Arrange each set of atoms in order of increasing IE₁:
(a) Sr, Ca, Ba (b) N, B, Ne (c) Br, Rb, Se (d) As, Sb, Sn

Answer:low lattice energy for the solid and a high hydration energy for its ions

Explanation:

The lattice energy is the energy that holds the ions in the crystal lattice together. This energy must also be supplied for the lattice to disintegrate into its component ions. When this energy is lower than the energy released during the hydration of ions, the lattice easily breaks apart releasing the ions which are now strongly hydrated and the ionic solid is said to be soluble in water. Hence, solubility is favoured by lower lattice energy and higher hydration energy.

Final answer:

The option that favors the solubility of an ionic solid in water is a low lattice energy for the solid and a high hydration energy for its ions.

Explanation:

The correct choice that would most favor the solubility of an ionic solid in water is:

a low lattice energy for the solid and a high hydration energy for its ions

This is because low lattice energy allows for easier breakup of the ionic lattice, while high hydration energy promotes the attachment of water molecules to the ions, increasing solubility.

Answer: 10 ml of 200 mM [tex]CaCl_2[/tex] is required and 30 ml of water is required.

Explanation:

According to the dilution law,

[tex]C_1V_1=C_2V_2[/tex]

where,

[tex]C_1[/tex] = concentration of stock solution = 200mM

[tex]V_1[/tex] = volume of stock solution = ?

[tex]C_2[/tex] = concentration of resulting solution= 50mM

[tex]V_2[/tex] = volume of another acid solution= 40 ml

[tex]200\times x=50\times 40[/tex]

[tex]x=10ml[/tex]

Thus 10 ml of 200 mM [tex]CaCl_2[/tex] is required and (40-10) ml = 30 ml of water is to be added to make 40 ml of 50mM [tex]CaCl_2[/tex].

Final answer:

To make 40 ml of a 50 mM CaCl2 solution from a 200 mM stock, you need 10 ml of the stock solution and 30 ml of water.

Explanation:

To calculate how much of the 200 mM CaCl2 stock solution you need to dilute to get 40 ml of a 50 mM solution, you can use the dilution formula C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the final concentration, and V2 is the final volume. Plugging in the known values yields (200 mM) × V1 = (50 mM) × (40 ml), solving for V1 gives V1 = (50 mM × 40 ml) / (200 mM) = 10 ml. Therefore, to make 40 ml of a 50 mM CaCl2 solution, you need 10 ml of the 200 mM stock and to dilute it with 30 ml of water.

Answer:

Concentration of stock solution in g/L = 5.0g/L

Concentration of stock solution in mol/L = 0.00117 mol/L or 0.00117 M

Explanation:

Concentration in mol/L = (Concentration in g/L)/(Molar mass)

Mass of (NH₄)₃Fe(ox)₃.3H₂O stock = 125mg = 0.125g

Volume of stock solution required = 25mL = 0.025 L

Concentration in g/L = 0.125/0.025 = 5.0 g/L

Molar Mass of (NH₄)₃Fe(ox)₃.3H₂O = (NH₄)₃Fe(C₂O₄)₃.3H₂O = 428.0632 g/mol

Concentration in mol/L = 5/428.0632 = 0.00117 mol/L = 0.00117 M

Hope this helps!!!

Final answer:

Two resonance forms of the triphenylmethyl cation were provided, with the positive charge located at the central carbon atom and one of the ortho positions.

Explanation:

The triphenylmethyl cation is a carbocation formed by removing an electron from the triphenylmethyl radical. It has the formula C19H16+. To write two resonance forms, we need to show the positive charge at different positions in the structure. One possible resonance form is where the positive charge is at the central carbon atom, and the other form is where the positive charge is at one of the ortho positions. The resonance structures can be represented as follows:

C6H5--C+--C6H5

:    :  

C6H4--C6H5    C6H6--C6H4

Answer:

1. Co^2+ 1s2 2s2 2p6 3s2 3p6 3d7

2. N^3- 1s2 2s2 2p6

3. Ca^2+ 1s2 2s2 2p6 3s2 3p6

Explanation:

When Cobalt loses 2 electrons to become Co2+ it loses the electrons which are in 4s2, not the ones in 3d7 because the electrons in 4s2 have a high reactivity. Thus, when electrons are lost from Co atom, they are lost from the 4s orbital first because it is actually higher in energy when both 3d and 4s are filled with electrons.

