The nonvolatile, nonelectrolyte sucrose, C12H22O11 (342.3 g/mol), is soluble in water H2O.
Calculate the osmotic pressure (in atm) generated when 12.8 grams of sucrose are dissolved in 278 mL of a water solution at 298 K.

Answer:

1. LiOH : strong base

2. HF : weak acid

3. HCl : strong acid

4. NH3 : weak base

Ka expression: Ka = [A- ] * [H+] / [HA]

Explanation:

The Acidity Constant:

The acid dissociation constant, Ka, (or acidity constant, or acid ionization constant) is a measure of the strength of a weak acid (which is not completely dissociated):

HA ↔ A- + H +

HA is a generic acid that dissociates into A- (the conjugate base of the acid), and the hydrogen or proton ion, H +.

The dissociation constant Ka is written as the ratio of equilibrium concentrations (in mol / L):

Ka = [A- ] * [H+] / [HA]

When we write in square brackets, we refer to the concentration of that element.

Final answer:

LiOH is a strong base, HF is a weak acid with a Kₐ expression of Kₐ = [H⁺][F⁻]/[HF], HCl is a strong acid, and NH₃ is a weak base with a Kₐ expression for its conjugate acid NH₄⁺ as Kₐ = [NH₄⁺][OH⁻]/[NH₃].

Explanation:

Compounds can be identified as strong acids, weak acids, strong bases, or weak bases depending on their ability to dissociate in solution. The classification is based on the strength of the acids and bases, which is a measure of their tendency to donate or accept protons.

Answer:

Explanation:hiii

80 to 95 grams of carbohydrate should be consumed every hour for 4 hours post-exercise.

Carbohydrates are biomolecules that are made up of carbon, hydrogen, and oxygen atoms.

Examples of carbohydrates are starch, sugar, fiber.

Carbohydrate is the main component of our food which gives us energy.

If an athlete weighs 175 lb and has to exercise every four hours.

He will need a regular amount of carbohydrates to get energy.

Thus, the amount of carbohydrate required is 80 to 95 grams.

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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: 6kJ/mol

Explanation:

The full eclipse stereoisomer of ethanol has two H-H overlapped and one H-OH overlapped. The energy cost for H-H eclipse is 4kJ/mol or 1kcal/mol.

Total energy= 14kJ/mol

2(H-H eclipsed) = 2(4kJ/mol) = 8kJ/mol

Total Energy = H-H eclipsed + H-OH eclipsed

14kJ/mol = 8kJ/mol + H-OH eclipsed

14kJ/mol - 8kJ/mol = H-OH eclipsed

Therefore H-OH eclipsed = 6kJ/mol

Answer:

The concentration of the Cu(II) ion is 0.777M

Explanation:

Step 1: Data given

Volume of  1.70 M CuCl2 = 48.0 mL = 0.0480 L

Volume of 0.800 M (NH4)2S = 57.0 mL = 0.0570 L

Step 2: The balanced equation

CuCl2 (aq) + (NH4)2S (aq) → 2 NH4Cl (aq) + CuS (s)

Step 3: Calculate moles CuCl2

moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles

Step 4: Calculate moles (NH4)2S

moles (NH4)2S = 0.0570 L * 0.800 M = 0.0456 moles

Step 5: Calculate the limiting reactant

The ratio between CuCl2 and (NH4)2S is 1 : 1 so (NH4)2S is the limiting reactant . IT will completely be consumed (0.0456 moles).

CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles

Step 6: Calculate moles of CuS

For 1 mol CuCl2 we need 1 mol (NH4)2S to produce 2 moles of NH4Cl and 1 mol CuS

For 0.0456 moles we'll produce 0.0456 moles CuS

Step 7: Calculate moles of Cu(II)ion

There remains 0.0360 moles CuCl2.

