A 3.3×10-2 mg sample of a protein is dissolved in water to make 0.25 mL of solution. The osmotic pressure of the solution is 0.56 torr at 25°C. What is the molar mass of the protein?

Answer:

We can conclude that the Rf of that compound has a ratio of 4. It means that the solute has a ratio value of 4 times than that of solvent. As we can see that it has traveled 4 cm , this data is useful in determination of the compound in a mixture when compared with Rf values of other compounds.

Explanation:

Equation of the reaction:

H2(g) + F2(g) --> 2HF(aq)

1/2H2(g) + 1/2F2(g) --> HF(aq)

D(H-H) in H2 = 436 kJ/mol

D(F-F) in F2 = 158kJ/mol

ΔH bond breakage (dissociation):

1/2 mol H-H bonds = (1/2 X 436) kJ

= 218 kJ

1/2 mol F-F bonds = (1/2 X 158) kJ = = 80 kJ

Total = 218 + 80 = 298 kJ

ΔH bond formation:

1 mol H-F bonds = - 570 kJ

= DHreactant - DHproduct

ΔH°f = 298 kJ + -570 kJ = -272 kJ

Final answer:

The first four lines of the Balmer series involve electron transitions from 3p to 2s, 4p to 2s, 5p to 2s, and 6p to 2s orbitals.

Explanation:

The Balmer series involves electron transitions from higher energy levels to the second principal energy level (n=2), producing visible spectral lines. For the first four lines of the visible emission spectrum in the Balmer series, the allowed transitions must follow the selection rule Δl = ±1. Therefore, these transitions can only start from orbitals with l=1, which are the p orbitals.

For nfinal = 2, the corresponding ni initial energy levels for the first four visible emission lines are:

Answer:

d.All of the above are correct.

Explanation:

The curve of demand moves left or right continuously. Income, patterns and preferences, related products prices as well as the population size and composition are the key factors causing demand change.

The demand curve for gasoline can shift in response to changes in price, expected future price, and the number of sellers. Option b is the correct option.

The correct answer is d. All of the above are correct. The demand curve for gasoline reflects the relationship between the price and quantity demanded, so a change in the price of gasoline can shift the curve. The expected future price of gasoline can also influence current demand, as consumers may adjust their purchasing behavior based on their expectations. Additionally, the number of sellers of gasoline can impact market supply, which in turn affects the equilibrium price and quantity.

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Answer:

V NH3(g) = 1.833 L

Explanation:

balanced reaction:

assuming STP:

∴ V N2(g) = 3.52 L

∴ V H2(g) = 2.75 L

ideal gas:

∴ moles N2(g) = PV/RT

⇒ mol N2(g) = (1 atm)(3.52 L)/(0.082 atm.L/K.mol)(298 K)

⇒ mol N2(g) = 0.144 mol

∴ moles H2(g) = PV/RT

⇒ mol H2(g) = (1)(2.75)/(0.082)(298) = 0.113 mol (limit reagent)

∴ moles NH3(g) = (0.113 moles H2(g))(2 moles NH3 / 3 mol H2) = 0.075 mol

∴ V NH3(g) = RTn/P

⇒ V NH3(g) = ((0.082 atm.L/K.mol)(298 K)(0.075 mol))/(1 atm)

⇒ V NH3(g) = 1.833 L

Final answer:

From 3.52 L of nitrogen and 2.75 L of hydrogen, 1.83 L of ammonia can be produced, considering hydrogen as the limiting reactant based on the stoichiometry of the balanced chemical equation.

