A balloon filled with helium gas at 1.00 atm occupies 12.9 L. What volume would the balloon occupy in the upper atmosphere, at a pressure of 0.19 atm and a constant temperature? Enter your answer in the box provided.

Answer:

1.95 M

Explanation:

Considering

Moles of A = Moles of B

Or,

[tex]Molarity_{A}\times Volume_{A}=Molarity_{B}\times Volume_{B}[/tex]

Given  that:

[tex]Molarity_{A}=20.422\ M[/tex]

[tex]Volume_{A}=2\ mL[/tex]

[tex]Volume_{B}=2+12=14\ mL[/tex]

[tex]Molarity_{B}=?\ M[/tex]

So,  

[tex]20.422\times 2=Molarity_{B}\times 14[/tex]

Molarity of solution B = 2.92 M

Again Considering,

Moles of B = Moles of C

Or,

[tex]Molarity_{B}\times Volume_{B}=Molarity_{C}\times Volume_{C}[/tex]

Given  that:

[tex]Molarity_{B}=2.92\ M[/tex]

[tex]Volume_{B}=2\ mL[/tex]

[tex]Volume_{C}=2+1=3\ mL[/tex]

[tex]Molarity_{C}=?\ M[/tex]

So,  

[tex]2.92\times 2=Molarity_{C}\times 3[/tex]

Molarity of solution C = 1.95 M

The concentration of dye in Solution C is 1.945 M after performing the serial dilutions.

To find the concentration of dye in Solution C after serial dilutions, we can use the dilution equation which states that the concentration times the volume before dilution (C1V1) is equal to the concentration times the volume after dilution (C2V2), or C1V1 = C2V2.

For Solution B, the initial concentration is 20.422 M (C1) and the initial volume is 2 mL (V1). This solution is diluted with 12 mL water, so the final volume (V2) is 2 mL + 12 mL = 14 mL. Using the dilution equation, we calculate the concentration of Solution B (C2).

(20.422 M)(2 mL) = (C2)(14 mL)

C2 = (20.422 M)(2 mL) / (14 mL) = 2.917 M

Next, we perform a second dilution to create Solution C. We take 2 mL of Solution B and add 1 mL of water, giving a final volume of 3 mL. Again, we use the dilution equation to find the concentration of Solution C.

(2.917 M)(2 mL) = (C3)(3 mL)

C3 = (2.917 M)(2 mL) / (3 mL) = 1.945 M

Therefore, the concentration of dye in Solution C is 1.945 M, rounded to three significant figures.

Answer:

ΔG = 16.218 KJ/mol

Explanation:

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of [tex]N_2[/tex] = 100 kg  = 100000 g

Mass of [tex]H_2[/tex] = 100 kg = 100000 g

Molar mass of [tex]N_2[/tex] = 28 g/mole

Molar mass of [tex]H_2[/tex] = 2 g/mole

Molar mass of [tex]NH_3[/tex] = 17 g/mole

First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].

[tex]\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles[/tex]

[tex]\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]

So, 3571.43 moles of [tex]N_2[/tex] react with [tex]3571.43\times 3=10714.29[/tex] moles of [tex]H_2[/tex]

From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]NH_3[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]

So, 3571.43 moles of [tex]N_2[/tex] react to give [tex]3571.43\times 2=7142.86[/tex] moles of [tex]NH_3[/tex]

Now we have to calculate the mass of [tex]NH_3[/tex]

[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]

[tex]\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg[/tex]

Therefore, the mass of ammonia produced can be, 121.429 kg

Answer:

You need to do the following conversion to pass from 3M in 250 mL to g of sodium acetate

[tex]3 M (mol/L)*(1L/1000 mL)*(250 mL)*(82.03 g/1 mol)=61.52 g[/tex]

Explanation:

First, you need to dissolve 61.52 g of solid sodium acetate (MW 82.03 g/mol) in 200 ml of DI water. Then, using a volumetric flask add water to bring the total volume of the solution to 250 mL.