There will be 0.0456 moles CuS produced

Total moles Cu^2+ = 0.0816 moles

Step 8: Calculate concentration of Cu(II) ion

Concentration = moles / volume

Concentration = 0.0816 moles / (0.048+0.057)

Concentration = 0.777 M

The concentration of the Cu(II) ion is 0.777M

I hope this will help ;)

The structures for all constitutional isomers with the following molecular formulas a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl  c) CH₃-CH₂-Cl d) CH₂-CH-Cl .

The isomers are those compounds which contain same molecular formula but different molecular structures but the physical and chemical properties will be same like melting and boiling points.

The isomers for the formula will be,

(a) C6H14 = CH₃-CH₃-CH₃-CH₃-C₂H₂

(b) C2H5Cl = CH₃-CH₂-Cl

(c) C2H4Cl2 = CH₃-CH₂-Cl

(d ) C2H3Cl3 = CH₂-CH-Cl .

The compounds are same in molecular formula but the representation of structure is different.

Therefore,  a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl  c) CH₃-CH₂-Cl d) CH₂-CH-Cl . structures for all constitutional isomers with the following molecular formulas.

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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 :

(a) The symbol, group number, and period number of the element is, O, 16 and 2 respectively.

(b) The symbol, group number, and period number of the element is, P, 15 and 3 respectively.

Explanation :

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

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

To identify the block of the element after you get its electronic configuration:

(i) If the element belongs to s-block.

Group number = Number of valence electrons (or outermost shell electrons).

(ii) If the element belongs to p- block.

Group number = No. of valence electrons + 10 .i.e., 10 + np electrons + ns electrons.

(iii) If the element belongs to d-block.

Group no. = no. of electrons in (n-1) d subshell + no. of electron/s in ns shell.

(iv) If the element belongs to f-block, Group no. = 3.

(a) [He] 2s²2p⁴

From the given electronic configuration, we conclude that it has 6 valence electrons and belongs to p-block. So, it belongs to group number 16 (6+10).

The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=2). So, the period number is, 2.

Thus, the element which is present in 2nd period and 16 group number is, oxygen (O).

(b) [Ne] 3s²3p³

From the given electronic configuration, we conclude that it has 5 valence electrons and belongs to p-block. So, it belongs to group number 15 (5+10).

The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=3). So, the period number is, 3.

Thus, the element which is present in 3rd period and 15 group number is, phosphorous (P).

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:

A solution is red in color means it has absorbed other color wavelengths and reflects only red color wavelength. A red color wavelength would absorb wavelength corresponding to violet color. Hence, the wavelength should be fixed in the range of 400 nm to measure the absorbance. And not red light range ( 700 nm) to measure its absorbance."

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.

Answer:

a) P = 9.58 psi for  h=7.2 m

b) P=4.7 psi for h=5.94 m

Explanation:

Since the pressure Pon a static liquid level h is

P= p₀ + ρ*g*h

where p₀= initial pressure , ρ=density , g = gravity

then he variation of the liquid level Δh will produce a variation of pressure of

ΔP= ρ*g*Δh → ΔP/Δh =  ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m)  = 12/3.1 psi/m

if the liquid level is converted linearly

P = P₁ + ΔP/Δh*(h -h₁)

therefore choosing  P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m

P = 3 psi  + 12/3.1 psi/m *(7.2 m -5.5 m) = 9.58 psi

then P = 9.58 psi for  h=7.2 m

for P=4.7 psi

4.7 psi = 3 psi  + 12/3.1 psi/m *(h -5.5 m)

h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m

then P=4.7 psi for h=5.94 m

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.

Since, the options are not given the question is incomplete. The complete question is as follows:

A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What is this unknown molecule?

DNA

lipid

carbohydrate

RNA

protein

Answer: DNA

Explanation:

The DNA short for deoxyribonucleic acid is a genetic material that can be found in majority of the living beings on earth. It is composed of two strands that are coiled around each other. Each strand consists of nucleotides. Each nucleotide is composed of four nitrogen exhibiting nucleobases like guanine, cytosine, adenine and thymine along with the deoxyribose sugar and phosphate group. In DNA there are two groups of nitrogenous bases these includes the pyrimidines and purines. The pyrimidines are cytosine and thymine and the purines are guanine and adenine.  