Explanation:

The question involves a stoichiometric calculation based on the reaction between hydrogen and nitrogen gases to form ammonia. Given the balanced chemical equation N2(g) + 3H2(g) → 2NH3(g), we can determine how many litres of ammonia gas could form from 3.52 L of nitrogen gas and 2.75 L of hydrogen gas, assuming all gases are at the same temperature and pressure. Since the reaction consumes nitrogen and hydrogen in a 1:3 ratio to produce ammonia in a 2 moles product per 1 mole of nitrogen ratio, we first identify the limiting reactant. Here, hydrogen gas (H2) is the limiting reactant because we need 3 volumes of hydrogen for every volume of nitrogen, but we have less than that (2.75 L instead of 3.52*3 L). The amount of ammonia produced is therefore determined by the amount of hydrogen available. Since 3 volumes of H2 produce 2 volumes of NH3, 2.75 L of H2 would produce (2.75 L * (2/3)) = 1.83 L of NH3.

Answer: 1.07×10^-20microlitre

Explanation:

1cm3 = 1000microlitres

1.07×10^-23 cm3 of tungsten = 1.07×10^-23 x 1000 = 1.07×10^-20microlitre

The volume of a single tungsten atom in microliters is 1.07x10^-17 µL. This is found by multiplying the given volume in cubic centimeters by the conversion factor of 1,000,000 µL/cm³.

The volume of a single tungsten atom is 1.07×10-23 cm3. One cubic centimeter (cm3) is equal to 1,000,000 microliters (µL). To convert the volume from cubic centimeters to microliters, we need to multiply the original value by the conversion factor. Therefore, the volume of a tungsten atom in microliters will be 1.07×10-23 cm3 * 1,000,000 µL/cm3, which equals to 1.07×10-17 µL. Hence, the volume of a tungsten atom in microliters is 1.07×10-17 µL.

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Answer:

a) The element is Manganese (Mn)

b) The element is Zirconium (Zr)

Explanation:

The step by step analysis and explanation is as shown in the attachment

Answer:

Molarity is 0.99 M

Explanation:

5.21% by mass, is a sort of concentration which shows the mass of solute in 100 g of solution.

Molarity is a sort of concentration that indicates the moles of solute in 1 L of solution (mol/L)

Let's find out the volume of solution by density.

Solution density = Solution mass / Solution volume

1.15 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.15 g/mL → 86.9 mL

We must have the volume of solution in L, so let's convert it.

86.9 mL / 1000 = 0.0869 L

Now, we have to determine the moles of solute (urea)

5.21 g . 1 mol / 60 g = 0.0868 moles

Mol/L = Molarity → 0.0868 moles / 0.0869L  = 0.99 M

Answer:

[tex]\large \boxed{\text{1.00 mol/L}}[/tex]

Explanation:

Molar concentration = moles/litres

So, we need both the number of moles and the volume.

1. Volume

Assume a volume of 1 L.

That takes care of that.

2. Moles of urea

(a) Mass of solution

[tex]\text{ Mass of solution} = \text{1000 mL} \times \dfrac{\text{1.15 g solution}}{\text{1 mL}} = \text{1150 g solution}[/tex]

(b) Mass of urea

[tex]\text{Mass of urea} = \text{1150 g solution}\times \dfrac{\text{5.21 g urea}}{\text{100 g solution}} = \text{59.92 g urea}[/tex]

(c) Moles of urea

[tex]\text{Moles of urea} = \text{59.92 g urea} \times \dfrac{\text{1 mol urea}}{\text{60.06 g urea}} = \text{1.00 mol urea}[/tex]

3. Molar concentration

[tex]\text{Molar concentration} = \ \dfrac{\text{1.00 mol}}{\text{1 L}} = \textbf{1.00 mol/L}\\\text{The molar concentration of the urea is $\large \boxed{\textbf{1.00 mol/L}}$}[/tex]

Answer:

The required volume of the original sample required is 2 micro liter

Explanation:

assuming the original sample concentration is 1N

after  final dilution of 10-2 solution concentration becomes 0.01 N

normality of original sample  = 1 N

normality  of final solution = 0.01 N

volume of  original sample= ?

volume of   final solution  = 0.2 mL

Considering thef formula below :

N1V1 = N2V2

V1 = (N2V2)/N1

= (0.01*0.2)/1

= 0.002 mL

1 milli liter = 1000 micro liter

0.002 mL = 2 micro liter

The original sample required is 2 micro liter

To produce a final dilution of 10^-2 in a total volume of 0.2mL, 2 microliters of original sample are required.

