largest number of molecules

Final answer:

The compound with the largest number of molecules is formaldehyde with 2.00 x 10^22 molecules.

Explanation:

The substance with the largest number of molecules can be determined by calculating the number of moles of each substance and then using Avogadro's number to convert moles to molecules. The formula to calculate the number of moles is moles = mass / molar mass. By using this formula:

For benzene: moles = 1g / 78.11g/mol = 0.0128 moles

For formaldehyde: moles = 1g / 30.03g/mol = 0.0333 moles

For TNT: moles = 1g / 227.13g/mol = 0.0044 moles

For naphthalene: moles = 1g / 128.2g/mol = 0.0078 moles

For glucose: moles = 1g / 180.16g/mol = 0.0055 moles

Then, we can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules:

For benzene: molecules = 0.0128 mol x 6.022 x 10^23 molecules/mol = 7.72 x 10^21 molecules

For formaldehyde: molecules = 0.0333 mol x 6.022 x 10^23 molecules/mol = 2.00 x 10^22 molecules

For TNT: molecules = 0.0044 mol x 6.022 x 10^23 molecules/mol = 2.65 x 10^21 molecules

For naphthalene: molecules = 0.0078 mol x 6.022 x 10^23 molecules/mol = 4.68 x 10^21 molecules

For glucose: molecules = 0.0055 mol x 6.022 x 10^23 molecules/mol = 3.31 x 10^21 molecules

Therefore, the compound with the largest number of molecules is formaldehyde with 2.00 x 10^22 molecules.

Answer: The decreasing order of atomic size:

For a:  Pb > Sn > Ge

For b:  Sr > Sn > Te

For c:  Na > F > Ne

For d:  Mg > Na > Be

Explanation:

Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.

As moving from top to bottom in a group, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases. Thus, increasing the atomic radii of the atom.

As moving from left to right in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.

For the given options:

Germanium (Ge) is present in Group 14 and period 4 of the periodic table.

Lead (Pb) is present in Group 14 and period 6 of the periodic table.

Tin (Sn) is present in Group 14 and period 5 of the periodic table.

So, the decreasing order of atomic size is:

Pb > Sn > Ge

Tin (Sn) is present in Group 14 and period 5 of the periodic table.

Tellurium (Te) is present in Group 16 and period 5 of the periodic table.

Strontium (Sr) is present in Group 2 and period 5 of the periodic table.

So, the decreasing order of atomic size is:

Sr > Sn > Te

Fluorine (F) is present in Group 17 and period 2 of the periodic table.

Neon (Ne) is present in Group 18 and period 2 of the periodic table.

Sodium (Na) is present in Group 1 and period 3 of the periodic table.

So, the decreasing order of atomic size is:

Na > F > Ne

Beryllium (Be) is present in Group 2 and period 2 of the periodic table.

Magnesium (Mg) is present in Group 2 and period 3 of the periodic table.

Sodium (Na) is present in Group 1 and period 3 of the periodic table.

So, the decreasing order of atomic size is:

Mg > Na > Be

The decreasing order of atomic size are the following:

a:  Pb > Sn > Ge

b:  Sr > Sn > Te

c:  Na > F > Ne

d:  Mg > Na > Be

Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom. As move from top to bottom in a group,  addition of shell occurs and the atomic radii of the atom increases.

While on the other hand. as we move from left to right, atomic size of an atom decreases due to addition of nuclear charge in the nucleus.

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Answer:

Option (B)

Explanation:

In the field of mineralogy, cleavage is usually defined as a surface along which a mineral has the tendency to break. These surfaces are the plane of weaknesses that are formed due to deformation as well as metamorphism.

Fluorite is a mineral that has a hardness of 4, according to the Mohs hardness scale. It is comprised of four cleavage directions. This means that it has four sets of perfect cleavage, and this mineral often breaks into small pieces that are octahedron in shape.

Thus, the correct answer is option (B).

Answer: The molecular weight of unknown non-electrolyte is 61.75 g/mol

Explanation:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

[tex]\Delta T_f=\text{Freezing point of benzophenone}-\text{Freezing point of solution}[/tex]

To calculate the depression in freezing point, we use the equation:

[tex]\Delta T_f=iK_fm[/tex]

or,

[tex]\text{Freezing point of benzophenone}-\text{Freezing point of solution}=iK_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]where,

i = Vant hoff factor = 1 (for non-electrolyte)

[tex]K_f[/tex] = molal freezing point depression constant = 9.80°C/m

[tex]m_{solute}[/tex] = Given mass of unknown non-electrolyte = 1.0223 g

[tex]M_{solute}[/tex] = Molar mass of unknown non-electrolyte = ?

