What is the wavenumber of the radiation emitted when a hydrogen
atom makes a transition corresponding to a change in energy of
1.634 x 10-18 J?

Answer:

i=2 for HCl and i=1 for ethanol

Explanation:

Hello,

Since the hydrochloric acid is composed by sodium and chloride ions which are completely dissociated, two types of ions are present, that's why i becomes 2.

On the other hand, as long as the ethanol doesn't present dissociation in aqueous solution, the Van't Hoff factor becomes 1.

Best regards.

Answer:

Explanation:

Use the position of an element in the periodic table to deduce its valence.

The periodic table is a table that groups elements based on their periodic functions. A group is a vertical arrangement of elements. The group number shows the number of elements in the outer shell of the atoms.

The groups runs from 1 to 8. On the periodic table, elements that has more than 4 electrons in their outermost shell will have a valency of 8 less than the number of outermost electrons.

Write formulas for these hydrides without using subscripts, for example XH3. If no hydride forms, write "none".

for group I : NaH, KH,

    group II : MgH2, CaH2,

    group III: BH3, AlH3

  Group 8 do not form hydrides because of their inertness.

What is the formula of the hydride formed by sulfur?

the hydride of sulfur is H₂S

What is the formula of the hydride formed by potassium ?

the hydride of potassium is KH

Final answer:

The formula of the hydride formed by sulfur is H2S, while the formula for the hydride formed by potassium is KH, based on the valence of sulfur and potassium corresponding to their positions in the periodic table.

Explanation:

When determining the formulas for hydrides, the valence of the non-hydrogen element is crucial. Sulfur, which has a valence of 2, forms a hydride by combining with two hydrogen atoms, giving us the formula H2S. In contrast, potassium belongs to the alkali metals with a valence of 1, thus it combines with one hydrogen atom to form a hydride, resulting in the formula KH.

The periodic table helps us understand these valences due to an element's group number. Sulfur, located in group 16, typically forms compounds where it has two bonding sites, as seen with its hydride, hydrogen sulfide. Potassium, found in group 1, forms compounds by donating a single electron, resulting in a 1:1 ratio with hydrogen in potassium hydride.

Answer:

i need more to solve this

Explanation:

Options for full question:

(A) % oxygen is 32.1%

(B) % hydrogen is 4%

(C) total percent composition of all elements is approximately 100%

(D) % carbon is 15.9%

Answer:

Options B and C

Explanation:

Information Given;

Mass of sample - 50g

Mass of Oxygen - 32.1g

Mass of Hydrogen - 2g

Mass of Carbon - 15.9g

The percentage composition of an element in a compound is the mass percentage of the element present in the compound. It tells the mass percentage of each element present in a compound.

The formular is given as;

Percentage composition = (Mass of element / Mass of compound) * 100

For oxygen;

Percentage Composition = (32.1 / 50) * 100 = 0.642 * 100= 64.2%

For Hydrogen;

Percentage Composition = (2 / 50) * 100 = 0.04 * 100= 4%

For Carbon;

Percentage Composition = (15.9 / 50) * 100 = 0.318 * 100= 31.8%

Total percentage composition of all elements = 4% + 31.8% + 64.2% = 100%

Based on this,

Option A is incorrect

Option B is Correct.

Option C is Correct.

Option D is incorrrect

Answer:

b. 0.100M

Explanation:

The balanced chemical reaction is: NaOH + HA ⇒ H₂O + NaA

The NaOH and HA react in a 1:1 molar ratio, so at the equivalence point, the amount of NaOH added equals the amount of HA that was present in the solution.

The amount of NaOH that was added can be calculated and set equal to the amount of HA that must have been present to react with it.

n = CV = (0.200 mol/L)(12.5 mL) = 2.50 mmol NaOH = 2.50 mmol HA

Thus, there were 2.50 mmol of HA in 25.0 mL. The concentration can be calculated as follow:

C = n/V = (2.50 mmol)/(25.0mL) = 0.100 M

Answer:

Explanation:

1) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. The solubility of KCl at 370K is 42%mass -42kg of KCl in 100kg of water+KCl , thus, the amount of water added to 500kg of KCl is:

500 kg of KCl × [tex]\frac{100 kg Water+KCl}{42kgKCl}[/tex] = 1190 kg of water +KCl

1190 kg of water +KCl - 500 kg of KCl = 690 kg of water

2) The maximum amount that this solution could solubilize of KCl at 320K is:

1190 kg of water + KCl × [tex]\frac{31,5 kgKCl}{100kgWater+KCl}[/tex] = 375 kg

Thus, the mass of KCl crystals formed are:

500 kg of KCl - 375 Kg of KCl = 125 kg of KCl

I hope it helps!