Answer:

In order of basicity we have

1. Soda pop (Least basic Normally called acidic)

2. Orange juice

3. Milk

4. Blood (slightly basic)

5. Hand soap

6. Drain cleaner (Highly basic)

Explanation:

Orange juice; the pH of orange juice is in the 3.3 to 4.2 range

Milk; the pH of milk about 6.5 to 6.7

Blood; the blood pH is around 7.35 to 7.45

Hand soap with contents such as ammonium hydroxide is basic, its  pH is about 9-10

Drain cleaner contains baking soda or sodium bicarbonate which basic with a  pH of 12 to 14

Soda pop pH of soda pop is in the range of 2.34 to 3.10. It contains carbonated water with a pH of 3–4, making it mildly acidic.

Arranging the above listed in order of increasing basicity, we have

1. Soda pop

2. Orange juice

3. Milk

4. Blood

5. Hand soap

6. Drain cleaner

Final answer:

The solutions in increasing order of basicity are: orange juice, soda pop, milk, blood, hand soap, and drain cleaner. Indicators like phenolphthalein are used in chemistry to assess the basicity of solutions.

Explanation:

The simulation activity mentioned involves organizing various solutions by their basicity. Basicity refers to the ability of a solution to act as a base, which is typically measured on the pH scale where a pH greater than 7 indicates a basic (alkaline) nature. In increasing order of basicity, the solutions you've mentioned can be generally expected to arrange as follows:

To test basicity in the lab, indicators such as phenolphthalein may be used, which turn pink or fuschia in basic solutions. Basicity and acidity are fundamental concepts in chemistry, important in understanding the properties of substances and their reactions.

Explanation:

a) IE1 of Germenium.  ionization energy decreases down the group.

b) Here IE2 of Ga is higher because second valence electron of gallium is in s orbital whereas second valence electron of germenium is in p orbital.

c) IE3 of Ge is higher as this follows the trend.

d)IE4 of Ga is higher because the 4th valence electron of Ga is a core electron and the for Ge it is in s orbital and it takes higher energy to break a core electron than the orbital ones.

When comparing Ga and Ge, IE₁ and IE₂ of Ga are lower due to being earlier in the periodic table, but IE₄ of Ga is expected to be higher as it involves removing an electron from a full orbital. IE₃ for both may be similar due to both elements having three valence electrons. The correct answer is d) IE₄ of Ga or IE₄ of Ge.

The subject in question involves comparing first and successive ionization energies (IE) for elements in the periodic table, particularly gallium (Ga) and germanium (Ge). Ionization energy refers to the energy required to remove an electron from a gaseous atom or ion.

As general rules, (a) the first IE tends to increase across a period as the nuclear charge increases, and (b) IE increases for successive ionizations as the atom or ion becomes progressively more positively charged.

(a) IE₁ of Ga would be lower than IE₁ of Ge because Ga is to the left of Ge in the periodic table, and thus Ge has a higher nuclear charge and a stronger attraction to its valence electron.

(b) IE₂ of Ga would also be lower than IE₂ of Ge because the additional electron from Ga would come from the same valence shell as the first electron, while Ge's second ionization would remove an electron that experiences a stronger nuclear charge.

(c) As both Ga and Ge have three valence electrons, the IE₃ of Ga and Ge will both involve removing an electron from a similarly charged ion, but because Ga has a higher energy 4p subshell, its IE₃ might be slightly lower than Ge's, which is removing an electron from the 4s subshell (which after removing two electrons is now full and lower in energy).

(d) At the IE₄ level, we see a large jump in ionization energy for both elements as electrons are being removed from an energy level closer to the nucleus; however, IE₄ of Ga will generally be higher than IE₄ of Ge due to  of an electron from a full orbital in Ga, which has higher energy compared to the removal from a half-filled orbital in Ge.

These comparisons are based on the understanding that ionization energies increase both with increasing positive charge and with the removal of electrons closer to the nucleus.

The energy of a photon can be calculated using Planck's equation (E = hf). Given the radio wave frequency of 3.8 x 10^10 Hz and using Planck's constant (6.626 x 10^-34 J·s), the energy of one photon of this radiation is 2.52 x 10^-23 Joules.

To find the energy of one photon of radiation, we need to use Planck's formula: E = hf. Here, 'E' represents energy, 'h' is Planck’s constant (6.626 x 10^-34 J·s), and 'f' is frequency. Given that the frequency (f) of the wave is 3.8 x 10^10 Hz, we substitute these values into Planck's equation:

E = hf = (6.626 x 10^-34 J·s) (3.8 x 10^10 Hz) = 2.52 x 10^-23 J

So, the energy of one photon of this radiation is 2.52 x 10^-23 Joules.