Answer:

43.13Kg of melamine

Explanation:

The problem gives you the mass of urea and two balanced equations:[tex]CO(NH_{2})_{2}_{(l)}=HNCO_{(l)}+NH_{3}_{(g)}[/tex]

[tex]6HNO_{l}=C_{3}N_{3}(NH_{2})_{3}_{(l)}+3CO_{2}_{(g)}[/tex]

First we need to calculate the number of moles of urea that are used in the reaction, so:

molar mass of urea = [tex]60.06\frac{g}{mol}*\frac{1kg}{1000g}=0.06006\frac{Kg}{mol}[/tex]

The problem says that you have 161.2Kg of urea, so you take that mass of urea and find the moles of urea:

161.2Kg of urea[tex]*\frac{1molofurea}{0.06006Kgofurea}=[/tex]2684 moles of urea

Now from the stoichiometry you have:

2684 moles of urea[tex]*\frac{1molofHNCO}{1molurea}*\frac{1molofmelamine}{6molesofHNCO}[/tex] = 447 moles of melamine

The molar mass of the melamine is [tex]126.12\frac{g}{mol}[/tex] so we have:

[tex]447molesofmelamine*\frac{126.12g}{1molofmelamine}[/tex] = 5637.64 g of melamine

Converting that mass of melamine to Kg:

5637.64 g of melamine *[tex]\frac{1Kg}{1000g}[/tex] = 56.38 Kg of melamine, that is the theoretical yield of melamine.

Finally we need to calculate the mass of melamine with a yield of 76.5%, so we have:

%yield = 100*(Actual yield of melamine / Theoretical yield of melamine)

Actual yield of melamine = [tex]\frac{76.5}{100}*56.38Kg[/tex] = 43.13Kg of melamine

Using stoichiometry, the actual yield of Melamine from a reaction starting with 161.2 kg of urea with an overall yield of 76.5% can be found to be approximately 43.2 kg.

The question asks about the actual yield of Melamine, C3N3(NH2)3, from a reaction starting with 161.2 kg of urea, CO(NH2)2, with an overall yield of 76.5%. To understand this prediction, we must use the concept of stoichiometry, which is a method used in chemistry to calculate the quantities of reactants or products in a chemical reaction.

First, calculate the mole of urea used: it is the mass of the urea divided by the molar mass of urea. The molar mass of urea (CO(NH2)2) is approximately 60 g/mole. So, you have 161.2*1000/60 = approximately 2686.67 moles of urea.

From the balanced chemical equation, we can see that one mole of urea produces one mole of HNCO, and six moles of HNCO produce one mole of Melamine. Therefore, in theory, 2686.67/6 = 447.78 moles of Melamine can be produced.

However, the reaction yield is only 76.5%, so the actual yield of the Melamine will be less. Using the percentage yield, we can calculate the actual yield: 447.78 * 0.765 = approximately 342.65 moles.

Finally, to convert from moles to mass, multiply by the molar mass of melamine, which is approximately 126 g/mole: 342.65*126 = approximately 43173.9g or 43.2 kg.

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Answer: a) 0.08157 moles of [tex]Al^{3+}[/tex]

b) 0.24471 moles of  [tex]ClO_4^{-}[/tex]

c) 0.97884 moles of oxygen atoms

Explanation:-

The dissociation of the given compound is shown by the balanced equation:

[tex]Al(ClO_4)_3\rightarrow Al^{3+}+3ClO_4^{-}[/tex]

According to stoichiometry:

a) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 1 mole of [tex]Al^{3+}[/tex]

Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{1}{1}\times 0.08157=0.08157moles[/tex] of [tex]Al^{3+}[/tex]

b) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 3 moles of [tex]ClO_4^{-}[/tex]

Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{3}{1}\times 0.08157=0.24471moles[/tex] of  [tex]ClO_4^{-}[/tex]

c) 1 mole of [tex]Al(ClO_4)_3[/tex] produces 12 moles of oxygen atoms

Thus 0.08157 mole of [tex]Al(ClO_4)_3[/tex] produces=[tex]\frac{12}{1}\times 0.08157=0.97884moles[/tex] of oxygen atoms

Answer:

A) weight

Explanation:

The weight is the force that gravity makes in a body, and it is given by the equation:

W = mxg

Where W is the weight, m is the mass, and g is the gravity in the local. So, the length, the volume, and the mass are constant and don't depend on the local that is measured, but the weight depends on the local gravity.

Explanation:

(a)   Here, the assumption is that complete combustion of carbon is taking place. This means that there is no CO formation

Reaction equation when carbon is burning is as follows.