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

        Initial volume [tex](v_{1})[/tex] = 1.673 [tex]m^{3}[/tex]

          Final volume [tex](v_{2})[/tex] = 1.789 [tex]m^{3}[/tex]

As, the amount of water vapor condensed will be as follows.

                     [tex]\frac{(v_{2} - v_{1})}{v_{2}}[/tex]

                     = [tex]\frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}[/tex]

                     = [tex]\frac{0.116 m^{3}}{1.789 m^{3}}[/tex]

                     = 0.065 kg

Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.

Answer:

The answer is D. gamma rays

Explanation:

A radioactive atom can have three different types of emission:

alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.

Beta particles (β) = they have 5x[tex]10^{-4}[/tex] amu and they have an intermediate penetrating power

Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher

For this reason Gamma emissions (γ) has the smallest mass value.

Among the options provided, the gamma photon in radioactive emissions has the smallest mass (zero mass), which means that it's the correct answer to the question. Gamma radiation is energy emission without a corresponding mass.

In the context of radioactive emissions, each type has a different mass. The alpha particle is the heaviest and is comprised of two neutrons and two protons. Beta particles, which are electrons (or positrons in the case of beta-plus decay), have a smaller mass. Positrons have the same mass as electrons but with a positive charge.

However, the gamma photon, which is a type of electromagnetic radiation, has no rest mass at all. So, out of the given options, the gamma ray has the smallest (in this case, zero) mass. This is an example of energy being emitted without a corresponding mass in a radioactive process, which often occurs when the remaining nucleus is in an excited state post decay and moves to a lower energy level by emitting a gamma photon.

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Answer:

a) 63,0%

b) 54,4 Pounds HNO₃ per cubic foot of solution

c) 13,8 M

Explanation:

a) Weight percent is the ratio solute:solution times 100:

[tex]\frac{1,704 kg HNO_3}{2,704 kg Solution}[/tex] = 63,0%

b) Pounds HNO₃ per cubic foot of solution at 20 °c

Pounds HNO₃:

1,704 kg [tex]\frac{2,20462 pounds}{1 kg}[/tex] = 3,7567 pounds

Cubic foot:

2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex]x[tex]\frac{1 cubic foot }{28,3168 L}[/tex] = 0,069 ft³

Thus:[tex]\frac{3,7567 pounds}{0,069ft^3} =[/tex] = 54,4 Pounds HNO₃ per cubic foot of solution

c) Moles of HNO₃:

1704 g HNO₃ [tex]\frac{1 mol HNO_3}{63,01 g }[/tex] = 27,04 moles

Liters of solution:

2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex] = 1,96 L of solution

Molarity:

[tex]\frac{27,04 mol}{1,96 L}[/tex] = 13,8 M

Final answer:

The composition of the solution can be expressed as 63.02 wt% HNO3, 54.28 lb HNO3/ft3, and a molarity of 14.2 M at 20 °C.

Explanation:

To express the composition of a water solution with 1.704 kg of HNO3 per kg of H2O and a specific gravity of 1.382 at 20 °C, we can calculate the following:

Weight Percent HNO3

Weight percent (wt%) is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100. For 1.704 kg HNO3 in 1 kg of water, the total mass of the solution is 1.704 kg + 1 kg = 2.704 kg. The weight percent HNO3 is then ((1.704 kg) / (2.704 kg)) × 100 = 63.02 wt%.

Pounds HNO3 per Cubic Foot of Solution at 20 °C

Firstly, we need to convert the specific gravity to density in g/mL: 1.382 (Specific Gravity) × 1.000 g/mL (density of water) = 1.382 g/mL. To find the density in lb/ft3, we multiply by 62.43 lb/ft3 which is the conversion factor from g/mL to lb/ft3. Then, multiply the density of the solution by the weight percent of HNO3 to find pounds HNO3 per cubic foot of solution: (1.382 g/mL) × (62.43 lb/ft3) × (63.02%) = 54.28 lb HNO3/ft3.