[tex]W_{solvent}[/tex] = Mass of solvent (benzophenone) = 10.2685 g

Putting values in above equation, we get:

[tex]47.5-31.7=1\times 9.80\times \frac{1.0223\times 1000}{M_{solute}\times 10.2685}\\\\M_{solute}=\frac{1\times 9.80\times 1.0223\times 1000}{15.8\times 10.2685}=61.75g/mol[/tex]

Hence, the molecular weight of unknown non-electrolyte is 61.75 g/mol

The molecular weight of the unknown nonelectrolyte dissolved in benzophenone can be calculated using the formula for freezing point depression and the information provided in the question. After performing the necessary calculations, the molecular weight of the unknown compound is found to be 331 g/mol.

To calculate the molecular weight of the unknown compound, we use the formula for freezing point depression:

ΔT = iKfm

Where ΔT is the change in temperature, i is the van't Hoff factor which is 1 for nonelectrolytes, Kf is the cryoscopic constant for benzophenone, which, in this case, is 5.12°C kg/mol, and m is the molality of the solution. The change in temperature is the difference between the freezing points of pure benzophenone and the solution: 47.5°C - 31.7°C = 15.8°C.

The molality can be calculated by rearranging the formula:

m = ΔT / (iKf) = 15.8 / (1 * 5.12) = 3.086 mol/kg

Since we are given grams and want to convert to kilograms:

1.0223g of unknown compound / molar mass (unknown) = 3.086 mol/kg

Rearrange to find the molar mass (molecular weight) of the compound:

molar mass (unknown) = 1.0223g / 3.086 kg = 0.331 kg/mol = 331 g/mol

Hence the molecular weight of the unknown nonelectrolyte dissolved in benzophenone is 331 g/mol.

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Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

[tex]ln (\frac{C}{C_0})=-kt[/tex]

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as [tex]5.11 \times 10^{-5} \, s^{-1}[/tex]

So time t is given as

[tex]ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs[/tex]

So the concentration will become 15% of the initial value after 10.31 hrs.

Complete question:

The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M. How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?

Answer:

The time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours

Explanation:

The initial concentration of CH3NC is 3.00 X 10⁻²M =0.03M

The rate constant  K= 5.11 X 10⁻⁵s⁻¹

If the concentration of methyl isonitrile to drop to 15.0 %;

The new concentration of methyl isonitrile becomes 0.15 X 0.03 = 0.0045 M

The time taken to drop to 0.0045 M, can be calculated as follows:

[tex]t = -ln[\frac{(CH_3NC)}{(CH_3NC)_0}]/K[/tex]

[tex]t = (-ln[\frac{0.0045}{0.03}]/5.11 X 10^{-5})X(\frac{1 min}{60 s}) = 618.8 mins[/tex]

→ 618.8 mins X 1hr/60mins = 10.3 hours

Therefore, the time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours

Answer:

The mass of CaCO3 reacted  is 5.00 grams

Explanation:

Step 1 :Data given

Before the reaction, the container, including the CaCO3, had a mass of 24.20 g

After the reaction the container with product had a mass of only 22.00 g because the CO2 gas produced did not remain in the container.

Molar mass of CO2 = 44.01 g/mol

Molar mass CaCO3 = 100.09 g/mol

Step 2: The balanced equation

CaCO3 → CaO + CO2

Step 3: Calculate mass of CO2

Mass of CO2 = 24.20 grams - 22.00 grams

Mass of CO2 = 2.20 grams

Step 4: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.20 grams / 44.01 g/mol

Moles CO2 = 0.0500 moles

Step 5: Calculate moles CaCO3

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.0500 moles CO2 we need 0.0500 moles CaCO3

Step 6: Calculate mass CaCO3

Mass CaCO3 = moles CaCO3 * molar mass CaCO3

Mass CaCO3 = 0.0500 moles  * 100.09 g/mol

Mass CaCO3 = 5.00 grams

The mass of CaCO3 reacted  is 5.00 grams

Final answer:

2.20 grams of calcium carbonate (CaCO3) reacted, as calculated by the difference in mass before and after the decomposition reaction, which released carbon dioxide gas from the container.

Explanation:

The initial mass of the reaction container with calcium carbonate (CaCO3) was 24.20 g, while the final mass after the reaction was 22.00 g. The mass of CaCO3 that reacted can be found by subtracting the final mass from the initial mass, because the only mass lost would be the carbon dioxide (CO2) gas that escaped from the container.