              [tex]C + O_{2} \rightarrow CO_{2}[/tex]

Molar mass of C = 12 kg/kmol

Molar mass of [tex]CO_{2}[/tex] = 44 kg/kmol

Hence, according to the stoichiometry,

1 Kmol of C reacted = 1 kmol of [tex]CO_{2}[/tex] produced

Therefore, 12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced

It is given that coal contains 50 mass % Carbon

So, 1 kg of coal contains (0.50 × 1)kg Carbon

           Carbon in 1 kg Coal = 1 × 0.5 = 0.5 kg

As per the stoichiometry,

12 kg of C reacted = 44 kg [tex]CO_{2}[/tex] produced

0.5 kg of C reacted = x kg [tex]CO_{2}[/tex] produced

Therefore, value of x can be calculated as follows.

                    x = [tex]\frac{\text{44 kg of CO_{2} produced} \times \text{0.5 kg of C reacted}}{\text{12 kg of C reacted}}[/tex]

                       = 1.83 kg

This means that amount of [tex]CO_{2}[/tex] released is 1.83 kg.

(b)   It is assumed that coal contains 50 mass % carbon and 1 kg of coal burnt.

Since, it is given that energy density of coal is 24 Mj and efficiency of the power plant is 30%.

After burning 1 kg of coal amount of energy released = 24 Mj

Amount of energy converted to electricity = [tex]24 Mj \times 0.3[/tex] = 7.2 Mj

It is calculated that amount of [tex]CO_{2}[/tex] released per 1 kg of coal = 1.83 kg

Therefore, calculate the amount of [tex]CO_{2}[/tex] released in kg/Mj as follows.

        amount of [tex]CO_{2}[/tex] released in kg/Mj = [tex]\frac{\text{Amount of CO_{2} released in kg}}{\text{amount of energy converted to electricity}}[/tex]        

                        = [tex]\frac{1.83 kg}{7.2 mJ}[/tex]

                        = 0.2541 kg/Mj

Hence, the production of [tex]CO_{2}[/tex] in kg/MJ is 0.2541 kg/Mj.

To calculate the amount of CO2 produced from 1 kg of coal, first determine the mass of carbon, apply the conversion for the mass of CO2 produced per mass of carbon, and use the plant's efficiency and the coal's energy density to determine CO2 production per MJ.

The question involves consulting basic chemical stoichiometry and energy conversion to solve problems related to carbon dioxide emissions and energy efficiency in the context of coal combustion. Knowing the percentage of carbon in coal, we can determine the mass of carbon dioxide produced from burning coal. Additionally, we'll use the energy density of coal to calculate CO2 production per unit energy output of a coal-fired power plant.

To find the mass of CO2 produced from burning 1 kg of coal with 50% carbon content, we first calculate the mass of carbon that would burn:

The molecular weight of carbon is approximately 12 g/mol, and for CO2 it is approximately 44 g/mol. This implies that every 12 kg of carbon produces 44 kg of CO2. For 0.5 kg (or 500 g) of carbon, the mass of CO2 produced would be:

For the second part, to calculate the CO2 production in kg/MJ for a coal-fired power plant with 30% efficiency:

Answer: A. [tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]

B. [tex]2.5\times N[/tex] molecules of product.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

A. The balanced reaction for decomposition of potassium chlorate is:

[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]

B. According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

According to stoichiometry :

2 moles of reactant give 5 moles of products

Thus [tex]2\times 6.023\times 10^{23}[/tex] molecules of reactant give [tex]5\times 6.023\times 10^{23}[/tex] molecules of product

Thus N molecules of reactant give=  [tex]\frac{5\times 6.023\times 10^{23}}{2\times 6.023\times 10^{23}}\times N=\frac{5}{2}\times N[/tex] molecules of product.

Thus [tex]\frac{5}{2}\times N=2.5\times N[/tex] molecules of product are there.

Final answer:

The balanced chemical equation for the decomposition of potassium chlorate is 2 KClO3 → 2 KCl + 3 O2. If there are N molecules of potassium chlorate in the initial state, the number of product molecules of O2 can be calculated using the ratio of 2:3 in the equation.

Explanation:

The balanced chemical equation for the decomposition of potassium chlorate is:

2 KClO3 → 2 KCl + 3 O2

The stoichiometry of the reaction equation can be balanced by filling in the smallest possible integers as follows:

A. 2 KClO3 → 2 KCl + 3 O2

B. If there are N molecules of potassium chlorate in the initial state, the number of product molecules can be calculated as follows:

The ratio of KClO3 to O2 in the balanced equation is 2:3. So, for every 2 molecules of KClO3 that decompose, 3 molecules of O2 are formed. Therefore, the number of O2 molecules produced is given by the equation:

(3 / 2) N = (3 / 2) × N molecules of O2

Answer:

c. Kay's rule

Explanation:

Kay's rule -

The rule is used to determine the pseudo reduced critical parameters of mixture , with the help of using the critical properties of the components of a given mixture .