Molarity of HNO3 at 20 °C

Using the density of the concentrated HNO3 solution, 1.42 g/mL, and the given formula for molarity which is Molarity = [(%) (d)/ (Mw)] × 10, we substitute the values: Molarity = [(63.02) (1.42 g/mL) / (63.01)] × 10 = 14.2 M.

Explanation:

The given data is as follows.

          [tex]\Delta_{r} G[/tex] = -16.7 kJ/mol = [tex]-16.7 \times 10^{3}[/tex],       T = 298 K

          R = 8.314 J/mol K,       [tex]K_{eq}[/tex] = ?

Relation between [tex]\Delta_{r} G[/tex] and [tex]K_{eq}[/tex] is as follows.

                [tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]

Hence, putting the values into the above equation as follows.

                [tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]

                        [tex]-16.7 \times 10^{3} J/mol[/tex] = [tex]-8.314 J/mol K \times 298 K ln K_{eq}[/tex]

                     [tex]ln K_{eq}[/tex] = [tex]\frac{-16.7 \times 10^{3} J/mol}{-8.314 J/mol K \times 298 K}[/tex]    

                                         = 6.740

                            [tex]K_{eq}[/tex] = antilog (6.740)

                                            = 846

Thus, we can conclude that [tex]K_{eq}[/tex] for given values is 846.

Answer:

Unit conversion between Rankine and Kelvin is linear.

Unit conversion between degree Celsius and degree Fahrenheit is linear.

Explanation:

Relation between rankine and Kelvin is

[tex]R\ =\ \dfrac{9}{5}\ K[/tex]

So, the plot between Rankine and Kelvin is a straight line with zero intercept and has a slope having value [tex]\dfrac{9}{5}[/tex].

Relation between degrees Celsius and degrees Fahrenheit is given by

[tex]^{\circ}C\ =\ \dfrac{5}{9}\ (^{\circ}F\ -\ 32)[/tex]

So, the plot between degrees Celsius and degrees Fahrenheit is a straight line with slope [tex]\dfrac{5}{9}[/tex] and negative intercept of [tex]\dfrac{160}{9}[/tex].

Explanation:

1)

2)

Element are those substance which can not be split into simpler substance.they are made up of single atom.

Atom is the structural unit of the matter and smallest component of an element.

Molecules are group of atoms bonded together. The atom boned can be of more than one type.

Compounds are group of atoms of different elements bonded together.

3) Oxygen atoms in one molecule of following compounds:

a) [tex]NO_3[/tex]

1 × 3 = 3

There 3 oxygen atoms in the 1 molecule of [tex]NO_3[/tex].

b) [tex]Al(OH)_3[/tex]

3 × 1 = 3

There 3 oxygen atoms in the 1 molecule of [tex]Al(OH)_3[/tex].

c) [tex]Ca(NO_3)_2[/tex]

2 × 3 = 6

There 6 oxygen atoms in the 1 molecule of [tex]Ca(NO_3)_2[/tex].

d) [tex]Ba(OCN)_2[/tex]

2 × 1 = 2

There 6 oxygen atoms in the 1 molecule of [tex]Ba(OCN)_2[/tex].

Answer: The new volume of the gas will be 6.50 L

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=0.70atm\\V_1=3.8L\\P_2=0.409atm\\V_2=?L[/tex]

Putting values in above equation, we get:

[tex]0.70atm\times 3.8L=0.409atm\times V_2\\\\V_2=6.50L[/tex]

Hence, the new volume of the gas will be 6.50 L

Final answer:

To find the mass of the substance, multiply the given density (1.1 g/mL) by the given volume (12 mL) to get a mass of 13.2 grams, with the answer rounded to two significant figures.

Explanation:

To calculate the mass of a substance, you can use the formula mass = density × volume. Given a density of 1.1 g/mL and a volume of 12 mL, you simply multiply these two values together to find the mass.

