Radon-219 decays to radon-218 by releasing... a. a positron b.a muon O c. a neutron O d. an electron O e. a proton

Explanation:

The given data is as follows.

          [tex]\lambda[/tex] = 211 nm,          [tex]\sum = 6.17 \times 10^{3} mol/L/cm[/tex]

           l = 1 cm,             7% < Transmittance < 85%

Suppose the aqueous solution follows Lambert-Beer's law. Therefore,

                   Absorbance = [tex]-log \frac{\text{Percentage transmittance}}{100}[/tex]

Hence, for 7% transmittance the value of absorbance will be as follows.

                  Absorbance = [tex]-log \frac{7}{100}[/tex]

                           [tex]A_{1}[/tex] = 1.155

For 85% transmittance the value of absorbance will be as follows.

                 Absorbance = [tex]-log \frac{85}{100}[/tex]

                           [tex]A_{2}[/tex] = 0.07058

According to Lambert-Beer's law.

                  A = [tex]\sum \times l \times C[/tex]

where,       A = absorbance

                 [tex]\sum[/tex] = molar extinction coefficient

                 C = concentration

Therefore, concentration for 7% absorbance is as follows.

                    [tex]A_{1} = \sum \times l \times C_{1}[/tex]

                  [tex]C_{1}[/tex] = [tex]\frac{1.155}{6.17 \times 10^{3} \times 1}[/tex]

                                  = [tex]0.187 \times 10^{-3} mol/L[/tex]

                                  = 0.187 mmol/L

Concentration for 85% absorbance is as follows.

                   [tex]A_{2} = \sum \times l \times C_{2}[/tex]

                  [tex]C_{2}[/tex] = [tex]\frac{0.07058}{6.17 \times 10^{3} \times 1}[/tex]

                                  = [tex]0.01144 \times 10^{-3} mol/L[/tex]

                                  = 0.01144 mmol/L

Thus, we can conclude that linear range of phenol concentration is 0.01144 mmol/L to 0.187 mmol/L.

The linear range of phenol concentrations for the given conditions is from [tex]\( 1.14 \times 10^{-5} \) M to \( 1.87 \times 10^{-4} \) M[/tex].

To calculate the linear range of phenol concentrations, we need to use the Beer-Lambert law, which relates the absorbance (A) of a solution to its concentration (c), the molar absorptivity (µ), and the path length (l) of the cell:

[tex]\[ A = \varepsilon \cdot c \cdot l \][/tex]

The transmittance (T) is related to absorbance by the equation:

[tex]\[ T = 10^{-A} \][/tex]

Given that the molar absorptivity (µ) is [tex]\( 6.17 \times 10^3 \)[/tex] L/mol/cm and the path length (l) is 1 cm, we can rearrange the Beer-Lambert law to solve for concentration (c):

[tex]\[ c = \frac{A}{\varepsilon \cdot l} \][/tex]

First, we need to find the absorbance values corresponding to 85% and 7% transmittance:

For 85% transmittance:

[tex]\[ A = -\log(T) = -\log(0.85) \][/tex]

[tex]\[ A \approx 0.0706 \][/tex]

For 7% transmittance:

[tex]\[ A = -\log(T) = -\log(0.07) \][/tex]

[tex]\[ A \approx 1.1549 \][/tex]

Now we can calculate the concentration range:

For the lower limit of transmittance (7%):

[tex]\[ c_{min} = \frac{A_{min}}{\varepsilon \cdot l} = \frac{1.1549}{6.17 \times 10^3 \cdot 1} \] \[ c_{min} \approx 1.87 \times 10^{-4} \text{ M} \][/tex]

For the upper limit of transmittance (85%):

[tex]\[ c_{max} = \frac{A_{max}}{\varepsilon \cdot l} = \frac{0.0706}{6.17 \times 10^3 \cdot 1} \] \[ c_{max} \approx 1.14 \times 10^{-5} \text{ M} \][/tex]

Therefore, the linear range of phenol concentrations for the given conditions is from [tex]\( 1.14 \times 10^{-5} \) M to \( 1.87 \times 10^{-4} \) M[/tex].

Answer : The concentration of [tex]NH_4^+[/tex] ion is [tex]0.15E^1M[/tex]

Explanation :

First we have to calculate the moles of [tex](NH_4)_3PO_4[/tex].