Mass of CaCO3 that reacted = Initial mass - Final mass
= 24.20 g - 22.00 g
= 2.20 g

Therefore, 2.20 grams of calcium carbonate reacted, decomposing into calcium oxide (CaO) and releasing carbon dioxide gas according to the reaction CaCO3(s) → CaO(s) + CO2(g).

Answer:

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

    = 15.38 x 10⁻⁸

Hence, by definition,

pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813

Answer:

The answer to this is

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

Explanation:

To solve this we first list out the variables thus

Density of the water = 1.00 g/mL =1000 kg/m³

density of mercury  = 13.6 g/mL = 13600 kg/m³

Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals

Therefore from the equation for denstity we have

Density = mass/volume

Pressure = Force/Area  and for  a column of water, pressure = Density × gravity×height

Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa

This value of pressure should be supported by the column of water as follows

Pressure = 101396.16 Pa =  kg/m³×9.81 m/s² ×h

∴  [tex]h = \frac{101396.16}{(1000)(9.81)}[/tex] = 10.336 meters

The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters

Final answer:

A column of water about 10.33 meters high can be supported by standard atmospheric pressure, based on the relationship between pressure, fluid density, gravitational acceleration, and height of the fluid column.

Explanation:

To calculate the height of a column of water supported by standard atmospheric pressure, we can use the pressure exerted by a column of liquid formula, which is p = hρg where p is the pressure, h is the height of the liquid column, ρ (rho) is the density of the liquid, and g is the acceleration due to gravity. The standard atmospheric pressure is 101,325 Pa, and the density of water is 1.00 g/cm³ or 1000 kg/m³.

Since 1 atm is equivalent to the pressure exerted by a 760 mm column of mercury, and mercury is 13.6 times denser than water, a water column would need to be 13.6 times taller. Thus:

101325 Pa = (h)(1000 kg/m³)(9.81 m/s²)

h = 101325 Pa / (1000 kg/m³ × 9.81 m/s²) = 10.33 m

Therefore, a column of water about 10.33 meters high can be supported by standard atmospheric pressure.

Answer:

0.01 mol of H+.

Explanation:

Equation of reaction:

Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)

Molar concentration = number of moles/volume

= 1*0.01

= 0.01 mol.

Dissociation:

2HCl --> 2H+ + 2Cl-

By stoichiometry,

If 2 mole of HCl dissociates to guve 2 moles of H+

Number of moles of H+ = 0.01 mol.

Explanation:

Expression for the coefficient of thermal expansion is as follows.

           [tex]\alpha = \frac{1}{V}(\frac{\Delta V}{\Delta T})[/tex]

where,   V = initial volume

          [tex]\Delta V[/tex] = Final volume - initial volume

                      = (712.6 - 873.6) [tex]cm^{3}[/tex]

                      = -161 [tex]cm^{3}[/tex]

Now, we will calculate the change in temperature as follows.

          [tex]\Delta T[/tex] = Final temperature - Initial temperature

                       = (10 + 273) K - (70 + 273) K

                       = 283 K - 343 K

                       = -60 K

Substituting these values into the equation as follows.

     [tex]\alpha = \frac{1}{873.6} \times (\frac{161}{60}) K^{-1}[/tex]

                 = 0.00307 [tex]K^{-1}[/tex]

It is known that for non-ideal gases the value of alpha is 0.366% which is 0.00366 per Kelvin. As it is close to our result, hence the given sample of gas is a non-ideal gas.

Answer: pH of the solution is 12

Explanation:

NaOH —> Na+ + OH-

NaOH = 0.01M

Na+ = 0.01M

OH- = 0.01M

pOH = —log [OH-] = —log [0.01]

pOH = 2

pH + pOH = 14

pH = 14 — pOH = 14 — 2 = 12

pH = 12

Explanation:

(a)  Chemical reaction equation is given as follows.

           [tex]Ag^{+} + Cl^{-} \rightarrow AgCl(aq) \rightarrow K_{1}[/tex]

Also,  

         [tex]AgCl(s) \rightarrow Ag^{+} + Cl^{-} \rightarrow K_{3}[/tex]

Therefore, the net reaction equation is as follows.

         [tex]AgCl(s) \rightarrow AgCl(aq)[/tex]

Now, we will calculate the value of K for this reaction as follows.

          K = [tex]K_{1} \rightarrow K_{2}[/tex]

             = [tex]2.0 \times 10^{3} \times 1.8 \times 10^{-10}[/tex]

             = [tex]3.6 \times 10^{-7 }[/tex]

Hence, the value of K for the given reaction is  [tex]3.6 \times 10^{-7 }[/tex].