Mass = 1.1 g/mL × 12 mL = 13.2 g

The density has two significant figures and the volume has two as well, so our final answer will be reported with two significant figures, hence 13.2 grams is the mass.

Answer:

They hydrogen bonds exist at the covalents molecules, as water where the 0 is an atom electronegative. The hydrogen bonds are formed because the hydrogen  gives its electron to oxygen (d-), that's why we say, there is a dipole at the molecule. As the dipole has been formed, the hydrogen who gave the electron to oxygen will try to get another electro from other molecules. That is  how the hydrogen bonds are formed.

Explanation:

Answer: The volume of water required is 398 mL

Explanation:

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

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

We are given:

Mass of solute (manganese (II) nitrate tetrahydrate) = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Molarity of solution = 0.16 M

Putting values in above equation, we get:

[tex]0.16M=\frac{16g\times 1000}{251g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=398mL[/tex]

Hence, the volume of water required is 398 mL

Answer:

24.76 kJ/g

844.8 kJ/mol

Explanation:

The heat produced by the burning of the solid magnesium is equal to the heat absorbed by the calorimeter, which causes its temperature to increase.

First, we calculate the heat absorbed by the calorimeter, where C is the heat capacity and Δt is the temperature change.

Q = CΔt = (3024 J·°C⁻¹)(1.126 °C) = 3405 J

This is the same amount of heat that was produced by burning the magnesium

Now we can calculate the heat produced per gram of magnesium:

(3405 J)(1 kJ/1000 J) / (0.1375 g) = 24.76 kJ/g

We can convert grams to moles using the atomic weight of Mg (24.305 g/mol).

(24.76 kJ/g)(24.305 g/mol) = 844.8 kJ/mol

Answer:

NH₃ is the limiting reagent

Explanation:

Mass of ammonia = 132.0kg

Mass of carbon dioxide = 211.4kg

Mass of urea = 172.7kg

Unknown:

Limiting reagent = ?

Solution

 First, we write the balanced stoichiometeric equation:

     2NH₃ + CO₂ → CH₄N₂O + H₂O

The reactant that is present in short supply determines the amount of product that is formed in a reaction. This reactant is called the limiting reagent.

    To establish the limiting reagent, we need to go find out what is happening at the start of the reaction:

Convert the masses of the reactants to moles.

Number of moles of NH₃ = [tex]\frac{mass}{molar mass}[/tex]

   Molar mass of NH₃ = 14 + (3x1) = 17g/mol

   Number of moles of NH₃ =  [tex]\frac{132}{17}[/tex] = 7.765mole

Number of moles of CO₂ =  [tex]\frac{mass}{molar mass}[/tex]

    Molar mass of CO₂ = 12 + (2 x 16) = 44g/mol

     Number of moles of CO₂ =  [tex]\frac{211.4}{44}[/tex] = 4.805mole

From the reaction equation:

   2 moles of NH₃ reacted with 1 mole of CO₂

so 7.765 mole of NH₃ will require [tex]\frac{7.765}{2}[/tex]mole, 3.883 of CO₂

But we are given 4.805mole of CO₂.

Therefore, CO₂ gas is in excess and NH₃ is the limiting reagent.

The limiting reactant for the synthesis of CH₄N₂O given the data is ammonia, NH₃

2NH₃ + CO₂ —> CH₄N₂O + H₂O

Molar mass of NH₃ = 17 g/mole

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g = 0.034 Kg

Molar mass of CO₂ = 44 g/mole

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g = 0.044 Kg

SUMMARY

From the balanced equation above,

0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂

From the balanced equation above,

0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂

Therefore,

132 Kg of NH₃ will react with = (132 × 0.044) / 0.034 = 170.8 of Kg of CO₂

From the calculation made above, we can see that only 170.8 of Kg of CO₂ out of 172.7 Kg given is needed to react completely with 132 Kg of NH₃.

Thus, NH₃ is the limiting reactant

Learn more about stoichiometry:

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Explanation:

(a)  A number which is dimensionless and provides us an estimate of the degree of conversion which can be achieved in CSTR, that is, continuous stirred tank reactor is known as Damkohler number.

This number is denoted as Da.