To find out how many microliters of original sample are needed to generate a final dilution of 10^-2 in a total volume of 0.2 mL, you can use the formula: V1 = V2 × D2 / D1. Here, V1 is the volume of the original sample needed, V2 is the final volume required (which is 0.2 mL), D2 is the final dilution (which is 10^-2), and D1 is the original dilution (which is 1 for undiluted samples).

So, plugging these values into the formula gives: V1 = 0.2 mL × 10^-2 / 1 = 0.002 mL. Convert this volume from milliliters to microliters (1 mL = 1000 μL), so V1 = 0.002 mL * 1000 = 2 μL.

So, 2 microliters of original sample are required to produce a final dilution of 10^-2 in a total volume of 0.2 mL.

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Answer:

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]

Explanation:

Ionization Energy (IE):

It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.

If we look from left to right in a period, ionization energy increases due stability of valance shell.

From the data given to us:

IE₁ = 801

IE₂ = 2427

IE₃ = 3659

IE₄ = 25,022

IE₅ = 32,822

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]

Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.

Answer:

The PH of the mixture is 4.74

Explanation:

The number of millimoles of acetic acid is calculated using the formula:

No of millimoles= Molarity * Volume( in ml)

= 0.25M * 30ml = 7.5 moles

Number of millimoles of KOH is calculated using:

Number of millimoles = Molarity * Volume ( in ml)

=0.05M * 75ml

= 3.75 moles

The PH of the solution is derived using:

pH = pKa + log [salt] / acid

= [tex] -log [ 1.8 * 10^5 ] + log [ 3.75 mmoles/ 3.75 mmoles] [/tex]

=4.74

Answer:

a) f = (1.24 × 10^15) Hz and E = (8.214 × 10^-19) J

b) f = (1.36 × 10^15) Hz; E = (9.035 × 10^-19) J

Explanation:

a) The least energetic photons have the highest wavelength. That is, the wavelength of the least energetic photons is equal to the upperlimit of the wavelength inequality given.

λ = 242nm = 2.42 × 10⁻7 m

v = fλ; f = v/λ; v = 3×10^8 m/s

f = (3×10^8)/(2.42×10^-7)

f = (1.24 × 10^15) Hz

E = hf; h = planck's constant = (6.62607004 × 10^-34) Js

E = 6.626 × 10^-34 × 1.24 × 10^15

E = (8.214 × 10^-19) J

b) The photons with the least wavelength in the range provided are the most energetic ones.

λ = (2200 × 10^-10) m = (2.2 × 10^-7) m

v = fλ; f = v/λ; v = 3×10^8 m/s

f = (3×10^8)/(2.2×10^-7)

f = (1.36 × 10^15) Hz

E = hf; h = planck's constant = (6.62607004 × 10^-34) Js

E = 6.626 × 10^-34 × 1.36 × 10^15

E = (9.035 × 10^-19) J

QED!

In ozone formation, least energetic photons having wavelength 242 nm have frequency ≈ 1.24 x 10^15 Hz and energy ≈ 8.2 x 10^-19 J. The most energetic photons that ozone absorbs, with a wavelength of 2200 Å, have a frequency of ≈ 1.36 x 10^15 Hz and energy of ≈ 9.02 x 10^-19 Joules.

(a) The frequency (ν) of a photon is given by the formula: ν = c / λ, where c is the speed of light (3.00 x 10^8 m/s) and λ is wavelength. For the least energetic photons (wavelength of 242 nm), we would convert the wavelength to meters (242 nm = 242 x 10^-9 m). Applying the formula: ν = 3.00 x 10^8 m/s / 242 x 10^-9 m, we get ν ≈ 1.24 x 10^15  Hz.