(b)  As the reaction  is given as follows.

              [tex]AgCl(s) \rightarrow AgCl(aq)[/tex]

Therefore, when excess of AgCl(s) is added then the amount of [AgCl(aq)] present in equilibrium is as follows.

                   K = [AgCl(aq)] = [tex]3.6 \times 10^{-7 }[/tex]

Thus, the value of [AgCl(aq)] in equilibrium with excess AgCl(s) is [tex]3.6 \times 10^{-7 }[/tex].

(a) The equilibrium constant K for the reaction[tex]\(\text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)}\) is \( 3.6 \times 10^{-9} \).[/tex]

(b) The concentration of AgCl(aq) in equilibrium with excess AgCl(s) is [tex]\( 3.6 \times 10^{-9} \, \text{M} \).[/tex]

Part (a): Finding K for the reaction-

We are given the following equilibria and their constants:

1. [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad K_1 = 20\)[/tex]

2. [tex]\(\text{AgCl(aq)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_2 = 93\)[/tex]

3. [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_3 = 1.8 \times 10^{-10}\)[/tex]

We need to find K for the reaction:

[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)} \][/tex]

We can see that we need to combine these equilibria in a way that gets us from AgCl(s) to AgCl(aq)

Let's write the reactions again:

1. [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad K_1 = 20\)[/tex]

2. [tex]\(\text{AgCl(aq)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_2 = 93\)[/tex]

3. [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_3 = 1.8 \times 10^{-10}\)[/tex]

The relationship between these equilibria can be described as follows:

- [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad (K_3)\)[/tex]

- [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad (K_1)\)[/tex]

If we multiply K₃ by K₁, we will get the desired equilibrium:

[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad (K_3) \][/tex]

[tex]\[ \text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad (K_1) \][/tex]

Multiplying these two equilibria together, the intermediate [tex]\(\text{Ag}^+\)[/tex] and [tex]\(\text{Cl}^-\)[/tex] ions cancel out, giving us:

[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)} \][/tex]

The equilibrium constant for this reaction is:

[tex]\[ K = K_3 \times K_1 \][/tex]

Substituting the given values:

[tex]\[ K = (1.8 \times 10^{-10}) \times 20 \] \\[/tex]

[tex]\[ K = 3.6 \times 10^{-9} \][/tex]

So, the equilibrium constant K for the reaction [tex]\(\te{AgCl(s)} \rightleftharpoons \text{AgCl(aq)}\)[/tex] is [tex]\( 3.6 \times 10^{-9} \).[/tex]

Part (b): Finding AgCl(aq) in equilibrium with excess AgCl(s)

When AgCl(s) is in equilibrium with AgCl(aq), the equilibrium expression for this reaction is:

[tex]\[ K = [\text{AgCl(aq)}] \][/tex]

From Part (a), we found that:

[tex]\[ K = 3.6 \times 10^{-9} \][/tex]

Therefore, the concentration of [tex]\(\text{AgCl(aq)}\)[/tex] in equilibrium with excess [tex]\(\text{AgCl(s)}\)[/tex] is:

[tex]\[ [\text{AgCl(aq)}] = 3.6 \times 10^{-9} \, \text{M} \][/tex]

Answer:

In explanation

Explanation:

a. Selenium, atomic number 34

Argon is 18, adding the other electrons gives 34.

Other elements include: oxygen O , Sulphur S , Tellurium Te and Polonium Po

b. Hafnium, element 72

Xe is 54, adding the other electrons makes the total 72.

Other elements; Titanium Ti, Zirconium Zr, Rutherfordium Rf

c. Manganese, element 25

Other group members are Technetium Tc , Rhenium Re and Bohrium Bo

Answer:

(a) ml = 0, ±1, ±2

(b) ml = 0

(c) ml = 0, ±1, ±2, ±3, ±4

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So in our exercise,

(a) l = 2; equivalent with with sublevel d

-l ≤ ml ≤ l, ml = 0, ±1, ±2, equivalent with dxy, dxz, dyz, dx2-y2, dz2

(b) n = 1;

n = 1, only 01 level

l = 0, equivalent with sublevel s

ml = 0

(c) n = 4, l = 3.

l = 3, equivalent with sublevel f

ml = 0, ±1, ±2, ±3, ±4

Explanation:

1. 2,3-dimethylhexane

2. 2,4-dimethylhexane

3. 2,5-dimethylhexane

4. 3,4-dimethylhexane

Below are the structures of the isomers.