The equation for Kay's rule is as follows ,

PV = Z RT

Where Z = The compressibility factor of the mixture .

Hence from the given options , the correct answer is Kay's rule .

To estimate the pseudo-critical properties of mixtures based on pure-component critical constants, Kay's rule is used. It estimates the properties based on mole fractions of components, offering a simpler alternative to complex equations like Van der Waals or Peng-Robinson equations.

The equation that can be used to estimate the pseudo-critical properties of mixtures based on the pure-component critical constants is Kay's rule. This rule is a simple and common method for calculating the critical properties of mixtures, which assumes that the critical properties are proportional to the mole fractions of the components in the mixture. It is not as complex as the Van der Waals equation of state or the Peng-Robinson equation, which are more accurate but also more complicated.

For ideal solutions, Raoult's Law is often used to determine the vapor pressures and hence the equilibrium states of the components in a solution, given by the equation PA = XAPA, where PA is the partial pressure of component A, XA is the mole fraction of component A, and PAo is the vapor pressure of the pure component A.

Answer: Yes, [tex]HCl[/tex] is a strong acid.

acid = [tex]HCl[/tex] , conjugate base = [tex]Cl^-[/tex] , base = [tex]H_2O[/tex], conjugate acid = [tex]H_3O^+[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes [tex]HCl[/tex] is a strong acid as it completely dissociates in water to give [tex]H^+[/tex] ions.

[tex]HCl\rightarrow H^++Cl^-[/tex]

For the given chemical equation:

[tex]HCl+H_2O\rightarrow H_3O^-+Cl^-[/tex]

Here, [tex]HCl[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]Cl^-[/tex] which is a conjugate base.

And, [tex]H_2O[/tex] is gaining a proton, thus it is considered as a base and after gaining a proton, it forms [tex]H_3O^+[/tex] which is a conjugate acid.

Thus acid =  [tex]HCl[/tex]

conjugate base = [tex]Cl^-[/tex]

base = [tex]H_2O[/tex]

conjugate acid = [tex]H_3O^+[/tex].

Answer:

The diameter of the oil molecule is [tex]4.4674\times 10^{-8} cm[/tex] .

Explanation:

Mass of the oil drop = [tex]m=9.00\times 10^{-7} kg[/tex]

Density of the oil drop = [tex]d=918 kg/m^3[/tex]

Volume of the oil drop: v

[tex]d=\frac{m}{v}[/tex]

[tex]v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}[/tex]

Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.

Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m

1 cm = 0.01 m

Surface of the sphere is given as: a = [tex]4\pi r^2[/tex]

[tex]a=4\times 3.14\times (0.418 m)^2=2.1945 m^2[/tex]

Volume of the oil drop = v = Area × thickness

[tex]\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x[/tex]

[tex]x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm[/tex]

The thickness of the oil drop is [tex]4.4674\times 10^{-8} cm[/tex] and so is the diameter of the molecule.

The diameter of an oil molecule can be estimated by considering the oil slick as a cylinder and using the formula for the volume of a cylinder. From the given, we calculate the volume of the oil droplet. Dividing this volume by the area of the slick gives us the height of the cylinder, which represents the diameter of the oil molecule.

The oil slick forms a cylindrical shape with a height equal to the diameter of an oil molecule or two times the radius. We know the volume of the oil droplet, oil slick, and the diameter of the oil slick.

The volume of oil droplet (V) = Mass (m) / Density (ρ)
= 9 x 10⁻⁷ kg / 918 kg/m³
= 9.81 x 10⁻¹⁰ m³

The volume of the cylindrical oil slick (V) = πr²h, where r is the radius and h is the height of the cylinder.
So, h (height of the oil molecule) = V / (πr^2)
= 9.81 x 10⁻¹⁰ m³ / (3.14 x (0.418 m)²)
= 1.74 x 10⁻¹⁰ m or 174 pm (picometers)

Since the height of the cylinder is approximately equal to the diameter of an oil molecule, the approximate diameter of an oil molecule is 174 pm.

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The rate law for the reaction is rate = k[S2O8²⁻], and the rate of disappearance of I- can be determined using the stoichiometric coefficients.