The energy (E) of a photon is given by the formula: E = hν, where h is Planck’s constant (6.63 x 10^-34 Js). So, E = 6.63 x 10^-34 Js x 1.24 x 10^15 Hz, which gives us E ≈ 8.2 x 10^-19 Joules.

(b) For the most energetic of these photons, they have the shortest wavelength (2200 Å = 2200 x 10^-10 m). Using similar calculations as above: ν ≈ 1.36 x 10^15 Hz and E ≈ 9.02 x 10^-19 Joules.

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Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:

[tex]Na_2CO_3(aq.)+NiCl_2(aq.)\rightarrow 2NaCl(aq.)+NiCO_3(s)[/tex]

Ionic form of the above equation follows:

[tex]2Na^{+}(aq.)+CO_3^{2-}(aq.)+Ni^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow NiCO_3(s)+2Na^+(aq.)+2Cl^-(aq.)[/tex]

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

[tex]Ni^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow NiCO_3(s)[/tex]

Hence, the net ionic equation is written above.

The balanced molecular equation for the reaction of Na2CO3 and NiCl2 is Na2CO3(aq) + NiCl2(aq) ⟶ 2NaCl(aq) + NiCO3(s). The balanced net ionic equation, excluding the spectator ions, is CO32-(aq) + Ni2+(aq) ⟶ NiCO3(s).

The reaction between aqueous sodium carbonate (Na2CO3) and aqueous nickel(II) chloride (NiCl2) results in the formation of sodium chloride (NaCl) and nickel carbonate (NiCO3). Represented as molecular equation, it looks like this: Na2CO3(aq) + NiCl2(aq) ⟶ 2NaCl(aq) + NiCO3(s).

For the net ionic reaction, we exclude the spectator ions, which in this case are Na+ and Cl-. These ions remain unreacted in the solution. Therefore, the net ionic equation will be: CO32-(aq) + Ni2+(aq) ⟶ NiCO3(s).

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Answer: The abundance of Li-7 isotope is higher as compared to Li-6.

Explanation:

Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

We are given:

Two isotopes of lithium :

Li-6 and Li-7

Average atomic mass of lithium= 6.941

As, the average atomic mass of lithium is closer to the mass of isotope Li-7. This means that the relative abundance of Li-7 is higher as compared to Li-6.

Percentage abundance of Li-7> Percentage abundance of Li-6 isotope

The atomic weight of lithium is closer to the mass number of 7Li, indicating that 7Li is more abundant than 6Li in nature. Thus, the correct answer is A) The abundance of 7Li is greater than 6Li.

The atomic weight of lithium, being 6.941, is closer to 7 than 6. Consequently, this indicates that most naturally occurring lithium is of the heavier 7Li isotope. So, in terms of relative abundance, 7Li is indeed more prevalent than 6Li. This means the correct answer would be option A) The abundance of 7Li is greater than 6Li. Therefore, based on the atomic weight of lithium, we can conclude that the abundance of 7Li is greater than 6Li.

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Answer:

50 ml (5x TBE) + 540 ml (water)

Explanation:

To prepare 0.5x TBE solution from 5x TBE solution we need to use the following dilution formula:

C1 x V1 = C2 x V2,   where:

- C1, V1 = Concentration/amount (start), and Volume (start)

- C2, V2 = Concentration/amount (final), and Volume (final)

* So when we applied this formula it will be:

5 x V1 = 0.5 x 500

V1= 50ml

- To prepare 0.5x we will take 50ml from 5x and completed with 450ml water and the final volume will going to be 500ml.

Answer : The specify the l and ml values for n = 4 are:

At l = 0,  [tex]m_l=0[/tex]

At l = 1,  [tex]m_l=+1,0,-1[/tex]

At l = 2,  [tex]m_l=+2,+1,0,-1,-2[/tex]

At l = 3,  [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]

Explanation:

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.