Answer:

[tex]7.3\times 10^2\ days[/tex]

Explanation:

Given that:

Half life = [tex]9.8\times 10^3[/tex] days

[tex]t_{1/2}=\frac{\ln2}{k}[/tex]

Where, k is rate constant

So,  

[tex]k=\frac{\ln2}{t_{1/2}}[/tex]

[tex]k=\frac{\ln2}{9.8\times 10^3}\ days^{-1}[/tex]

The rate constant, k = 0.00007 days⁻¹

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

Given:

5 % is lost which means that 0.05 of [tex][A_0][/tex] is decomposed. So,

[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.05 = 0.95

t = ?

[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]

[tex]0.95=e^{-0.00007\times t}[/tex]

t = 732.76 days = [tex]7.3\times 10^2\ days[/tex]

It will take approximately [tex]\( 7.25 \times 10^2 \)[/tex] days for benzoyl peroxide to lose 5% of its potency when refrigerated.

To determine the time, it takes for benzoyl peroxide to lose 5% of its potency, we can use the first-order reaction kinetics formula:

[tex]\[ N(t) = N_0 \times e^{-kt} \][/tex]

where:

[tex]\( N(t) \)[/tex] is the amount of substance remaining after time [tex]\( t \)[/tex],

[tex]\( N_0 \)[/tex] is the initial amount of substance,

[tex]\( k \)[/tex] is the rate constant,

[tex]\( t \)[/tex] is the time.

The half-life [tex]\( t_{1/2} \)[/tex] is related to the rate constant [tex]\( k \)[/tex] by the equation:

[tex]\[ t_{1/2} = \frac{\ln(2)}{k} \][/tex]

Given the half-life [tex]\( t_{1/2} = 9.8 \times 10^3 \)[/tex] days, we can solve for [tex]\( k \)[/tex]:

[tex]\[ k = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{9.8 \times 10^3} \][/tex]

Now, we want to find the time [tex]\( t \)[/tex] when 95% of the substance remains, so [tex]\( N(t) = 0.95N_0 \)[/tex]. Plugging this into the first-order reaction formula:

[tex]\[ 0.95N_0 = N_0 \times e^{-kt} \][/tex]

[tex]\[ 0.95 = e^{-kt} \][/tex]

Taking the natural logarithm of both sides:

[tex]\[ \ln(0.95) = -kt \][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[ t = -\frac{\ln(0.95)}{k} \][/tex]

Substituting [tex]\( k \)[/tex] with the expression involving the half-life:

[tex]\[ t = -\frac{\ln(0.95)}{\ln(2)/t_{1/2}} \][/tex]

[tex]\[ t = -\frac{\ln(0.95) \cdot t_{1/2}}{\ln(2)} \][/tex]

[tex]\[ t = -\frac{\ln(0.95) \cdot 9.8 \times 10^3}{\ln(2)} \][/tex]

[tex]\[ t \approx -\frac{\ln(0.95) \cdot 9.8 \times 10^3}{0.693} \][/tex]

[tex]\[ t \approx -\frac{-0.051293 \cdot 9.8 \times 10^3}{0.693} \][/tex]

[tex]\[ t \approx \frac{0.5027 \times 10^3}{0.693} \][/tex]

[tex]\[ t \approx 7.25 \times 10^2 \][/tex]

Answer:

(a) ν = 3.1 × 10¹³ s⁻¹

(b) λ = 3.467 μm

Explanation:

We can solve both problems using the following expression.

c = λ × ν

where,

c: speed of light

λ: wavelength

ν: frequency

(a)

c = λ × ν

ν = c / λ

ν = (3.000 × 10⁸ m/s) / (9.6 × 10⁻⁶ m)

ν = 3.1 × 10¹³ s⁻¹

(b)

c = λ × ν

λ = c / ν

λ = (3.000 × 10⁸ m/s) / (8.652 × 10¹³ s⁻¹)

λ = 3.467 × 10⁻⁶ m

λ = 3.467 × 10⁻⁶ m (10⁶ μm/ 1 m)

λ = 3.467 μm

Answer:

A. 1.79kg

Explanation:

The value of the mass of a particular substance could be obtained if there is adequate information about the volume of that particular substance and the density of that substance.

Mathematically expressed, the mass of a substance is the product of the density and the volume .

In this particular question, the density is 0.01932kg/cm3 while the volume is 92.5cm3

The mass is thus = 92.5 * 0.01932 = 1.7871kg which is approximately 1.79kg

The mass of a 92.5 cm3 sample of gold is approximately 1.79 kg.