Mathematically,     Da = [tex]\frac{\text{reaction rate}}{\text{convection rate}}[/tex]

             Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]

Now, for first order system, Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]

                        = [tex]\frac{k \times CA_{o} \times V}{v \times CA_{o}}[/tex] = Tk

where,      rA = rate of reaction

                V = volume of reactor

                [tex]Fa_{o}[/tex] = molar flow rate of component A

                 k = rate constant

               [tex]CA_{o}[/tex] = initial concentration of A

                  v = volumetric flow rate of A

                  T = residence time

(b)   Since, from a given Damkohler number we can figure out the possible conversion of CSTR, that is, continuous stirred tank reactor.

So, if we have a low Damkohler number then the system will give us a less conversion formula. As the conversion is as follows.

                       X = [tex]\frac{Da}{Da + 1}[/tex]

Hence, we can conclude that [tex]Da \leq 0.1[/tex] will give less than 10% conversion as calculated by using above formula.

Final answer:

The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. This means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.

Explanation:

The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. It is defined as the ratio of the characteristic time for a chemical reaction to occur to the characteristic time for a system to transport material through itself.

A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. In practical terms, this means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.

For example, if we have a packed bed reactor (a reactor where reactants flow through a bed of solid catalyst particles), a low Damkohler number implies that the reaction is so fast that the reactants are fully converted into products even before they can flow through the bed, resulting in high conversion levels.

Answer:

Mole fraction of ethanol is 0.363.

Mole fraction of methanol is 0.387.

Explanation:

Mole fraction of water =[tex]\chi_1=0.250[/tex]

Mole fraction of ethanol =[tex]\chi_2[/tex]

Mole fraction of methanol = [tex]\chi_3[/tex]

[tex]\chi_1+\chi_2+\chi_3=1[/tex]

[tex]\chi_2+\chi_3=1-\chi_1=0.750[/tex]

[tex]\chi_2+\chi_3=0.750[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]

[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]

[tex]n_1+n_2+n_3=1[/tex]

Moles of water = [tex]n_1=0.250 mol[/tex]

Moles of ethanol= [tex]n_2[/tex]

Moles of methanol= [tex]n_3[/tex]

[tex]n_2+n_3=0.750 mol[/tex]

Mass of the mixture = M

Mass of water, [tex]m_1[/tex]

[tex]=n_1\times 18.0 g/mol=0.250 mol\times 18.0 g/mol=4.5 g[/tex]

Fraction of water by mass = 0.134

[tex]\frac{n_1\times 18.0 g/mol}{M}=0.134[/tex]

[tex]M=\frac{0.250 mol\times 18.0 g/mol}{0.134}=33.58 g[/tex]

Mass of ethanol = [tex]m_2[/tex]

Mass of methanol = [tex]m_3[/tex]

[tex]m_1+m_2+m_3=M[/tex]

[tex]4.5 g+m_2+m_3=33.58 g[/tex]

[tex]m_2+m_3=29.08 g[/tex]..[1]

[tex]\frac{m_2}{46.0 g/mol}+\frac{m_3}{32.0 g/mol}=0.750 mol[/tex]

[tex]16m_1+23m_3=552[/tex]..[2]

On solving [1] and [2]:

[tex]m_2 = 16.70, m_3= 12.38 g[/tex]

Mole fraction of ethanol =[tex]chi_2[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]

[tex]=\frac{\frac{16.70 g}{46.0 g/mol}}{1 mol}= 0.363[/tex]

Mole fraction of methanol = [tex]chi_3[/tex]

[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]

[tex]=\frac{\frac{12.38 g}{32.0 g/mol}}{1 mol}= 0.387[/tex]

Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

This means that in 100 mL of solution 12.9  L of alcohol is present.

Volume of alcohol = v = 12.9 L

Mass of the ethanol = m

Density of the ethanol ,d= [tex]0.789 g/cm^3=0.789 g/mL[/tex]

[tex]1 cm^3=1 mL[/tex]

[tex]m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g[/tex]

Mass of water = M

Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL

Density of water = D=1.00 g/mL

[tex]M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g[/tex]

Mass percent

[tex](w/w)\%=\frac{m}{m+M}\times 100[/tex]

[tex]\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%[/tex]

Molality :

[tex]m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}[/tex]

M = 87.1 g = 0.0871 kg (1 kg =1000 g)

[tex]=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}[/tex]

[tex]m=2.5403 mol/kg[/tex]

To calculate the concentration of ethanol in wine, multiply the volume percent of ethanol by the density of ethanol. Then divide by the mass of the solution and multiply by 100 to get the mass percent.