[tex]\text{Moles of }(NH_4)_3PO_4=\text{Concentration of }(NH_4)_3PO_4\times \text{Volume of solution}=0.50M\times 0.06L=0.03mole[/tex]

The balanced chemical reaction will be:

[tex](NH_4)_3PO_4\rightleftharpoons 3NH_4^++PO_4^{3-}[/tex]

From the reaction we conclude that,

1 mole of [tex](NH_4)_3PO_4[/tex] dissociate to give 3 moles of [tex]NH_4^+[/tex] ion and 1 mole of [tex]PO_4^{3-}[/tex] ion

So,

0.03 mole of [tex](NH_4)_3PO_4[/tex] dissociate to give [tex]3\times 0.03=0.09[/tex] moles of [tex]NH_4^+[/tex] ion and 0.03 mole of [tex]PO_4^{3-}[/tex] ion

Now we have to calculate the concentration of [tex]NH_4^+[/tex] ion.

[tex]\text{Concentration of }NH_4^+=\frac{\text{Moles of }NH_4^+}{\text{Total volume}}[/tex]

[tex]\text{Concentration of }NH_4^+=\frac{0.09mole}{0.06L}=1.5M=0.15E^1M[/tex]

Therefore, the concentration of [tex]NH_4^+[/tex] ion is [tex]0.15E^1M[/tex]

Spontaneous reactions result in an increase in entropy, aligning with the second law of thermodynamics.

The true statement about the relationship between entropy and spontaneity is: Spontaneous reactions tend to lead to higher entropy. This concept is encapsulated by the second law of thermodynamics, which states that all spontaneous changes cause an increase in the entropy of the universe. The increase in entropy relates to how energy and matter become more spread out and disordered over time in a spontaneous process, such as a chemical reaction or heat flow between two objects.

Answer:

[tex]CaCl_2[/tex]

Explanation:

Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is [tex][Ar]4s^2[/tex].

Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is [tex][Ne]3s^23p^5[/tex].

Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Ca         Cl

2            1

Cross multiplying the valency, We get, [tex]CaCl_2[/tex]

Thus, the formula of calcium chloride is [tex]CaCl_2[/tex].

Calcium Chloride, with the formula CaCl2, is composed of one calcium (Ca2+) ion and two chloride (Cl-) ions to ensure electrical neutrality.

The compound Calcium Chloride is formed by the elements calcium (Ca) and Chloride (Cl). From the periodic table, we can see that calcium (Ca) forms +2 ions and Chloride (Cl) forms -1 ions. As compounds must be electrically neutral, the formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to balance out the charges. Thus, for every calcium ion, we need two chloride ions to balance out the charges. So the formula for this compound is CaCl2.

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Answer:

The number of mol corresponding to 2,20 g Cl2 is 0,062.

Explanation:

As we know that the molar mass of chlorine is 35.45 g/mol and in this case we have 2.20 grams of gas, if we divide the mass of the chlorine over the molar mass we will have the moles of our compound.

moles = mass / molar mass

moles = 2,20 g / 35,45 g/mol

moles = 0,062

Answer:

1.728 mol /(L*min)

Explanation:

Hello,

In the attached photo, you'll find the numerical procedure for your question.

- Take into account that the negative sign is eligible for reagents and positive for products.

Best regards!

Answer:

2 molecular ions are produced and they are charged negatively

Explanation:

If an ion loses an electron then its charge is +1, if it gains an electron then its charge is -1

3 molecular ions lost 2 electrons each, each ion is charged +2

Total charge = 3(+2) = +6

2 molecular ions gained 3 electrons each, each ion is charged -3

Total charge = 2(-3) = -6

1 molarcular ion gained 2 electrons, ion is charged -2

Net charge of the reactants is:

+6 -6 -2 = -2

The net charge of the reactants must be the same as the net charge of the products.

In the products you have 2 neutral molecules (with charge 0) and molecular ions with a charge of magnitude e.

Since the net charge in the reactants is negative, the net charge in the products is also negative

If the number of molecular ions in the products is x then its total charge is:

x*(-1) = -2

x = 2

Explanation:

An ionic bond is generally formed by a metal and a non-metal.

For example, lithium is an alkali metal with atomic number 3 and its electronic distribution is 2, 1.

And, chlorine is a non-metal with atomic number 17 and its electronic distribution is 2, 8, 7.

So, in order to complete their octet lithium needs to lose an electron and chlorine needs to gain an electron.