As we are given, n = 4 then the value of l and ml are,

l = 0, 1, 2, 3

At l = 0,  [tex]m_l=0[/tex]

At l = 1,  [tex]m_l=+1,0,-1[/tex]

At l = 2,  [tex]m_l=+2,+1,0,-1,-2[/tex]

At l = 3,  [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈  +  25/2O₂  →   8CO₂  +  9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8)  = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg  1.12×10¹⁴ g / 1000 =  1.12×10¹¹ kg

It will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.

Given: Length = 0.102 km = 0.102 km  × 1000 m/km = 102 m

Width = 0.069 km = 0.069 km  × 1000 m/km = 69 m

The area of the field: Area = Length × Width

                                    Area = 102 m × 69 m = 7038 sq. meters

The area by the cost of the grass turf per square meter:

Cost = Area × Cost per square meter

Cost = 7038 sq. meters  ×$9.75/sq. meter = $68,618.50

Therefore, it will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.

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Answer:

32.04°C will be the final temperature of the solution.

Explanation:

Moles of potassium chloride = 0.200 mol

MAs sof KCl= 0.200 mol × 74.5 g/mol= 14.9 g

Enthalpy of solvation of potassium nitrate =

[tex]\Delta H_{solv}=17.1 kJ/mol[/tex]

Energy released when 0.200 moles of KCl is dissolved in water = Q

[tex]Q=17.1kJ/mol\times 0.200 mol=3.42 kJ=3420 J[/tex]

(1 kJ = 1000 J)

Heat released on dissolving 0.200 moles of KCl is equal to heat absorbed by water = Q

Mass of solution , m= 80.0 g +14.9 g = 94.9 g

Specific heat of water = c = 4.184 J/g°C

Initial temperature of the water = [tex]T_1=23.4^oC[/tex]

Final temperature of the water = [tex]T_2=?[/tex]

[tex]Q=m\times c\times (T_2-T_1)[/tex]

[tex]3420 J=94.9g\times 4.184 J/g^oC\times (T_2-23.4^oC)[/tex]

[tex]T_2=32.04^oC[/tex]

32.04°C will be the final temperature of the solution.

Answer:

3030.2 g/mol is the molar mass for our polymer

Explanation:

Formula for Osmotic pressure → π = M . R .T

where π is pressure (atm)

M is molarity mol/L

R, the Universal Gases Constant (0.082 L.atm/mol.K)

T, Absolute T° (T°C + 273)

25°C + 273 = 298 K

Let's replace the values

0.174 atm = M . 0.082l.atm/mol.K . 298K

0.174 atm / (0.082l.atm/mol.K . 298K) = M

7.12×10⁻³ mol/L

As molarity is mol/L, and we have the volume of solution (in mL we must convert to L) we can find out the moles of our polymer that corresponds to the mass we used.

260mL . 1L/1000mL = 0.260L

7.12×10⁻³ mol/L = mol / 0.260L

7.12×10⁻³ mol . 0.260 = mol → 1.85×10⁻³

These moles refers to te 5.61 g of solute, to if we want to determine the molar mass, we should do:

g/mol → 5.61 g / 1.85×10⁻³ mol = 3030.2 g/mol

Final answer:

The molar mass of the polymer is approximately 28.93 g/mol.

Explanation:

To calculate the molar mass of the polymer, we can use the osmotic pressure formula:

π = MRT

Where π is the osmotic pressure, M is the molar mass, R is the ideal gas constant, and T is the temperature in Kelvin.

Given that the osmotic pressure is 0.174 atm and the temperature is 25.0 °C (which needs to be converted to Kelvin), we can rearrange the formula to solve for M:

M = π / (RT)

Now we can plug in the values:

M = 0.174 atm / (0.0821 L·atm/mol·K * 298 K)

M = 0.00706 mol / 0.024438 L

M = 28.93 g/mol

Therefore, the molar mass of the polymer is approximately 28.93 g/mol.

Answer:

Paramagnetism is dependent on the unpaired electron in the last orbital . In this regard, Cu(I) chloride is paramagnetic whereas Cu(II) chloride is not.