According to the given situation, a molecule from a new organism consists of adenine, cytosine thymine and guanine these all are nitrogenous bases which can be found in DNA.

Answer: The mass of estrogen that must be added is 2.83 grams

Explanation:

The equation used to calculate relative lowering of vapor pressure follows:

[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{solute}[/tex]

where,

[tex]\frac{p^o-p_s}{p^o}[/tex] = relative lowering in vapor pressure

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

[tex]\chi_{solute}[/tex] = mole fraction of solute = ?

[tex]p^o[/tex] = vapor pressure of pure ethanol = 54.68 mmHg

[tex]p_s[/tex] = vapor pressure of solution = 54.11 mmHg

Putting values in above equation, we get:

[tex]\frac{54.68-54.11}{54.68}=1\times \chi_{\text{estrogen}}\\\\\chi_{\text{estrogen}}=0.0104[/tex]

This means that 0.0104 moles of estrogen are present in the solution

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

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

Moles of estrogen = 0.0104 moles

Molar mass of estrogen = 272.4 g/mol

Putting values in above equation, we get:

[tex]0.0104mol=\frac{\text{Mass of estrogen}}{272.4g/mol}\\\\\text{Mass of estrogen}=(0.0104mol\times 272.4g/mol)=2.83g[/tex]

Hence, the mass of estrogen that must be added is 2.83 grams

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:

Explanation:

The solution has been attached

Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

Atomic mass of the elements are:

Na: 22.989 u

S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

= (17.240 g)/(158.105 g/mol) = 0.1090 mole.

Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

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: Least Acidic:  A

Moderate Acidic: C

Most Acidic: B

Explanation:

Given compounds' acidity cannot be determined as the molecular formulas seem incorrect. Normally, acidity is determined by factors such as the presence of hydrogen atoms and their ability to be donated influenced by bond polarity and molecular structure.

The acidity of the given compounds cannot be determined as the given molecular formulas (h5ch17p19b1, h5ch17p19a1, h5ch17p19c1) appear to be incorrect or non-standard. Typically, the acidity of a compound is influenced by factors like the presence of hydrogen atoms, how easily these can be donated (as determined by bond polarity and structure of the molecule), and the stability of the conjugate base after a hydrogen atom has been donated.

In common terminology, acidity refers to the ability of a substance to donate a proton (H+) in a chemical reaction. The traditional scale for measuring acidity is the pH scale, where lower pH values indicate higher acidity.

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Answer:

Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.

Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.

Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.

Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.

The atomic orbital and radius increases by gaining electron linearly so even electron numbered atoms are diamagnetic while the odd electron numbered atoms are paramagnetic.

Running through the first 18 elements one can observe that there is an alternative odd number of electrons and an even number proofing that that half of the first 18 elements shows paramagnetism and diamagnetism respectively.

Explanation:

The given configurations correspond to the elements Cadmium (Cd) and Nickel (Ni) respectively. Cd is in group 12, period 5 and Ni is in group 10, period 4.

The requested electron configurations correspond to specific elements in the periodic table.

(a) The configuration [Kr] 5s²4d¹⁰ corresponds to the element Cadmium (Cd). Its symbol is Cd, it is in group 12, and period 5. The partial valence-level orbital diagram is as follows:

(b) The configuration [Ar] 4s²3d⁸ corresponds to the element Nickel (Ni). Its symbol is Ni, it is in group 10, and period 4. The partial valence-level orbital diagram is as follows:

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The configurations [Kr] 5s²4d¹⁰ and [Ar] 4s²3d⁸ represent Cadmium and Nickel respectively. Cadmium is in the 12th group, 5th period, and Nickel is in the 10th group, 4th period. Both are transition metals with distinct chemical reactions.