Hence, both of then on chemically combining together results in the formation of an ionic compound that is, lithium chloride (LiCl).

An ionic compound is formed by LiCl because lithium has donated its valence electron to the chlorine atom.  

For example, [tex]O_{2}[/tex] is a covalent compound as electrons are being shared by each oxygen atom.

Answer:

Osmotic pressure is: 2,01 atm

Explanation:

Osmotic pressure is the minimum pressure that you needs to be apply to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. The formula of osmotic pressure  is:

π = M×R×T

Where:

M is molar concentration of dissolved species (units of mol/L). 0,078M

R is ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹, ).

T is the temperature on the Kelvin scale. 41°C +273,15 = 314,15 K

Replacing you have:

π = 0,078M×0.08206 L atm mol⁻¹ K⁻¹×314,15 K

π = 2,01 atm

I hope it helps!

Answer:  The reaction order with respect to A is m

Explanation:

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

For the given reaction:

[tex]aA+bB\rightleftharpoons cC+dD[/tex]  

[tex]Rate=k[A]^m[B]^n[/tex]

In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction.

Hence, this is not considered as an elementary reaction.

Order with respect to A = m

Order with respect to B = n

Overall order = m+n

Thus order with respect to A is m.

In a rate law equation, the reaction order with respect to a reactant, for instance, A, is represented by the corresponding exponent in that equation. It defines how the rate of the chemical reaction alters with the concentration change of that reactant.

In a rate law equation, the reaction order with respect to a certain reactant, say, A, is represented by the exponent 'm' in the equation rate=k[A]m[B]n. This shows how the rate of the chemical reaction changes with the change in concentration of the reactant A. The higher 'm' is, the more sensitive the reaction rate is to the concentration of reactant A. It could be zero, indicating that the reaction rate is independent of the concentration of A; one, indicating a first-order dependence, and so on.

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Answer : The moles of Ne gas are 46.25 moles.

Explanation :

using ideal gas equation

[tex]PV=nRT[/tex]

where,

P = pressure of gas = 45 atm

V = volume of gas = 24.3 L

T = temperature of gas = [tex]15.0^oC=273+15.0=288K[/tex]

R = gas constant = 0.0821 L.atm/mole.K

n = moles of gas = ?

Now put all the given values in the ideal gas equation, we get:

[tex](45atm)\times (24.3L)=n\times (0.0821L.atm/mole.K)\times (288K)[/tex]

[tex]n=46.25mole[/tex]

Therefore, the moles of Ne gas are 46.25 moles.

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

Answer:

The final concentration is 0.140 M

Explanation:

n₁=M₁.V₁ (First equation)

n₁=1.50 M

V₁=55 mL

M₂=n₂/V₂ (Second Equation)

V₂=278 mL

n₁=n₂

M₂=M₁.(V₁/V₂) (Third Equation)

n₃=M₃.V₃ (Forth Equation)

V₃=139 mL

M₃=M₂

n₃=M₁.(V₁/V₂).V₃ (Forth Equation)

M₄=n₄/V₄ (Fifth Equation)

n₄=n₃

M₄=M₁.(V₁/V₂).(V₃/V₄) (Sixth equation)

V₄=V₃+155 mL (Seventh Equation)

M₄=M₁.(V₁/V₂).(V₃/(V₃+155 mL))

M₄=1.50 M . (55mL / 278 mL) . ((139mL)/(139mL+155mL))

M₄=1.50 M . (55mL / 278 mL) . (139mL/294mL)

M₄=0.140 M

Answer:

a) D = 33.44 Lbmol/h

⇒ B = 62.56 Lbmol/h

b) D = 16.848 Kmol/h

⇒ B = 28.152 Kmol/h

Explanation:

global balance:

∴ F = 100 Lbmol/h

balance per component:

A: 0.4*F = 0.9*D + 0.1*B = 0.4*100 = 40 Lbmol/h..............(2)

B: 0.6*F = 0.1*D + 0.9*B = 0.6*100 = 60 Lbmol/h..............(3)

from (2):

⇒ 0.9*D = 40 - 0.1*B

⇒ D = ( 40 - 0.1*B ) / 0.9............(4)

(4) in (3):

⇒ 0.1*((40-0.1*B)/0.9) + 0.9*B = 60

⇒ B = 62.56 Lbmol/h............(5)

(5) in (1):

⇒ D = 100 - B

⇒ D = 37.44 Lbmol/h

∴ Lbmol = 0.45 Kmol

⇒ B = 62.56 Lbmol/h * ( 0.45 Kmol/ Lbmol ) = 28.152 Kmol/h

⇒ D = 37.44 Lbmol/h * ( 0.45 Kmol/h ) = 16.848 Kmol/h

Answer:

a) Schmidt number

Explanation:

Prandtl number in heat transfer is analogues to Schmidt number in mass transfer.