Explanation:

Paramagnetism is the property of materials/components which makes them attracted them weekly to the magnetic field.

It is related to electronic configuration, such that  it depends on the unpaired electron in the last orbital possess the property.

On basis of this property, Cu(I) chloride is paramagnetic while Cu(II) chloride is non paramagnetic. This is because Cu(I) chloride contains an unpaired electron in the last orbital whereas Cu(II) chloride does not have any unpaired electron.

Answer:

Explanation:

Paramagnetism is a type of magnetism whereby materials are weakly attracted to an externally applied magnetic field and then form internal, induced magnetic fields in the direction of the applied magnetic field. They are attracted to magnetic fields and have magnetic moment induced by the applied field is linear in the field strength. Paramagnetic materials include elements such as Oxygen,

Aluminium etc. and maybe some compounds like FeO etc.

Paramagnetism occurs due to the presence of unpaired electrons in an atom, so atoms with incompletely filled atomic orbitals are paramagnetic, there are exceptions such as copper exist and this is due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets. They have a magnetic permeability slightly greater than 1. External magnetic field causes the electrons spin to align parallel to the field hence, causing a net attraction. Paramagnetic materials include aluminium, oxygen, titanium, and iron oxide (FeO).

From the example,

Cu(I) and Cu(II)

Electronic configuration

Cu(I) - [Ar] 3d10

Cu(II) - [Ar] 3d9

[Ar] - 1s2 2s2 2p6 3s2 3p6 4s2

Therefore, Cu(I) is Paramagnetic while Cu(II) is not Paramagnetic (diamagnetic).

Answer:

n = 3 for similar blue light

Explanation:

The principle applied here is energy levels and energy changes. There are different energy levels depending on the value of the integer as explained by Max planck - a german physicist in 1900, Max planck claimed that electrons in an atom were presumed to be oscillating with a frequency f, then there enrrgy will be given by the plancks equation ; E =hf, where h is the plancks constant.

In general energy of each level can be written as E =nhf

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

[tex]Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})[/tex]

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

[tex]121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042[/tex]

[tex]120.9038x+122.9042\left(100-x\right)=12176.01[/tex]

Solving for x, we get that:

x = 57.2 %

Thus, percentage abundance of 121 Sb is = 57.2 %

percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %

Considering the definition of atomic mass, isotopes and atomic mass of an element,

First of all, the atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.

On the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.

In this case, 121 Sb and 123 Sb are the naturally isotopes of antimony. 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an atomic mass of 122.9042 u.

Being the average atomic mass of antomony 121.7601 u, the average mass of antimony can be calculated as:

average mass of antimony= percent natural abundance of 121 Sb× atomic mass of 121 Sb + percent natural abundance of 123 Sb× atomic mass of 123 Sb

Being percent natural abundance of 123 Sb= 1 - percent natural abundance of 121 Sb, and substituting the corresponding values, you get:

121.7601 u= percent natural abundance of 121 Sb× 120.9038 u + (1 - percent natural abundance of 121 Sb)× 122.9042 u

Solving:

121.7601 u= percent natural abundance of 121 Sb× 120.9038 u + 1× 122.9042 u - percent natural abundance of 121 Sb× 122.9042 u

121.7601 u= percent natural abundance of 121 Sb× (-2.0004 u) + 122.9042 u

121.7601 u - 122.9042 u= percent natural abundance of 121 Sb× (-2.0004 u)

-1.1441= percent natural abundance of 121 Sb× (-2.0004 u)

-1.1441÷ (-2.0004 u) = percent natural abundance of 121 Sb

0.5719 ×100 % = percent natural abundance of 121 Sb

57.19% = percent natural abundance of 121 Sb

So:

percent natural abundance of 123 Sb= 1 - percent natural abundance of 121 Sb

percent natural abundance of 123 Sb= 1 - 0.5719

percent natural abundance of 123 Sb= 0.4281

percent natural abundance of 123 Sb= 0.4281× 100

percent natural abundance of 123 Sb= 42.81%

Finally, the percent natural abundance of 121 Sb is 57.19% and the percent natural abundance of 123 Sb is 42.81%

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Answer: (a) seven orbitals, (b). 3 orbitals, (c). 5 orbitals and (d). 4 orbitals.