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Answer:

The molality of HCl solution is 16.24 mol/kg.

The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.

The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.

Explanation:

formula used:

[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]

1) Mass percentage of the HCl solution = 37.2%

This means that in 100 grams of solution 37.2 grams of HCl is present.

Mass of HCl (solute)= 37.2 g

Mass of water(solvent) = 100 g - 37.2 g = 62.8 g = 0.0628 kg (1g = 0.001 kg)

Mole of HCl = [tex]\frac{37.2 g}{36.465 g/mol}=1.020 mol[/tex]

[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]

[tex]m=\frac{1.020 mol}{0.0628 kg}=16.24 mol/kg[/tex]

The molality of HCl solution is 16.24 mol/kg.

2) Mass percentage of the [tex]HC_2H_3O_2[/tex] solution = 99,8%

This means that in 100 grams of solution 99.8 grams of [tex]HC_2H_3O_2[/tex] is present.

Mass of [tex]HC_2H_3O_2[/tex](solute)= 99.8  g

Mass of water(solvent) = 100 g - 99.8 g = 0.2 g = 0.0002 kg (1g = 0.001 kg)

Mole of [tex]HC_2H_3O_2[/tex] = [tex]\frac{99.8 g}{60.05g/mol}=16.50 mol[/tex]

[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]

[tex]m=\frac{16.50 mol}{0.0002 kg}=82,500 mol/kg[/tex]

The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.

3) Mass percentage of the [tex]NH_3[/tex] solution = 28.0%

This means that in 100 grams of solution 28.0 grams of [tex]NH_3[/tex] is present.

Mass of [tex]NH_3[/tex](solute)= 37.2 g

Mass of water(solvent) = 100 g - 28.0 g = 72.0 g = 0.072 kg (1g = 0.001 kg)

Mole of [tex]NH_3[/tex]= [tex]\frac{28.0g}{17 g/mol}=2mol[/tex]

[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]

[tex]m=\frac{2 mol}{0.072kg}=27.78 mol/kg[/tex]

The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.

Final answer:

To calculate the molalities of HCl, HC2H3O2, and NH3, convert the weight percentage to mass of solute, then convert that to moles using the formula weight, and divide by the mass of the solvent in kilograms. Use the provided densities to infer the mass of water solvent and proceed with calculations respecting significant figures.

Explanation:

Calculating Molalities of Commercial Reagents

To calculate the molalities of commercial reagents, you'll first need to determine the moles of solute and then divide that by the mass of the solvent in kilograms. Below is the process for each reagent provided:

For HCl (hydrochloric acid), we use the weight % to find the mass of HCl per 100g of solution, convert that mass to moles using the formula weight, and then divide by the solvent mass (water) in kilograms to find the molality.

In the case of HC2H3O2 (acetic acid), we apply a similar approach, starting with the weight % to calculate the mass of acetic acid in 100g of solution, convert to moles using its formula weight, and then divide by the mass of solvent to find molality.

For NH3 (aqueous ammonia), we begin by taking the 28.0 weight % value to get the mass of NH3 in 100g of solution, converting this mass to moles using NH3's formula weight, and finding the molality by dividing moles over kilograms of solvent.

Molality (°m) is defined as the number of moles of solute per kilogram of solvent and is especially useful because it doesn’t change with temperature. Using the provided densities and weights, we can assume the solvent used is predominantly water and calculate the weight of the water to use in the molality equation. The molar mass information provided, such as the formula weight of HCl (36.465 g/mol), acts as a conversion factor between grams and moles of solute.

At each step, remember to account for significant figures based on the precision of the given data.

Answer:

[tex]\lambda=2.61\times 10^{-9}\ m[/tex] = 261 nm

Silver is not a good choice.

Explanation:

[tex]E=\frac {h\times c}{\lambda}[/tex]

Where,  

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] is the wavelength of the light

Given that:- Energy = [tex]7.59\times 10^{-19}\ J[/tex]

[tex]7.59\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]7.59\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]

[tex]\lambda=2.61\times 10^{-9}\ m[/tex] = 261 nm

Visible range has a spectrum of 380 to 740 nm

So, Silver is not a good choice.

The maximum wavelength needed to remove an electron from silver is approximately 262 nm. Silver is not a good choice for a photocell that uses visible light.