Prandtl number in heat transfer is the ration of momentum diffusivity to the heat diffusivity.

[tex]P_r = \frac{\nu}{\alpha}[/tex]

Whereas, Schmidt number in mass transfer is the ratio of momentum diffusivity to the mass diffusivity.

[tex]S_c= \frac{\nu}{\nu_{AB}}[/tex]

Answer:

One urea molecule can make hydrogen bonding with 8 water molecules

Explanation:

Urea can form up to 4 hydrogen bonds with water molecules.

The maximum theoretical number of water molecules with which one urea molecule can hydrogen bond is 4.

This is because urea has 4 hydrogen bond donor sites (NH groups) and 2 hydrogen bond acceptor sites (carbonyl oxygen).

Each hydrogen bond requires one hydrogen bond donor and one hydrogen bond acceptor, so one urea molecule can form up to 4 hydrogen bonds with water molecules.

Answer:

True

Explanation:

*For polar and associated substances, methods based on four should be used  four or more parameters, like analytical equation of state

*The term "analytical equation of state" implies that the function

It contains powers of v not greater than four.

*Most expressions are of the cubic type and are grouped into

the so-called cubic equations of state.

*Cubic EoS calls are very popular in simulation of

processes due to its robustness and its simple extension to mixtures.

*They are based on the van der Waals state equation of more than

100 years.

Answer:

18840 lbf

Explanation:

Considering the physical equation:

[tex]F = PA[/tex]

It is possible to determine the force exerted by the piston if the maximum pressure developed ([tex]P_{max}[/tex]) and the area ([tex]A[/tex]) are previously known.

The area of the piston can be known through:

[tex]A = \pi *r^{2}[/tex]

Being [tex]r = \frac{d}{2} =  1~in[/tex]

Then

[tex]A = \pi * r^{2} = 3,14~in^{2}[/tex]

On the other hand, physical units of pressure in PSI are [tex]\frac{lbf}{in^{2}}[/tex], where the force is given in lbf.

Finally, the exerted force is:

[tex]F = 6000~PSI \cdot 3,14~in^{2} = 18840~lbf[/tex]

Answer:

86.71 g of glyoxylic acid monohydrate, and 348.56 g of sodium glyoxylate monohydrate.

Explanation:

In order to solve this problem we need to use the Henderson–Hasselbalch equation:

pH= pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

In this case [A⁻] is the concentration of sodium glyoxylate, and [HA] is the concentration of glyoxylic acid.

Using the Henderson–Hasselbalch (H-H) equation, the given pH and pka we can determine a relationship between [A⁻] and [HA]:

3.85 = 3.34 +  [tex]log\frac[{A^{-}] }{[HA]}[/tex]

0.51 = [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

[tex]10^{0.51} =\frac{[A^{-}] }{[HA]} \\3.24=\frac{[A^{-} ]}{[HA]}[/tex]

Also from the problem, we know that

[A⁻] + [HA] = 1.60 M

We rearrange that equation to

[A⁻] = 1.60 M - [HA]

And replace the value of  [A⁻] in the H-H equation and solve for [HA]:

[tex]3.24=\frac{1.60M-[HA]}{[HA]}\\3.24*[HA]=1.60-[HA]\\4.24*[HA]=1.60\\0.377 M = [HA][/tex]

We substract 0.377 M to 1.60 M in order to calculate [A⁻].

1.60 M - 0.377 M= 1.223 M = [A⁻]

Lastly we calculate the mass of each reagent, using the concentration, volume and molecular weights:

2.50 L * 0.377 M * 92 g/mol = 86.71 g glyoxylic acid monohydrate.

2.50 L * 1.223 M * 114 g/mol = 348.56 g sodium glyoxylate monohydrate.

Why does the problem ask that we use the monohydrate forms? Because that's the available reagent in the laboratory.