The mass of an object can be calculated by multiplying its density by its volume. In this case, the density of gold is given as 0.01932 kg/cm3, and the volume of the gold sample is 92.5 cm3. To find the mass, we can use the formula:

mass = density * volume

Substituting the values, we have:

mass = 0.01932 kg/cm3 * 92.5 cm3

Calculating this, we find that the mass of the gold sample is approximately 1.79 kg.

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A) The higher level that the electron reached from ground state is; n = 5

B) The intermediate level that the electron reached from ground state is; n = 3

C) The wavelength of the second photon emitted is; λ = 103 nm

A) We are given;

Wavelength of photon absorbed by ground state H atoms; λ_g = 94.91 nm = 94.91 × 10⁻⁹ m

Formula to get the higher level is Rydberg's formula;

1/λ = R(1/n₁² - 1/n₂²)

where;

R is rydberg constant = 1.097 × 10⁷ m⁻¹

Thus;

1/(94.91 × 10⁻⁹) = 1.097 × 10⁷(1/1² - 1/n₂²)

0.9605 = 1 - 1/n₂²

1/n₂² = 1 - 0.9605

1/n₂² = 0.0395

n₂ = √(1/0.0395)

n₂ ≈ 5

B) We want to find the intermediate level where wavelength = 1281 nm = 1281 × 10⁻⁹ m

Thus;

1/(1281 × 10⁻⁹) = 1.097 × 10⁷(1/n₂² - 1/5²)

0.0712 = 1/n₂² - ¹/₂₅

1/n₂² = 0.0712 + ¹/₂₅

1/n₂² = 0.1112

n₂ = √(1/0.1112)

n₂ ≈ 3

C) Formula for energy of photon is;

E = hc/λ

where;

h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s

c is speed of light = 3 × 10⁸ m/s

Thus;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1281 × 10⁻⁹)

E = 1.552 × 10⁻¹⁹ J

The energy at ground state is;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(94.91 × 10⁻⁹)

E = 20.944 × 10⁻¹⁹ J

Thus;

Energy of second photon = (20.944 × 10⁻¹⁹) - (1.552 × 10⁻¹⁹)

Energy of second photon = 19.352 × 10⁻¹⁹ J

Wavelength of second photon emitted is;

λ = hc/E

λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/19.352 × 10⁻¹⁹

λ = 103 nm

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The electron moves to the third energy level after the initial photon absorption, then drops to the fourth energy level releasing a photon of 1281 nm wavelength, and finally back to the ground state with a recorded emission of 97.08 nm wavelength.

To solve this, we can use the Rydberg formula for hydrogen: 1/λ = R (1/n1^2 - 1/n2^2), where λ is the wavelength of the light, R is the Rydberg constant (1.096776 x 10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level.

(a) The photon absorbed takes the electron to a higher energy level, so we use the wavelength 94.91 nm, and n1 = 1 (ground state). This calculates to n2 = 2.32, meaning the electron moves to roughly the third energy level as energy levels are only in whole numbers.

(b) The first emitted photon has wavelength 1281 nm, which brings the electron to an intermediate level. By substituting the values, we calculate n1 = 3 and get n2 = 4. So, the intermediate energy level is 4.

(c) For the second photon emitted, we know the electron goes from n = 4 to the ground state n = 1. Rearranging the Rydberg formula, we find λ = 97.08 nm, which is the wavelength of the second photon emitted.

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Answer:   It is important to wet the filter paper in the Buchner funnel first with cold re crystallization solvent before the re crystallization mixture being filtered to minimize gaps around the edges of the filter paper which can prevent mechanical impurities from passing through. This gives better filtration where most impurities can be filtered. Furthermore, it provides good vacuum.

Answer: The number of seconds in a solar year is [tex]3.2\times 10^7s[/tex]

Explanation:

We are given some conversion factors:

1 minute = 60 seconds

1 hour = 60 minutes

1 solar day = 24 hours

1 solar year = 365.24 solar days

Calculating the number of seconds in 1 solar year by using the conversion factors, we get:

[tex]\Rightarrow (\frac{60s}{1min})\times (\frac{60min}{1hr})\times (\frac{24hr}{1\text{solar day}})\times (\frac{365.24\text{solar days}}{1\text{ solar year}})\\\\\Rightarrow (\frac{31556736s}{1\text{ solar year}})[/tex]

There are 31556736 seconds in 1 solar year.