To calculate the concentration of ethanol in wine, we can use the mass percent formula. Mass percent is calculated by dividing the mass of the solute (ethanol) by the mass of the solution (wine), and multiplying by 100. The mass of ethanol can be found by multiplying the volume percent of ethanol (12.9%) by the density of ethanol (0.789 g/cm³). The density of wine is typically close to 1 g/cm³. So, to find the concentration of ethanol in terms of mass percent, we can follow these steps:

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Answer:

16-fold dilution.

Explanation:

A serial dilution is any dilution where the concentration decrease by the same quantity in each successive step. So, for a 2-fold dilution, the concentration decrease 1/2, it means that if we have a sample with 10 M of concentration, after a 2-fold dilution it will be 5 M. For the next step it will be 1/2 of 5= 2.5 M, and successively.

Then, we just multiply the factor for each dilution. After 4 serial dilutions:

1/2 x 1/2 x 1/2 x 1/2 = 1/16

So, it would be a 16-fold dilution in the end.

Answer:

a. The mothballs gradually vaporize in a closet. Physical change

b. Hydrofluoric acid attacks glass, and is used to etch calibrations marks on glass laboratory utensils. Chemical change

c. A French chef making a sauce with brandy is able to burn of the alcohol from the brandy leaving just the brandy flavoring. Chemical change

d. Chemistry majors sometimes et holes in the cotton jeans they wear to lab, because of acid spills. Chemical change

e. A piece of egg boiled in water for 20 minutes. Chemical change

Explanation:

In a physical change, there is no change in the chemical composition of the substance, the change is only on the physical properties. For example, state changes (a).

In a chemical change instead, usually there is a combination of two or more substances that combine to form a new one different from the originals. For example, chemical reactions like combustion (c) or protein denaturalization (e). The attacking of hydrofluoric acid to glass (b) or the acid pill attacking cotton are chemical reactions too (d).

Physical change or a chemical change are given as:

a. The moth balls gradually vaporize in a closet -physical change.

b. Hydrofluoric acid attacks glass and is used to etch calibration marks on glass laboratory utensils-chemical change.

c. A French chef making a sauce with brandy is able to burn off the alcohol from the brandy, leaving just the brandy flavoring- a chemical change.

d. Chemistry majors sometimes etch holes in the cotton jeans they wear to the lab because of acid spills- physical change.

A piece of egg boiled in water for 20 minutes-  chemical change.

a. In a closet, moth balls progressively vaporise: This represents a physical transformation. The moth balls sublimate, going straight from a solid state to a gaseous state, undergoing a phase change from a solid to a gas.

b. Glass is attacked by hydrofluoric acid, which is also used to etch calibration marks on glass laboratory utensils: This refers to a chemical transformation. The glass and hydrofluoric acid interact, causing a chemical reaction that etches the surface of the glass.

c. A French cook who uses brandy in a sauce can burn out the alcohol, leaving only the brandy flavouring: This explains a chemistry alteration. Alcohol is burned off through a chemical reaction, specifically alcohol combustion, which transforms ethanol (alcohol) into carbon dioxide and water vapour.

d. Because of acid spills, chemistry majors occasionally etch holes in the cotton pants they wear to the lab: This sentence depicts a physical alteration. The fibres in the jeans melt or break down as a result of acid spills, creating holes.

e. A chemical change is described by a piece of egg that was boiled in water for 20 minutes. The denaturation of proteins and the coagulation of the egg white are two distinct chemical reactions that occur inside the egg as a result of boiling.

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Answer:

The 8-hour TWA exposure for the employee is 101 ppm and it exceeds the OSHA PEL of 100 ppm for ethylbenzene.

Explanation:

The TWA for 8 hours is calculated by the sum of airbone concentrations multiplied by the time it has been exposed to that period. The total is divided by 8 which refers to the 8 hours total the employee has been exposed.

TWA = (125x2+88x2+112x2.5+70x1.5)/8.