Explanation:

In order to solve this question we need to know how to explain the behaviour of electrons in atoms,and what we need to know is what is called the quantum numbers. There are four different kinds of quantum numbers and they are;

(1). Principal quantum numbers: the principal quantum number is denoted by the letter 'n'. It is used to describe the orbitals' energy. It has the values of n=1,2,3,4,...

(2). The spin quantum numbers: the spin quantum numbers is denoted by m(s). The 's' in the parenthesis is in subscript. It has the values of +1/2 and -1/2.

(3). Azimuthal quantum numbers: this is denoted by ℓ and it is used to explain orbital angular momentum and orbital shape. It has the values of ℓ= 0,1,2,3,....n-1.

Note that => ℓ = 0; we have a s-subshell,sphere shape.

ℓ = 1; p-subshell, dumb bell shape.

ℓ=2; d- subshell, double dumb bell shape.

ℓ= 3; f - subshell, multiple lobes.

(4). Magnetic quantum number: it is denoted by m(l) where the 'l' in the parenthesis is in subscript.

===> NOTE: there are (2ℓ + 1 ) orbitals in a subshell, also, there are n^2 number of orbitals in a shell.

Having known all that above, let us jump right in to the solution.

(a). From above we can see that; there are (2ℓ + 1 ) orbitals' in a subshell, also, f= ℓ= 3.

Therefore, the number of orbitals in 5f = 2 ℓ + 1 = (2×3) + 1 = 6+1 = 7 orbitals for 5f.

(b). 4p, the numbers of orbitals in 4p is; p= ℓ= 1=> 4 ℓ + 1 = (2×1) + 1 = 2+1 = 3 orbitals for 4p.

(c). 5d, the numbers of orbitals' in 5d is; d= ℓ= 2 = (2×2) + 1 = 4 + 1 = 5 orbitals for 5d.

(d). For n= 2, the numbers of orbital is ; n^2. Where the n given is 2. Therefore, 2^2= 2×2 = 4 orbitals in n=2.

An orbital refers to a region in space where electrons can be found.

An orbital refers to a region in space where there is a high probability of finding an electron. Orbitals that posses the same amount of energy are called degenerate orbitals.

The number of orbitals in an atom that can have the following designations are shown below;

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Answer: The nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.

Explanation:

Nuclear reaction are defined as the reactions in which nucleus of an atom is involved.

Positron emission is defined as the emission process in which positron particle is emitted. In this process, a proton gets converted to neutron and an electron neutrino particle.

[tex]_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e[/tex]

The chemical equation for the reaction of He-3 with a proton follows:

[tex]_2^3\textrm{He}+_1^1\textrm{H}\rightarrow _2^4\textrm{He}+_{+1}^0e[/tex]

Hence, the nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.

The nuclear equation where a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron is written as ³He + ¹H → ⁴He + e⁺. This ensures that mass and charge are conserved in the reaction.

To complete the nuclear chemical equation where a helium-3 nuclide (³He) transforms into a helium-4 nuclide (⁴He) by absorbing a proton (¹H) and emitting a positron (e⁺), we must ensure that both mass and charge are conserved in the reaction. In this case, the equation can be represented as:

³He + ¹H → ⁴He + e⁺

The mass number on the left side of the equation is 3 (from helium-3) plus 1 (from the proton), totaling 4, which matches the mass number of helium-4 on the right side of the equation. The atomic number (number of protons) is also conserved through this reaction: 2 (from helium-3) + 1 (from the proton) equals 2 (from helium-4) + 1 (from the emitted positron), with positrons having a positive charge but no atomic number associated.