Hence, the number of seconds in a solar year is [tex]3.2\times 10^7s[/tex]

Answer:

40mL of glycerol are needed to make a 20% v/v solution

Explanation:

This problem can be solved with a simple rule of three:

20%  v/v is a sort of concentration. In this case, 20 mL of solute are contained in 100 mL of solution.

Therefore, in 100 mL of solution you have 20 mL of solvent (glycerol)

In 200 mL, you would have,  (200 .20)/ 100 = 40 mL

Answer:

0.0634 M.

Explanation:

Equation of the neutralisation reaction:

CH3COOH + NaOH--> CH3COONa + H2O

Number of moles = molar concentration * volume

= 0.1002 * 0.01581

= 0.00158 mol.

By stoichiometry,

1 mole of acetic acid reacts with 1 mole of NaOH

Number of moles of acetic acid = 0.00158 mol.

Concentration in 2nd dilution = moles/ volume

= 0.00158/0.25

= 0.00634 M

At 25 ml,

Concentration =

C1 * V1 = C2 * V2

= (0.00634 * 0.25)/0.025

= 0.0634 M.

Given:

Neutralization reaction's equation:

Now,

The number of moles will be:

= [tex]Molar \ concentration\times Volume[/tex]

= [tex]0.1002\times 0.01581[/tex]

= [tex]0.00158 \ mol[/tex]

and,

The concentration of second dilution will be:

= [tex]\frac{Moles}{Volume}[/tex]

= [tex]\frac{0.00158}{0.25}[/tex]

= [tex]0.00634 \ M[/tex]

hence,

The concentration at 25 mL will be:

→ [tex]C_1\times V_1 = C_2\times V_2[/tex]

or,

→         [tex]C_2 = \frac{C_1\times V_1}{V_2}[/tex]

                 [tex]= \frac{0.00634\times 0.25}{0.025}[/tex]

                 [tex]= 0.0634 \ M[/tex]

Thus the above response is right.  

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Answer:

a) Thymol represents a minor concentration.

b) Of the ingredients listed, the ones that qualify as being contained in trace amounts are the ones with concentrations less than 0.01%, 100ppm or 0.1mg/L. And in this question, none of the given concentrations is less than 0.01%. Maybe one of the other constituents whose concentrations weren't given is in trace amounts.

c) The mouthwash brought for analysis is the sample.

d) Thymol is the analyte.

e) The matrix is every component of the mouthwash sample apart from the analyte constituent, thymol.

Menthol, eucalyptol, methyl salicylate, caramel, water, alcohol in addition to sodium benzoate, color, poloxamer 407 and benzoic acid are all member substances of the matrix.

f) The blank can contain every constituent apart from the analyte constituent, thymol.

Explanation:

a) The composition range for major, minor and trace components are given thus.

1-100% Major

0.01-1% Minor

1 ppb to 100ppm Trace

<1ppb ultra trace

In Chemistry terms,

•Major constituents: Substances in concentrations over 1mg/L

•Minor constituents: Substances in concentrations between 1mg/L and 0.1 mg/L

•Trace constituents: Substances in concentrations under 0.1 mg/L

Hence, thymol with a concentration of 0.064% represents a minor constituent.

b) Of the ingredients listed, the ones that qualify as being contained in trace amounts are the ones with concentrations less than 0.01%, 100ppm or 0.1mg/L. And in this question, none of the given concentrations is less than 0.01%. Maybe one of the other constituents whose concentrations weren't given is in trace amounts.

c) In chemical analysis or Analytical chemistry, a sample is a portion of material selected from a larger quantity of material.

Therefore, for this question, the mouthwash is the sample.

d) In Analytical Chemistry, an analyte or analyte component is a substance or chemical constituent that is of interest in an analytical procedure.

For this question, the analyte is thymol; the constituent the analyser is interested in.

e) In chemical analysis, matrix refers to the components of a sample other than the analyte of interest.

Therefore, the matrix is every component of the mouthwash sample apart from the analyte constituent, thymol.

Menthol, eucalyptol, methyl salicylate, caramel, water, alcohol in addition to sodium benzoate, color, poloxamer 407 and benzoic acid are all member substances of the matrix.

f) A blank solution is a solution containing little to no analyte of interest, so, the blank can contain every constituent apart from thymol.

Hope this helps!

Answer:

a) F⁻ < Cl⁻ < Br⁻

b) Mg²⁺ < Na⁺ < F⁻

c) Cr³⁺ < Cr²⁺

Explanation:

The ions in (a) part of the question belong to a halogen group. F, Cl, and Br are present in periods 2, 3, and 4 respectively. As we move down the group the size of atoms increases hence their ions will be in the same order. (ion from top to bottom of group 7)

The ions in (b) part of the question are isoelectronic. The relative size of such species can be estimated by the charge on their nucleus. Lower the nucleus charge greater will be the size of the ion.