To find the maximum wavelength needed to remove an electron from silver, we can use the work function of silver, which is Φ = 4.73 eV. The threshold wavelength for observing the photoelectric effect in silver can be calculated using Equation 6.16, which is λ = hc/Φ. Substituting the given values, we have λ = (1240 eV⋅nm) / (4.73 eV), which gives us a threshold wavelength of approximately 262 nm. Since visible light ranges from 400 to 700 nm, silver is not a good choice for a photocell that uses visible light.

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To determine the final temperature after ice is dropped into water, we analyze the energy transfer phases: ice warming, melting, and temperature equalization, applying specific heat capacities and latent heat of fusion.

To find the final temperature after dropping a cube of ice into water, we must consider the energy exchanges that occur: the ice warming up to 0°C, melting, and the resulting water warming up or cooling down to reach thermal equilibrium with the water already in the glass. For this, we use the concept of specific heat capacity and the latent heat of fusion for water.

The specific heat capacity of ice is about 2.062 J/(g°C), and the specific heat capacity of liquid water is 4.184 J/(g°C). The latent heat of fusion of ice is roughly 334 J/g. By applying the conservation of energy, we equate the heat lost by the warmer substance (water) to the heat gained by the colder substance (ice and the water resulting from melted ice), taking into account phase changes.

However, without numerical calculation in this particular example, the key takeaway is understanding the phases of energy transfer: warming up the ice, melting the ice, and then equalizing the temperature of the mixture. The initial temperatures of the substances and their masses play crucial roles in determining the final temperature.

Answer:

The answer is 12.35

Explanation:

From the question we are given that the concentration of [tex]H^{+}[/tex] is [tex]8.1 * 18^{-6}M[/tex]

 Generally The rate equation is given as

                                                           [tex]K_{w} = [H^{+} ][OH^{-} ][/tex]

and [tex]K_{w}[/tex] the rate constant has a value [tex]1 * 10^{-14}[/tex]

     Substituting and making [[tex]OH^{-}[/tex]] the subject we have

                                                 [tex][OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}[/tex]

                                                  [tex][OH ^ {-}] = 1.235 * 10^{-9}M[/tex]

                            Multiply the value by [tex]10^{10}[/tex] as instructed from the question we have  

                       Answer =   [tex]1.235 * 10 ^{-9} * 10^{10} = 12.35[/tex]

Hence the answer in 2 decimal places is 12.35

The concentration of OH¯ in the solution multiplied by 10¹⁰ is 12.35

The concentration of hydroxide ion [OH¯] can be obtained as follow:

K = [H⁺][OH¯]

Equilibrium constant (K) = 1×10¯¹⁴

1×10¯¹⁴ = 8.1×10¯⁶ × [OH¯]

Divide both side by 8.1×10¯⁶

[OH¯] = 1×10¯¹⁴ / 8.1×10¯⁶

[OH¯] = 1.235×10¯⁹

Multiply by 10¹⁰

[OH¯] = 1.235×10¯⁹ × 10¹⁰

[OH¯] = 12.35

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The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

C+3D→CD₃

Based on this mechanism, determine the rate law for the overall reaction.

Answer: The rate law for the reaction is [tex]\text{Rate}=k'[C][D]^2[/tex]

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

[tex]C+3D\rightarrow CD_3[/tex]

The intermediate reaction of the mechanism follows:

Step 1:  [tex]C+D\rightleftharpoons CD;\text{ (fast)}[/tex]

Step 2:  [tex]CD+D\rightarrow CD_2;\text{(slow)}[/tex]

Step 3:  [tex]CD_2+D\rightarrow CD_3;\text{(fast)}[/tex]

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

[tex]\text{Rate}=k[CD][D][/tex]           ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

[tex]K=\frac{[CD]}{[C][D]}[/tex]

[tex][CD]=K[C][D][/tex]

Putting the value of [CD] in equation 1, we get:

[tex]\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2[/tex]  

Hence, the rate law for the reaction is [tex]\text{Rate}=k'[C][D]^2[/tex]

Answer:

The molar mass of the compound is 720.8 g/mol

Explanation:

Let's apply the colligative property of Osmotic pressure to solve this.

Formula is π = M . R . T

where π is pressure (atm)

M is molarity (mol/L)

R, Universal Constant Gases

T, Absolute T° ( T° in K = T° in C + 273)

Let's replace the data:

8.44 Torr = M . 0.082 L.atm/mol.K . 298K

As we have the pressure in Torr, we must convert to atm, to work properly.