Answer: The molality of potassium hydroxide solution is 0.608 m

Explanation:

We are given:

3.301 mass % of potassium hydroxide solution.

This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution

Mass of solvent = Mass of solution - Mass of solute (KOH)

Mass of solvent = (100 - 3.301) g = 96.699 g

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

[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams})}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute (KOH) = 3.301 g

[tex]M_{solute}[/tex] = Molar mass of solute (KOH) = 56.1 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent = 96.699 g

Putting values in above equation, we get:

[tex]\text{Molality of KOH}=\frac{3.301\times 1000}{56.1\times 96.699}\\\\\text{Molality of KOH}=0.608m[/tex]

Hence, the molality of potassium hydroxide solution is 0.608 m

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P =   ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

[tex]P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g[/tex]   ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

[tex]P_{(a)= 790kg/m^3*9.8m/s^2*10[/tex]           ----------- equation (2)

if we equate equation 1 and 2 together; we have

[tex]P_{(g)= P_{(a)[/tex]

[tex]1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm[/tex]

Making [tex]h_g[/tex] the subject of the formula, we have :

[tex]h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}[/tex]

[tex]h_g[/tex] = 6.269 cm

The difference in the height denoted  by ΔH can therefore be calculated as:

ΔH [tex]= H_a-H_g[/tex]

ΔH [tex]= 10cm - 6.269cm[/tex]

ΔH = 3.731 cm

ΔH ≅ 3.73 cm           (to two significant figures)

Answer:

50.0 μL

Explanation:

When a dilution is done, the mass of the solute (in this case the ampicillin) remains constant, following the Lavoiser's law that the mass is conserved. The mass is the concentration (C) multiplied by the volume (V), so if 1 is the stock solution, and 2 is the bacterial culture after the addition of the antibiotic:

m1 = m2

C1*V1 = C2*V2

C1 = 100 mg/mL = 100000 μg/mL (1 mg = 1,000μg)

C2 = 100 μg/mL

V2 = 50 mL + V1 = 50000μL + V1 (V1 in μL)

100000*V1 = 100*(50000 + V1)

1000V1 = 50000 + V1

999V1 = 50000

V1 = 50.0 μL

Answer:

0.0076kg

Explanation:

To get the mass, we use the relation among density, mass and volume.

Mass = density * volume

Here mass? , density = 0.779g/ml , volume = 9.76ml

Mass = 9.76 * 0.779 = 7.60g

Answer is wanted in kg so we divide by 1000. This is 7.60/1000 = 0.0076kg

Answer:

Explanation:

The structural repeating unit  of beta sheet is 7 anstrom/2 aminoacids. So,

  [tex]\frac{80.5 Angstrom}{3.5} = 23 aminoacids[/tex]

If a segment of a single chain  in an antiparallel beta sheet has a length of 80.5 angstrom, then there will be 23 residues in this segment.

To find the number of residues in an 80.5 Å long segment of an antiparallel beta-pleated sheet, divide the total length by the length of one residue (3.5 Å per residue), which gives approximately 23 residues.

The student asked how many residues are in a segment of a single chain in an antiparallel beta-pleated sheet with a length of 80.5 Å. To determine the number of amino acid residues in the segment, we can use the typical amino acid residue length in a ß-pleated sheet, which is approximately 3.5 Å per residue in an extended conformation. This measurement considers the distance hydrogen bonds can span between the carbonyl oxygen and amino hydrogen along the peptide chain in the secondary structure.

By dividing the total length of the chain (80.5 Å) by the length of one residue (3.5 Å), you can calculate the number of residues:

    Number of residues = Total length ÷ Length per residue

    Number of residues = 80.5 Å ÷ 3.5 Å/residue

    Number of residues ≈ 23 residues

This calculation does not account for slight variations that may occur in different proteins or specific contexts, but it provides a general estimate for the number of amino acids in a segment of a beta sheet.