Nuclear charge of Mg²⁺ = no. of protons = 12

Nuclear charge of Na⁺   = no. of protons = 11

Nuclear charge of F⁻     = no. of protons =  9

The ions in (c) part are the two oxidized states of chromium. In such cases, higher the number of nuclear charge smaller will be the ion.

Answer:

The volume of NaOH required is 80mL

pH of the resulting solution is 7 since reaction produces a normal salt

Explanation:

First, we generate a balanced equation for the reaction as shown below:

NaOH + HCl —> NaCl + H2O

From the equation,

nA = 1

nB = 1

From the question,

Ma = 2.0 mol/L

Va = 20 mL

Mb = 0.5 mol/L

Vb =?

MaVa / MbVb = nA /nB

2x20 / 0.5 x Vb = 1

0.5 x Vb = 2 x 20

Divide both side by 0.5

Vb = (2 x 20)/0.5

Vb = 80mL

The volume of NaOH required is 80mL

Since the reaction involves a strong acid and a strong base, a normal salt solution will be produced which is neutral to litmus paper. So since the salt solution is neutral to litmus paper then the pH of the resulting solution is 7

The pH of a solution prepared by mixing HCl and NaOH, we need to determine the concentration of the resulting solution. Since HCl and NaOH react in a 1:1 ratio to form water and salt (NaCl), we can use the concept of neutralization to find the resulting concentration.

Moles of HCl = volume (in L) × molarity = 0.020 L × 0.30 mol/L = 0.006 mol

Moles of NaOH = volume (in L) × molarity = 0.015 L × 0.60 mol/L = 0.009 mol

Since HCl and NaOH react in a 1:1 ratio, the limiting reagent is HCl. This means that all of the HCl will react with NaOH, and we will have an excess of NaOH. Therefore, the number of moles of HCl remaining will be zero.

Total volume = 20.0 mL + 15.0 mL = 35.0 mL = 0.035 L

Concentration of resulting solution = (Moles of HCl + Moles of NaOH) / Total volume

= (0.006 mol + 0.009 mol) / 0.035 L

= 0.429 mol/L

pH = -log[H+]

pH = -log(0.429)

≈ 0.366

Therefore, the pH of the solution prepared by mixing 20.0 mL of 0.30 M HCl with 15.0 mL of 0.60 M NaOH is approximately 0.366.

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Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

Answer:

238,485 Joules

Explanation:

The amount of energy required is a summation of heat of fusion, capacity and vaporization.

Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)

m (mass of water) = 75 g

Lf (specific latent heat of fusion of water) = 336 J/g

C (specific heat capacity of water) = 4.2 J/g°C

∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

Lv (specific latent heat of vaporization of water) = 2,260 J/g

Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J

Answer : The electric potential on proton is, 27.1 V

Explanation :

To calculate the electric potential we are using the formula.

[tex]V=\frac{1}{4\pi \epsilon_0}\frac{q}{r}[/tex]

where,

V = electric potential

q = charge on proton = [tex]1.6\times 10^{-19}C[/tex]

r = radius = [tex]0.530\times 10^{-10}m[/tex]

[tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]

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

[tex]V=(8.99\times 10^9)\times \frac{1.6\times 10^{-19}}{0.530\times 10^{-10}}[/tex]

[tex]V=27.1V[/tex]

Thus, the electric potential on proton is, 27.1 V

The best NMR spectrum that corresponds to p-bromoaniline can be determined by analyzing the chemical shift values and multiplicity of the peaks. Tab 2 is the correct spectrum that matches the expected chemical shifts for p-bromoaniline.

The NMR spectrum that corresponds best to p-bromoaniline can be determined by analyzing the chemical shift values and multiplicity of the peaks. In the NMR spectrum of p-bromoaniline, we would expect to see a peak for the bromine atom at a lower chemical shift value and a peak for the amino group at a higher chemical shift value.

The chemical shift values for p-bromoaniline would be different from those of the other molecules given in the choices. By comparing the chemical shift values and multiplicity pattern, we can identify the correct spectrum that matches the expected chemical shifts for p-bromoaniline.

Based on the given information, the best NMR spectrum that corresponds to p-bromoaniline would be Tab 2. This tab shows the expected chemical shift values and multiplicity pattern for p-bromoaniline.

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