To prepare a 1.60 M buffer at pH 3.85 using glyoxylic acid and sodium glyoxylate, you will need 60.95 g of glyoxylic acid and 296.93 g of sodium glyoxylate.

To calculate the grams of glyoxylic acid and sodium glyoxylate needed to prepare a 1.60 M buffer at pH 3.85, we first need to determine the mole ratios of the components in the buffer. The Henderson-Hasselbalch equation can be used to calculate these ratios:

pH = pKa + log([A-]/[HA])

In this case, the pKa of glyoxylic acid is given as 3.34. We can rearrange the equation to solve for the ratio of [A-]/[HA]:

log([A-]/[HA]) = pH - pKa = 3.85 - 3.34 = 0.51

10^0.51 = [A-]/[HA]

[A-]/[HA] = 3.295

Since we know the total volume of the buffer is 2.50 L and the molar concentration is 1.60 M, we can use the mole ratio to calculate the moles of glyoxylic acid and sodium glyoxylate needed:

[HA] = (1.60 M) / (1 + 3.295) = 0.327 M

[A-] = 3.295 * [HA] = 3.295 * 0.327 M = 1.076 M

Now we can use the molar masses of glyoxylic acid and sodium glyoxylate to calculate the mass needed:

Mass of glyoxylic acid = (0.327 M) * (2.50 L) * (74.04 g/mol) =  60.95 g

Mass of sodium glyoxylate = (1.076 M) * (2.50 L) * (110.07 g/mol) =  296.93 g

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Answer:

Risk management

Explanation:

Risk communication is exchange of the advice, real-time information and the opinions between the experts and the people which are facing threats to their economic, health, or well-being.

Risk assessment is overall process where the hazards and the risk factors which have potential to cause the harm is identified.

Risk perception is subjective judgement which people/ experts make about characteristics and the severity of the risk.

Risk management is evaluation, identification and prioritization of the risks followed by the coordinated and the economical application of the resources to monitor, minimize and control probability of the unfortunate events.

Thus, Option D is correct because in risk management, a Quantitative Solution is given to decrease risk level like to recycle the plastic.

Answer:

Risk management.

Explanation:

Hello

In this case, it is convenient for us to define the four types of risks regarding to manufacturing processes as follows:

- Risk communication is an interactive process in which the just the exchange of information and opinion about a particular risk among risk specialists and other interested organisms. This is not the answer as it implies a requirement not the aforesaid exchange of information.

- Risk assessment useful to describe the widespread procedure to identify hazards and risk factors and their consequences and to both analyze and evaluate the risk regarding to such hazard. This is not the answer as a requirement is not procedure indeed.

- Risk perception involves how one both thinks and feels about the faced risks. This is not the answer as a requirement is not something coming from a thought or feeling.

- Risk management accounts for the identification, evaluation, and prioritization of risks followed by a coordinated and economical application of resources to minimize, monitor, and control the probability or impact of harmful facts or to maximize the realization of opportunities. This is the answer as a requirement is eligible for the identification and a subsequent application of a resource as the industries must recycle the specified amount via varied industrial processes (recirculation, storage and others).

Best regards.

Answer:

The statement is true

Explanation:

Any physical quantity has a unique dimensional representation and when we change the dimensions of any physical quantity it looses it's original form. while as unit conversion is applicable to physical quantities of same nature as speed can be converted from kilometers per hour to miles per hour or meters per second, changing the units of any physical quantity into different unit's but of same dimensional formula only changes a magnitude of the physical quantity as in the above case the speed will be speed no matter whatever be the units.

since [tex]\frac{m^3}{sec}[/tex] are not the units of velocity but of volumetric rate of flow  we cannot convert this quantity into velocity by only unit conversion

Answer:

The increasing order of Bronsted acids is:

(b) < (c) < (e) < (d) < (a)

Explanation:

A Bronsted acid is a substance who donated protons.

[tex]AH + H_{2}O[/tex] ⇄ [tex]A^{-} + H_{3}O^{+}[/tex]

The acidity of a compound is determined by the acidity constant and every factor that stabilizes the product of the acid reaction (the anion) will contribute to increase the acidity of the compound.

First, it is necesary to identify the donor group:

(a) is -SO3H

(b) is -OH

(c), (d) and (e) is -COOH

For these groups the order of acidity is

-SO3H > -COOH > -OH

The strongest is the group -SO3H because it comes from a strong inorganic acid (H2SO4). Second is the group -COOH, carboxylic acids are the strongest organic acids because of the resonance structure of the anion at acid equilibrium, this equilibrium shifts the reaction to products favoring deprotonation. Third is the group of -OH because alcohols are weak acids.