The element with the electron configuration [Kr] 5s²4d¹⁰ is Cadmium (Cd). It belongs to the 12th group and 5th period. Its valence electron configuration diagram shows that there are 2 electrons in the 5s subshell and 10 electrons in the 4d subshell.

On the other hand, the element with the electron configuration [Ar] 4s²3d⁸ is Nickel (Ni). This element belongs to the 10th group and 4th period. Its valence electron configuration diagram shows it has 2 electrons occupying the 4s subshell and 8 electrons in the 3d subshell.

These identified elements, Cadmium and Nickel, represent their unique chemical and physical properties. For instance, they are both transition metals and behave similarly in many chemical reactions.

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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:(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: The precipitate will not be formed in the above solution.

Explanation:

The chemical equation for the reaction of potassium bromide and lead acetate follows:

[tex]2KBr(aq.)+Pb(CH_3COO)_2(aq.)\rightarrow PbBr_2(s)+2CH_3COOK(aq.)[/tex]

We are given:

Concentration of KBr = 0.013 M

Concentration of lead acetate = 0.0035 M

1 mole of KBr produces 1 mole of potassium ions and 1 mole of bromide ions

So, concentration of bromide ions = 0.013 M

1 mole of lead (II) acetate produces 1 mole of lead (II) ions and 2 moles of acetate ions

So, concentration of lead (II) ions = 0.0035 M

The salt produced is lead (II) bromide

The equation follows:

[tex]PbBr_2(s)\rightleftharpoons Pb^{2+}(aq.)+2Br^-(aq.)[/tex]

The expression of [tex]Q_{sp}[/tex] for above equation follows:

[tex]Q_{sp}=[Pb^{2+}]\times [Br^-]^2[/tex]

Putting values of the concentrations in above expression, we get:

[tex]Q_{sp}=(0.0035)\times (0.013)^2\\\\Q_{sp}=5.92\times 10^{-7}[/tex]

We know that:

[tex]K_{sp}[/tex] for lead (II) bromide = [tex]4.67\times 10^{-6}[/tex]

There are 3 conditions:

As, the [tex]Q_{sp}[/tex] is less than [tex]K_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.

Hence, the precipitate will not be formed in the above solution.

Answer:1s22s22p63s2- Magnesium

1s22s22p63s23p3- phosphorus

1s22s22p63s23p64s23d104p6- krypton

1s22s2- Beryllium

1s22s22p6- neon

1s22s22p3- nitrogen

Explanation:

The identity of an element is deducible from its electronic configuration by looking out for the outermost shell configuration and counting the total number of electrons present. for example; phosphorus 15 electrons and an outermost shell configuration of ns2 np3. Any electronic configuration written above which reflects these characteristics must belong to phosphorus.

The question is about identifying elements based on given electron configurations and then matching these configurations to known ones. The elements matching the given configurations have the same number of electrons in their outermost energy level, indicating the possibility of similar chemical properties.

The question is regarding the identification of elements based on their electron configurations and comparing them to known configurations. The electron configuration describes the distribution of electrons in an atom's electron shells and subshells.

For instance, the given first electron configuration (1) 1s22s22p63s2 corresponds to the element Neon. The element that would match this configuration is one whose electron configuration is 1s22s22p6, which is also for Neon in a neutral state.

For the second case (1s22s22p63s23p64s23d104p6), it corresponds to the element Krypton. The element that would match this configuration is also Krypton in a neutral state. These elements are said to have similar chemical properties because they have the same number of electrons in their outermost energy level, making their bonding patterns similar.

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Answer: 2.5g

Explanation:

5 % sucrose solution means that 5 % of the weight of the solution is sucrose.

If 1 liter of water weighs 1000 grams. To prepare 5% sucrose solution 5/100 x 1000 = 50 grams. Since 1 liter equals 1000ml, thus a 5 % solution has 50 grams of solute dissolved in one liter.