8.44 Torr . 1 atm/ 760 Torr = 0.0111 atm

0.0111 atm = M . 0.082 L.atm/mol.K . 298K

0.0111 atm / (0.082 L.atm/mol.K . 298K) = M  → 4.54×10⁻⁴ mol/L

So molarity is the moles of solute (mass (g) / molar mass) / volume (L)

Let's convert the volume to L → 279.6 mL . 1L / 1000 mL = 0.2796 L

4.54×10⁻⁴ mol/L . 02796 L = 1.27×10⁻⁴ moles

This moles are represented by the 91.6 mg, so let's convert the mass of solute from mg to g

91.6 mg . 1 g / 1000 mg = 0.0916 g

Molar mass → g/mol → 0.0916 g / 1.27×10⁻⁴ moles → 720.8 g/mol

Explanation:

(a)   The given data is as follows.

           Speed of electron (u) = [tex]5.5 \times 10^{4} m/s[/tex]

According to De Broglie's formula,

                wavelength, [tex]\lambda = \frac{h}{mu}[/tex]

where,   h = Planck's constant = [tex]6.626 \times 10^{-34} Js[/tex]

              m = mass of electron = [tex]9.11 \times 10^{-31} kg[/tex]

Hence, we will calculate the wavelength as follows.

           [tex]\lambda = \frac{h}{mu}[/tex]

                      = [tex]\frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31}kg \times 5.5 \times 10^{4} m/s}[/tex]

                      = [tex]0.132 \times 10^{-7}[/tex] m

                      = [tex]13.2 \times 10^{-9}[/tex] m

It is known that for any microscope, smallest object that can be observed is equal to [tex]\frac{1}{2}[/tex] the wavelength of of the radiation  smallest object observable with an electron microscope.

Hence,     [tex]\frac{13.2 \times 10^{-9}}{2}[/tex]

                = [tex]6.6 \times 10^{-9}[/tex] m

               = 6.6 nm          (as 1 m = [tex]10^{-9} nm[/tex])

Therefore, the smallest object observable with an electron microscope will be 6.6 nm.

(b)  At [tex]3.0 \times 10^{7} m/s[/tex], the wavelength will be calculated as follows.

              wavelength, [tex]\lambda = \frac{h}{mu}[/tex]

                                  = [tex]\frac{6.626 \times 10^{-34} Js}{9.11 \times 10^{-31} kg \times 3.0 \times 10^{7} m/s}[/tex]

                                  = [tex]24.2 \times 10^{-12}[/tex] m

As, for any microscope, smallest object that can be observed is equal to [tex]\frac{1}{2}[/tex] the wavelength of of the radiation  smallest object observable with an electron microscope.

                = [tex]\frac{24.2 \times 10^{-12}}{2}[/tex]

                = [tex]12.1 \times 10^{-12} m \times 10^{9}nm/m[/tex]

                = 0.0121 nm

Therefore, at [tex]3.0 \times 10^{7} m/s[/tex]  the smallest object observable with an electron microscope is 0.0121 nm.

The smallest object observable with an electron microscope can be calculated using the formula: Size of object = Wavelength of electrons / 2. Plugging in the values and solving for the size of the object gives: Size of object = 3.37 x 10^-12 m. For a speed of 3.0 x 10^7 m/s, the calculation would be: Size of object = 1.22 x 10^-10 m.

The smallest object observable with an electron microscope can be calculated using the formula:

Size of object = Wavelength of electrons / 2

Using the given speed of 5.5 x 10^4 m/s, we can calculate:

Size of object = (h / (m * v)) / 2, where h is Planck's constant and m is the mass of an electron.

Plugging in the values and solving for the size of the object gives:

Size of object = (6.626 x 10^-34 J s / (9.109 x 10^-31 kg * 5.5 x 10^4 m/s)) / 2

Size of object = 3.37 x 10^-12 m

For a speed of 3.0 x 10^7 m/s, the calculation would be:

Size of object = (6.626 x 10^-34 J s / (9.109 x 10^-31 kg * 3.0 x 10^7 m/s)) / 2

Size of object = 1.22 x 10^-10 m

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Answer: hydroxyl group

Explanation:

When 3 mol of phosphine reacts, (3/4) mol of phosphorus and 4.5 mol of hydrogen are produced. The number of moles of phosphine, phosphorus, and hydrogen at any time can be determined using the stoichiometry of the reaction. The time needed for 3 mol of phosphine to react can be found by integrating the given rate equation.

To determine the amount of phosphorus and hydrogen produced when 3 mol of phosphine reacts, we can use the stoichiometry of the reaction. From the balanced equation, we see that for every 4 mol of PH3 that reacts, we get 1 mol of P4 and 6 mol of H2. Therefore, when 3 mol of PH3 reacts, we will produce (3/4) mol of P4 and (3/4) x 6 = 4.5 mol of H2.