Further, between (c), (d) and (e) it is necessary to analyze the molecule substituents. The -Cl substituent is an electronegative group that stabilizes the anion of the equilibrium because of its inductive effect. Thus, the molecule with the -Cl near to de acidic group will be more acidic. Therefore the order of acidity between these three compounds will be: (d) > (e) > (c)

Summarised all, the final answer is (b) < (c) < (e) < (d) < (a)

In increasing order of acidity, the compounds are b. CH3CH OH, a. CH3CH2SO3H, c. CH3CH2COOH, d. CH3CHCICOOH, e. CICH CH2COOH.

The acidity of a compound is determined by the stability of its conjugate base. The more stable the conjugate base, the stronger the acid. In this case, we can rank the compounds in increasing acidity as follows:

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Answer:

Spring constant: [tex]\rm 447\; N \cdot m^{-1}[/tex].

Work done: [tex]\rm 0.0438\; J[/tex].

Explanation:

Convert all values to SI units.

Assume that the spring-mass system is vertical and is placed on the surface of the earth. The gravitational acceleration constant will be equal to [tex]\rm 9.81\; N\cdot kg^{-1}[/tex].

Gravitational pull on the weight:

[tex]W = m\cdot g = \rm 0.638\; kg \times 9.81\; N\cdot kg^{-1} = 6.25878\; N[/tex].

That's also the size of the force on the spring. [tex]F = \rm 6.25878\; N[/tex].

The spring constant is the size of the force required to deshape the spring (by stretching, in this case) by unit length.

[tex]\displaystyle k_{\text{spring}} = \frac{F}{\Delta x} = \rm \frac{6.25878\; N}{0.014\; m} = 447.056\; N\cdot m^{-1}[/tex].

Assume that there's no energy loss in this process. The work done on the spring is the same as the elastic potential energy that it gains:

[tex]\begin{aligned} &\text{EPE} \\=& \frac{1}{2}k\cdot x^{2} \\=&\rm \frac{1}{2} \times 447.056\; N\cdot m^{-1} \times (0.014\; m)^{2}\\ =& \rm 0.0438\; J\end{aligned}[/tex].

Answer: The moles of hydrochloric acid is [tex]1.0346\times 10^{-4}\mu mol[/tex]

Explanation:

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

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Or,

[tex]\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}[/tex]

We are given:

Molarity of solution = 0.5173 M

Volume of solution = [tex]200\mu L[/tex]

Putting values in above equation, we get:

[tex]0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol[/tex]

Hence, the moles of hydrochloric acid is [tex]1.0346\times 10^{-4}\mu mol[/tex]

Answer:

2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.

Explanation:

We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.

Now we devise the following reasoning:

If          18 g of phthalic acid are dissolved in 100 mL of water at 99 °C

Then   0.5 g of phthalic acid are dissolved in X mL of water at 99 °C

X = (0.5 × 100) / 18 = 2.77 mL of water

Final answer:

To dissolve 500 mg of phthalic acid, a student could use a minimum of approximately 2.78 mL of boiling water, considering the solubility of 18 g per 100 mL at 99 ℃.

Explanation:

The question asks for the smallest volume of boiling water (at 99 ℃) needed to dissolve 500 mg of phthalic acid, with the solubility of phthalic acid provided as 0.54 g at 14 ℃ and 18 g at 99 ℃. To find this volume, we start by converting the mass of phthalic acid to be dissolved (500 mg or 0.5 g) to grams because the solubility is given in grams. Knowing the solubility of phthalic acid at 99 ℃ is 18 g per 100 mL, we calculate the volume required for 0.5 g as follows:

Solubility equation: (Volume of water) x (Solubility of compound in g/100mL) = Mass of compound to be dissolved

Thus, (Volume of water) x (18 g/100 mL) = 0.5 g

Rearranging for Volume of water gives: Volume of water = (0.5 g) / (18 g/100 mL) = 2.78 mL

Therefore, the student could use a minimum of approximately 2.78 mL of boiling water to dissolve 500 mg of phthalic acid.