To prepare 5% sucrose solution in 50mls

=5/100 x 50

= 0.05 x 50

= 2.5g

Therefore to prepare 5% sucrose solution in 50mls we dissolve 2.5g of sucrose in 50ml of water

Final answer:

To prepare a 5% by mass sucrose solution with 50.0 ml of distilled water, approximately 2.63 g of sucrose is required, which is calculated using the percent by mass formula and assumes that water's density is 1 g/ml.

Explanation:

The student's question asks to calculate the mass of sucrose necessary to prepare a 5% by mass sucrose solution using 50.0 ml of distilled water. The concept involved here is the percent by mass calculation which is used in preparing solutions in chemistry. To calculate the mass of the sucrose needed, one must use the formula:

Percent by mass = (mass of solute / mass of solution) × 100%

Given that we want a 5% sucrose solution, we can rearrange the formula to solve for the mass of sucrose:

Mass of sucrose = (Percent by mass × mass of solution) / 100%

First we need to convert the volume of water to mass, assuming the density of water is approximately 1 g/ml:

Mass of water (solvent) = 50.0 ml × 1 g/ml = 50.0 g

The total mass of the solution will be the mass of the water plus the mass of the sucrose, which we can call 'x':

Mass of solution = mass of water + x

Now, plugging in the known values and solving for 'x' gives us:

x = (5% × (50.0 g + x)) / 100%

Solving this equation for 'x' yields:

0.05 × 50.0 g + 0.05x = x

2.5 g + 0.05x = x

2.5 g = x - 0.05x

2.5 g = 0.95x

x = 2.5 g / 0.95

x = 2.6316 g

Therefore, the mass of sucrose necessary to make a 5% sucrose solution with 50.0 ml of distilled water is approximately 2.63 g.

Answer:

0.327 M is the concentration of hydrochloric acid in the final solution.

Explanation:

Case 1:

716.7 ml of 3.86 M HCl. Using a volumetric pipet, you take 119.56 ml of that solution and dilute it to 969.88 ml in a volumetric flask.

Before dilution :

[tex]M_1=3.86 M, V_1=119.56 ml[/tex]

After dilution :

[tex]M_2=?, V_2=969.88 ml[/tex]

[tex]M_1V_1=M_2V_2[/tex] ( dilution )

[tex]M_2=\frac{M_1V_1}{V_2}=\frac{3.86 M\times 119.56 mL}{969.88 mL}[/tex]

[tex]M_2=0.4758 M[/tex]

Case 2:

Now you take 100.00 ml of case 1 solution and dilute it to 145.62 ml in a volumetric flask

Before diluting it further  :

[tex]M_2=0.4758 M, V_1=100.00 ml[/tex]

After dilution :

[tex]M_3=?, V_3=145.62 ml[/tex]

[tex]M_3V_3=M_2V_2[/tex] ( dilution )

[tex]M_3=\frac{M_2V_2}{V_3}=\frac{0.4758 M\times 100.00 mL}{145.62 mL}[/tex]

[tex]M_3=0.327 M[/tex]

0.327 M is the concentration of hydrochloric acid in the final solution.

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:

ΔSmix,m = 5.7628 J/K.mol

Explanation:

mix: A + B

∴ nA = x mol A

∴ nB = y mol B

⇒ n mix = x + y = 1 mol

∴ P total = 1 bar

∴ T: constant

entropy of gases when mixing:

∴ XA = x/1 = x

∴ XB = y/1 = y

⇒ ΔSmix = - x*R*Lnx - y*R*Lny

assuming: x = y = 0.5 mol

⇒ ΔSmix = - (0.5)(R)(- 0.693) - (0.5)(R)(- 0.693)

⇒ ΔSmix = (0.3465)(R) + (0.3465)(R)

⇒ ΔSmix = (0.6931)(R)

∴ R = 8.314 J/K.mol

⇒ ΔSmix,m = (0.6931)(8.314 J/K.mol)

⇒ ΔSmix,m = 5.7628 J/K.mol