Let's denote the number of moles of phosphine as nPH3, the number of moles of phosphorus as nP4, and the number of moles of hydrogen as nH2. At any time t, when 3 mol of phosphine has reacted, the corresponding number of moles of phosphorus and hydrogen can be determined using the stoichiometry of the reaction as mentioned above.

To determine how long it would take for 3 mol of phosphine to have reacted, we can use the given rate equation: dCPH3/dt = -3.715 × 10-6 CPH3. Since we have the initial concentration of phosphine (CPH3) as 5 kmol/m3, we can integrate the rate equation to find the time needed for 3 mol of phosphine to react.

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Answer:

12.45

Explanation:

There is some info missing. I think this is the original question.

A chemist dissolves 169 mg of pure barium hydroxide in enough water to make up 70 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Round your answer to 2 significant decimal places.

First, we will calculate the molarity of barium hydroxide.

M = mass / molar mass × liters of solution

M = 0.169 g / 171.34 g/mol × 0.070 L

M = 0.014 M

Barium hydroxide is a strong base that dissociates according to the following equation.

Ba(OH)₂ → Ba²⁺ + 2 OH⁻

The molar ratio of Ba(OH)₂ to OH⁻ is 1:2. The concentration of OH⁻ is 2 × 0.014 M = 0.028 M

The pOH is:

pOH = -log [OH⁻] = - log 0.028 = 1.55

The pH is:

pH = 14 - pOH = 14 - 1.55 = 12.45

Final answer:

a) The ratio of lb-mole O2 react/lb-mole NO formed is 1.25. b) The oxygen feed rate corresponding to 40% excess O2 is 175.0 kmol/h. c) Ammonia is the limiting reactant, with the oxygen being in excess by 5.85%. The extent of reaction is 36.17 mol NO produced, with a mass of 1085.73 g.

Explanation:

a. To calculate the ratio (lb-mole O2 react/lb-mole NO formed), we can use the stoichiometry of the reaction. From the balanced equation, we can see that for every 5 moles of O2, we get 4 moles of NO. So, the ratio is 5/4 or 1.25 lb-moles O2 react/lb-mole NO formed.

b. To find the oxygen feed rate corresponding to 40% excess O2, we need to calculate the stoichiometric amount of O2 required and then add 40% of that amount. Since the reaction requires 5 moles of O2 for every 4 moles of NO, the stoichiometric amount of O2 required is (5/4) * 100.0 kmol NH3/h = 125.0 kmol O2/h. Adding 40% of that amount gives 125.0 kmol O2/h + (40/100) * 125.0 kmol O2/h = 175.0 kmol O2/h.

c. To determine the limiting reactant, we need to compare the amounts of ammonia and oxygen given. The molecular weight of ammonia (NH3) is 17 g/mol and oxygen (O2) is 32 g/mol. So, the mass of 50.0 kg of ammonia is 50.0 kg * (1000 g/kg) / (17 g/mol) = 2941.18 mol. The mass of 100.0 kg of oxygen is 100.0 kg * (1000 g/kg) / (32 g/mol) = 3125.0 mol. Comparing these amounts, we can see that ammonia is the limiting reactant as there is less of it. The percentage by which the other reactant (oxygen) is in excess can be calculated as [(3125.0 mol - 2941.18 mol) / 3125.0 mol] * 100 = 5.85 %. The extent of reaction and mass of NO produced can be determined using the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of NO produced, 5 moles of O2 are consumed. So, the extent of reaction is (3125.0 mol - 2941.18 mol) / 5 = 36.17 mol NO produced. The mass of NO produced can be calculated as 36.17 mol * (30.01 g/mol) = 1085.73 g.

a) Ratio (lb-mole O2 react/lb-mole NO formed) is 1.25 lb-mole O₂ / lb-mole NO.

b) Oxygen feed rate is 175.0 kmol O₂/h.

c) Limiting reactant is NH₃; Percentage excess of O₂ is 10.1%; Extent of reaction is 2.94 kmol NH₃; Mass of NO produced is 88.23 kg

a. Calculate the ratio (lb-mole O₂ react/lb-mole NO formed).

b. If ammonia is fed to a continuous reactor at a rate of 100.0 kmol NH₃/h, what oxygen feed rate (kmol/h) would correspond to 40.0% excess O₂?

c. If 50.0 kg of ammonia and 100.0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced (kg) if the reaction proceeds to completion.

Statements 1 is correct.