The OSHA PEL is a known number for every compound and it can be find in PEL tables. In the case of ethylbenezene, it is 100 ppm.

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}[/tex]

where,

[tex]\Delta S[/tex] = change in molar entropy

n = number of moles = 1.0 mol

[tex]T_f[/tex] = final temperature = 300 K

[tex]T_i[/tex] = initial temperature = 273 K

[tex]C_{p,m}[/tex] = heat capacity of chloroform = [tex]91.47+7.5\times 10^{-2}(T/K)[/tex]

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

[tex]\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}[/tex]

[tex]\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}[/tex]

[tex]\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)[/tex]

[tex]\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)[/tex]

[tex]\Delta S=8.626+2.025[/tex]

[tex]\Delta S=10.651J/K.mol[/tex]

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

Answer: Molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 0.1015 M

[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex]solution = 200 ml

[tex]M_2[/tex] = molarity of dilute [tex]HNO_3[/tex] solution = ?

[tex]V_2[/tex] = volume of  dilute [tex]HNO_3[/tex]  solution = (2300 +200 )ml = 2500 ml

Putting in the values we get:  

[tex]0.1015M\times 200=M_2\times 2500[/tex]

[tex]M_2=8.12\times 10^{-3}M[/tex]

Thus the molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]

Answer:

The correct order is: Fluorine; Oxygen; Nitrogen; Carbon; Hydrogen; Sodium.

Explanation:

The electronegativity in the periodic table increases to the right in a period and up in a group. We can figure out the electronegativity of each element according to its electron configuration:

Carbon: [He] [tex]2s^{2} 2p^{2}[/tex]

Fluorine: [He] [tex]2s^{2} 2p^{5}[/tex]

Hydrogen: [tex]1s^{1}[/tex]

Nitrogen: [He] [tex]2s^{2} 2p^{3}[/tex]

Oxygen: [He] [tex]2s^{2} 2p^{4}[/tex]

Sodium: [Ne] [tex]3s^{1} [/tex]

The period of a chemical element is given by the last energy level of electron configuration.

The group of a chemical element is given by the amount of electrons in the last energy level of electron configuration.

Therefore,

Carbon: Period 2 Group 4

Fluorine: Period 2 Group 7

Hydrogen: Period 1 Group 1

Nitrogen: Period 2 Group 5

Oxygen: Period 2 Group 6

Sodium: Period 3 Group 1

Electronegativity increases when the number of the group increases.

Electronegativity increases when the number of the period decreases.

In conclusion, Fluorine has the greatest number of group and Sodium has the lowest number of group (also it has a greater number of period than hydrogen, so it is less electronegative than hydrogen)

F>O>N>C>H>Na

Final answer:

To prepare a 100 mM acetic acid solution, calculate the moles needed, measure the glacial acetic acid based on its density and volume, and dilute with distilled water to the final volume. Mix well for homogeneity.

Explanation:

The question involves preparing a solution of acetic acid and relates to the subjects of molarity and solution preparation in chemistry. The solution is to be made to a specified concentration using a direct dilution from a more concentrated stock.

Steps to Prepare 100 mM Acetic Acid Solution

First, we calculate the number of moles of acetic acid required for 100 mL of a 100 mM solution using the molar mass of acetic acid (MW 60.05). Second, we use the density of glacial acetic acid (1.05 g/mL) and the volume needed (from the calculated moles and molarity) to determine how much of the glacial acetic acid to measure out. Finally, we add this measured amount of acid to a beaker and dilute to the mark with distilled water (ddH2O) to achieve the desired final concentration. Mixing with a stir bar ensures a homogeneous solution.

To calculate the mole fraction of acetic acid in a solution, we need the number of moles of acetic acid and the number of moles of water in the solution. This can be accomplished by converting the mass of each component to moles using their respective molar masses.

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

Answer:

0.9

Explanation:

The pka represents the force by which the molecules need to dissociate for the acids ,

Hence , lower the pka stronger will be the acid and that therefore will  dissociate completely and vice versa , for a weak acid higher the pka .

And in case of a base , it will be completely reversed , lower pKa , weaker base ,

and higher pKa , stronger base .

From the data of the question ,

0.9 is the lowest value of the pKa , hence , weakest base .