Answer:

[CH₃COO⁻]  [H⁺] pH

0,1 M  0,0025 M  6,30

0,1 M  0,005 M  6,02

0,1 M  0,01 M  5,70

0,1 M  0,05 M  4,74

0,01 M  0,0025 M  5,22

0,01 M  0,005 M  4,75

0,01 M  0,01 M  3,38

0,01 M  0,05 M  1,40

Explanation:

The equilibrium of sodium acetate is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵

Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.

For  [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:

[CH₃COO⁻] = 0,1 M - x

[H⁺] = 0,0025 M - x

[CH₃COOH] = x

The expression for this equilibrium is:

Ka = [tex]\frac{[CH3COO^-] [H^+] }{[CH3COOH]}[/tex]

Replacing:

1,8x10⁻⁵ = [tex]\frac{[0,1-x] [0,0025-x] }{[x]}[/tex]

Thus:

0 = x²-0,102518x +2,5x10⁻⁴

Solving:

x = 0,100 ⇒ No physical sense

x = 0,0024995

Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷

pH = - log [H⁺] = 6,30

Following the same procedure changing both  [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:

[CH₃COO⁻]  [H⁺] pH

0,1 M  0,0025 M  6,30

0,1 M  0,005 M  6,02

0,1 M  0,01 M  5,70

0,1 M  0,05 M  4,74

0,01 M  0,0025 M  5,22

0,01 M  0,005 M  4,75

0,01 M  0,01 M  3,38

0,01 M  0,05 M  1,40

I hope it helps!

The pH values for the [tex]0.1 \ M[/tex] sodium acetate solution when [tex]0.0025 \ M[/tex], [tex]0.005\ M, 0.01\ M,[/tex] and [tex]0.05\ M[/tex] [tex]HCl[/tex] are added are approximately [tex]6.35, 6.03, 5.70,[/tex] and [tex]4.75[/tex] respectively. For the 0.01 \ M sodium acetate solution, the pH values are approximately [tex]5.23, 4.75, 2,[/tex] and [tex]1.30[/tex] respectively.

The Henderson-Hasselbalch equation is given by:

[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \][/tex]

For acetic acid ([tex]CH_3COOH[/tex]) and its conjugate base acetate ([tex]CH_3COO^-[/tex]), the [tex]pKa[/tex] is approximately [tex]4.75[/tex] at [tex]25^\circ C[/tex]

Let's calculate the pH for each case:

1. For the [tex]0.1 \ M[/tex] sodium acetate solution:

a. When [tex]0.0025 \ M \ HCl[/tex] is added:

The concentration of acetate ions remains [tex]0.1 \ M[/tex], while the concentration of acetic acid formed by the reaction of [tex]HCl[/tex] with acetate ions is [tex]0.0025 \ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.1}{0.0025} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (40) \][/tex]

[tex]\[ \text{pH} = 4.75 + 1.60 \][/tex]

[tex]\[ \text{pH} = 6.35 \][/tex]

b. When [tex]0.005\ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.095 \ M[/tex], and the concentration of acetic acid is [tex]0.005\ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.095}{0.005} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (19) \][/tex]

[tex]\[ \text{pH} = 4.75 + 1.28 \][/tex]

[tex]\[ \text{pH} = 6.03 \][/tex]

c. When [tex]0.01 \ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.09 \ M[/tex], and the concentration of acetic acid is[tex]0.01\ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.09}{0.01} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (9) \][/tex]

[tex]\[ \text{pH} = 4.75 + 0.95 \][/tex]

[tex]\[ \text{pH} = 5.70 \][/tex]

d. When [tex]0.05 \ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.05 \ M[/tex], and the concentration of acetic acid is [tex]0.05 \ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.05}{0.05} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (1) \][/tex]

[tex]\[ \text{pH} = 4.75 \][/tex]

2. For the [tex]0.01 \ M[/tex] sodium acetate solution:

a. When [tex]0.0025\ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.0075 \ M[/tex], and the concentration of acetic acid is [tex]0.0025 \ M[/tex].