Explanation:

Pentanal - is an aldehyde have the molecular formula C₅H₁₀O, is soluble in organic solvent and it is derived from Pentane.

3-pentanone is a ketone have the molecular formula C₅H₁₀O.

1,3,5-pentanetriol is an alcohol have the molecular formula C₅H₁₂O₃.

All the 3 molecules have 5 carbon atoms and they don't have equal number of hydrogen atoms.

Pentanal and pentanone, both have carbonyl groups.

3-pentanone is slightly soluble in water, whereas 1,3,5-pentanetriol is mostly soluble in water, since it contains 3-OH groups forms hydrogen bond with water.

All three molecules are derived from Pentane and not from Pentyne.

So statement 1 is correct.

Answer:

a. glucose

Explanation:

Glucose is the compound that would be most soluble in water because it has regions of hydrogen-oxygen polar bonds, making it hydrophilic. Cholesterol, large proteins, and triglycerides are nonpolar and not very soluble in water.

Water is considered a polar solvent, and as such, substances that dissolve in water are usually polar or ionic compounds. Glucose, a, would be the most soluble in water because it has regions of hydrogen-oxygen polar bonds, making it hydrophilic. Cholesterol, b, is a nonpolar molecule and therefore not very soluble in water. Large proteins, c, can have both polar and nonpolar regions, so their solubility in water depends on the specific protein. Triglycerides, d, are nonpolar and not soluble in water.

Answer:

g = 1.11x10⁻⁵Ω.

Explanation:

The membrane conductance (g) can be calculated by dividing membrane current (I) through the driving force (Vm - E) as follows:  

[tex] g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} [/tex]

where Vm: is the membrane potential and [tex]E_{ion}[/tex]: is the equilibrium potential for the ion or reversal potential.  

The equilibrium potential for the ion can be calculated using the Nernst equation:

[tex] E_{ion} = \frac{RT}{zF}*Ln(\frac{[ion]_{out}}{[ion]_{ins}}) [/tex]

where R: is the gas constant = 8.314 J/K*mol, F: is the Faraday constant = 96500 C/mol, T: is the temperature (K), z: is the ion's charge, [ion]out and [ion]ins: is the concentration of the ion outside and inside, respectively.    

[tex] E_{ion} = \frac{(8.314 J*K^{-1}*mol^{-1})((15 + 273)K)}{(+1)(96500 C*mol^{-1})}*Ln(\frac{[500mM]}{[70mM]}) = 48.78 mV [/tex]  

Now, we can calculate the  membrane conductance (g) using equation (1):

[tex] g_{ion} = \frac{I_{ion}}{V_{m} - E_{ion}} = \frac{-318*10^{-9} A}{20*10^{-3} V - 48.78*10^{-3} V} = 1.11*10^{-5} \Omega [/tex]

Therefore, the membrane conductance is 1.11x10⁻⁵Ω.

I hope it helps you!        

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

To calculate the edge length for given density of metal, we use the equation:

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]

where,

[tex]\rho[/tex] = density = [tex]10.28g/cm^3[/tex]

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]

a = edge length of unit cell =?

Putting values in above equation, we get:

[tex]10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm[/tex]

Conversion factor used:  [tex]1cm=10^{10}pm[/tex]  

Hence, the edge length of the unit cell is 314 pm

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

[tex]R=\frac{\sqrt{3}a}{4}[/tex]

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

[tex]R=\frac{\sqrt{3}\times 314}{4}=135.9pm[/tex]

Hence, the radius of the molybdenum atom is 135.9 pm

The question is incomplete, here is the complete question:

371. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.118 atm at 25°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer: The molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=iMRT[/tex]

or,

[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 0.118 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of protein = 371. mg = 0.371 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of protein = ?

Volume of solution = 5.00 mL

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

T = temperature of the solution = [tex]25^oC=[25+273]K=298K[/tex]

Putting values in above equation, we get:

[tex]0.118atm=1\times \frac{0.371\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.371\times 1000\times 0.0821\times 298}{0.118\times 5}=1538.4g/mol=1.54\times 10^3g/mol[/tex]

Hence, the molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]

Answer:

a) Yes, these bands of signals overlap.

b) FM has broader transmission band.