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.0075}{0.0025} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (3) \][/tex]

[tex]\[ \text{pH} = 4.75 + 0.48 \][/tex]

[tex]\[ \text{pH} = 5.23 \][/tex]

b. When [tex]0.005 \ M \ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0.005 \ M[/tex], and the concentration of acetic acid is[tex]0.005 \ M[/tex]

[tex]\[ \text{pH} = 4.75 + \log \left( \frac{0.005}{0.005} \right) \][/tex]

[tex]\[ \text{pH} = 4.75 + \log (1) \][/tex]

[tex]\[ \text{pH} = 4.75 \][/tex]

c. When [tex]0.01 \ M\ HCl[/tex] is added:

The concentration of acetate ions is now [tex]0 M[/tex] (fully reacted), and the concentration of acetic acid is [tex]0.01 \ M[/tex]. The solution is no longer buffered, and the pH is that of a [tex]0.01 \ M[/tex] acetic acid solution.

[tex]\[ \text{pH} = -\log (0.01) \][/tex]

[tex]\[ \text{pH} = 2 \][/tex]

d. When [tex]0.05 \ M \ HCl[/tex] is added:

The solution is overwhelmed by the strong acid, and the pH is determined by the excess [tex]HCl[/tex]. The pH is approximately that of a [tex]0.05 M HCl[/tex] solution.

[tex]\[ \text{pH} = -\log (0.05) \][/tex]

[tex]\[ \text{pH} = 1.30 \][/tex]

Explanation:

It is given that peak (maximum) emission of star is 300 nm.

And, it is known that according to Wein's displacement relation between wavelength and temperature is as follows.

                    [tex]\lambda_{max} \times T[/tex] = 2898 micrometer.K

Hence, putting the given values into the above formula as follows.

               [tex]\lambda_{max} \times T[/tex] = 2898 micrometer.K

               [tex]300 \times 10^{-9} m \times T[/tex] = 2898 \times 10^{-6} meter.K

                 T = [tex]\frac{2898000}{300}[/tex] K

                     = 9660 K

Thus, we can conclude that the estimated surface temperature of star is 9660 K.

Using Wien's Law, the estimated surface temperature of a star with a peak emission at 300 nm is approximately 9660 Kelvin, considering a 2% precision.

To estimate the surface temperature of a star based on its peak emission wavelength, we use Wien's Law, which is a principle in physics that describes the relationship between the peak wavelength of emission from a blackbody and its temperature. Wien's Law can be written as:

λmax * T = b

where λmax is the peak wavelength in meters, T is the temperature in Kelvin, and b is Wien's displacement constant, approximately 2.897 × 10-3 m·K.

To find the temperature, we rearrange the formula to solve for T:

T = b / λmax

Converting the peak wavelength from nanometers to meters (300 nm = 3 × 10-7 m), we can calculate the temperature of the star as follows:

T = 2.897 × 10-3 m·K / 3 × 10-7 m
= 9660 Kelvin

Allowing for a 2% precision, the estimated temperature range of the star is approximately 9660K ± 2%.

The molecular size of succinic acid is larger than that of the decomposed product which is evident from the production of a gas and a solid with a different melting point upon heating.

In the process described where succinic acid is heated and gives off a gas, then leaves behind a solid that melts at a different temperature, suggests a chemical reaction has occurred, possibly a decomposition reaction. The succinic acid molecules are likely larger than the molecules of the white solid produced because decomposition breaks larger molecules into smaller ones. This change in melting point and the evolution of gas support that the molecular structure of the succinic acid has been altered, creating a substance with new properties, including a new melting point that typically involves smaller molecules compared to the original substance.

Answer:

Covalent bonds

Explanation:

Covalent bonds

These types of chemical bonds are between the atoms of same or comparable electronegtivity .

The formation of bond is to attain stability of the compound formed .

Hence , octet rule plays a major rule .

For example ,

The bond between the atoms of H₂ is a covalent bond , since , one atom of H have only 1 electron in its valence shell .

So it shares its one electron with another Hydrogen atom to attain stability , and forms a covalent bond .

Explanation:

When the water droplets are placed in oily surface, oil being non-polar, thus there will not be adhesive force between water and oil.

Thus, surface tension is the factor which is responsible for shape of the water droplets to be spherical. The water molecules are pulled into the spherical shape by the action of the cohesive forces which is acting on the surface layer. As, the medium is totally different outside, the water molecules tend to get the minimum area and thus, forming spherical droplets.

Answer:

a) 40,75 atm

b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

                                     PV = nRT

where:

Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.

The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the intermolecular forces and the space occupied by the gas

[tex]\frac{Pv}{RT} = 1 + \frac{B}{v}[/tex]

where:

a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

[tex]\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}[/tex]

and v = 3 L/5 moles = 0,6 L/mol

[tex]P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm[/tex]