Explanation:

a) Frequency range of TV channels = 59.5 to 215.8 MHz

Wavelength range of FM radio = 2.78 m to 3.41 m

Frequency of the wave = [tex]\nu [/tex]

Wavelength of the wave = [tex]\lambda [/tex]

Speed of light = c = [tex]3\times 10^8 m/s[/tex]

[tex]\nu =\frac{c}{\lambda }[/tex]

Frequency of wave with, [tex]\lambda = 2.78 m[/tex]

[tex]\nu =\frac{3\times 10^8 m/s}{2.78 m}=1.079\times 10^8 Hz[/tex]

[tex]1.079\times 10^8 Hz=1.079\times 10^8\times 10^{-6} MHz=107.9 MHz[/tex]

Frequency of wave with, [tex]\lambda = 3.41m[/tex]

[tex]\nu =\frac{3\times 10^8 m/s}{3.41 m}=8.798\times 10^7 Hz[/tex]

[tex]8.798\times 10^7 Hz=8.798\times 10^7\times 10^{-6} MHz=87.98 MHz[/tex]

Frequency range of FM = 87.89 to 107.9 MHz

Frequency range of TV channels = 59.5 to 215.8 MHz

Yes, these bands of signals overlap.

b)  

Frequency range of AM = 550 to 1600 kHz

1 kHz = 0.001 MHz

550 to 1600 kHz = [tex]550\times 0.001 MHz[/tex] to [tex]1600\times 0.001 MHz[/tex]

= 0.55 MHz to 1.6 MHz

Frequency range of AM = 0.55 to 1.6 MHz

Frequency range of FM = 87.89 to 107.9 MHz

FM has broader transmission band.

Answer:

the complete question is found in the attachment

Explanation:

the complete explanation is found in the attachment

A subsidy from the water district to local plant nurseries would lower the cost of production, enabling them to offer more drought-tolerant plants. This would shift the supply curve to the right on a graph, representing an increase in supply. Assuming demand remains steady, this could result in a lower price and increased quantity of plants.

In the economics of supply and demand, a subsidy given to local plant nurseries would likely increase the supply of drought-tolerant plants like purple sage. As the cost of production decreases due to the subsidy, nurseries can afford to produce and sell more plants, shifting the supply curve to the right.

On a graph showing supply and demand curves, you would drag the supply curve to the right to represent this change. This movement should ideally cause a decrease in the price of the plants and increase in quantity, assuming demand stays consistent.

This scenario illustrates the basic economic principle that if you reduce the cost of production (in this case by providing a subsidy), producers are able to supply more of the product, leading to increases in quantity and potential decreases in price. This is an effective tool used often by governments and organizations to influence market dynamics and promote specific goods or services.

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Answer: aldehyde

Explanation:

the aldehyde group (—CHO) is always at the extreme

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.

But when we move from top to bottom in a group then there occurs an increase in size of the atoms. Hence, ionization energy decreases along a group.

(a)  As Sb, Sn and I are all period 5 elements. Hence, these elements are arranged in order of increasing [tex]IE_{1}[/tex] as follows.

                             Sn < Sb < I

(b)  As Sr, Ca, and Ba are all elements of group 2a. Hence, these elements are arranged in order of increasing [tex]IE_{1}[/tex] as follows.

                            Ba < Sr < Ca

Answer:

Explanation:

It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.

When conducting liquid-liquid extractions, it is better to use an organic solvent that is lighter than water. Using a lighter organic solvent allows for easier phase separation and more efficient extraction of the organic compound from the aqueous phase.

When conducting liquid-liquid extractions, it is better to use an organic solvent that is lighter than water. This preference is based on the fact that organic solvents that are lighter than water form a separate layer on top of the aqueous phase, allowing for easier phase separation. This means that the organic compound can be more efficiently extracted from the aqueous phase into the organic layer, minimizing the number of transfer steps required.

For example, if you are extracting an organic compound from water, using an organic solvent like diethyl ether (density ≈ 0.71 g/mL) would be more effective compared to using an organic solvent like chloroform (density ≈ 1.478 g/mL) that is heavier than water.

The molarity of the solution when 23.640 g of Mn(ClO4)2 · 6 H2O is added to 200.0 mL of water is 0.3014 M, calculated by dividing the moles of solute by the volume of solution in liters.

To calculate the molarity of a solution when a certain amount of solute is added to a known volume of water, you must first determine the number of moles of solute. For the molarity of the resulting solution when 23.640 g of Mn(ClO4)2 · 6 H2O (molar mass approximately 392.13 g/mol) is dissolved in 200.0 mL of water, the calculation is as follows:

To calculate the moles of Mn(ClO4)2 · 6 H2O: 23.640 g / 392.13 g/mol ≈ 0.06028 mol.

To find molarity (M), which is moles of solute per liter of solution: 0.06028 mol / 0.200 L = 0.3014 M.