The rate law for the reaction of peroxydisulfate ion (S2O8²⁻) with iodide ion (I⁻) can be determined by using the initial rates data. In the given data, we can see that the initial rate varies linearly with the concentration of S2O8²⁻ and is independent of the concentration of I⁻. This means that the rate law for this reaction is rate = k[S2O8²⁻].

Using the given rate law, we can determine the rate of disappearance of I⁻ when [S2O8²⁻] = 1.8×10⁻²M and [I⁻] = 5.0×10⁻²M. Since the stoichiometric coefficient of I⁻ in the balanced chemical equation is 3, the rate of disappearance of I⁻ is 3 times the rate of appearance of S2O8²⁻. Therefore, the rate of disappearance of I⁻ is (3)(1.4×10⁻⁵M/s) = 4.2×10⁻⁵M/s.

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Explanation:

The given data is as follows.

            Volume of tank = 4 [tex]m^{3}[/tex]

             Density of water = 1000 [tex]kg/m^{3}[/tex]

Since, the tank is initially half-filled. Hence, the volume of water in the tank is calculated as follows.

                        [tex]\frac{1}{2} \times 4 = 2 m^{3}[/tex]

Also, density of a substance is equal to its mass divided by its volume. Therefore, initially mass of water in the tank is as follows.

                    Mass = [tex]Density \times initial volume[/tex]

                              = [tex]1000 \times 2[/tex]

                              = 2000 kg

Whereas mass of water in tank when it is full is as follows.

                     Mass = [tex]Density \times final volume[/tex]

                               = [tex]1000 \times 4[/tex]

                               = 4000 kg

So, net mass of the fluid to be filled is as follows.

                  Net mass to be filled = Final mass - initial mass

                                                      = 4000 kg - 2000 kg

                                                      = 2000 kg

Mass flow rate [tex](m_{in})[/tex] = 6.33 kg/s

Mass flow rate [tex](m_{out})[/tex] = 3.25 kg/s

       Time needed to fill tank = [tex]\frac{\text{net mass to be filled}}{\text{net difference of flow rates}}[/tex]

                                       = [tex]\frac{2000 kg}{m_{in} - m_{out}}[/tex]

                                       = [tex]\frac{2000 kg}{6.33 kg/s - 3.25 kg/s}[/tex]

                                       = 649.35 sec

Thus, we can conclude that 649.35 sec is taken by the tank to overflow.

Answer: The empirical and molecular formula for the given organic compound is [tex]C_2S[/tex] and  [tex]C_{16}S_8[/tex] respectively.

Explanation:

We are given:

Percentage of C = 42.83 %

Percentage of S = 57.17 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.83 g

Mass of S = 57.17 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.83g}{12g/mole}=3.57moles[/tex]

Moles of Sulfur = [tex]\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=\frac{57.17g}{32g/mole}=1.79moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.79 moles.

For Carbon = [tex]\frac{3.57}{1.79}=1.99\approx 2[/tex]

For Sulfur = [tex]\frac{1.79}{1.79}=1[/tex]

The ratio of C : S = 2 : 1

Hence, the empirical formula for the given compound is [tex]C_2S_1=C_2S[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

We are given:

Mass of molecular formula = 448.70 g/mol

Mass of empirical formula = 56 g/mol

Putting values in above equation, we get:

[tex]n=\frac{448.70g/mol}{56g/mol}=8[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(8\times 2)}S_{(8\times 1)}=C_{16}S_8[/tex]

Thus, the empirical and molecular formula for the given compound is [tex]C_2S[/tex] and  [tex]C_{16}S_8[/tex]

Final answer:

The empirical formula of 'sulflower' is CS2, found by using the given percentages to calculate the simplest whole number ratio of moles of carbon to sulfur. Dividing the compound's molar mass by the empirical formula's molar mass indicates that the molecular formula of 'sulflower' is C6S12.

Explanation:

To determine the empirical formula of the molecule known as 'sulflower', we can start by using the given percentages to find the moles of each element in the compound. Assuming we have 100 grams of the substance, we would have 57.17 grams of sulfur (S) and 42.83 grams of carbon (C). Using their molar masses (atomic sulfur is 32 g/mol and carbon is 12 g/mol), we can calculate the moles of each:

To find the empirical formula, we then find the simplest whole number ratio by dividing by the smallest number of moles, which gives us:

So, the empirical formula is CS2. To determine the molecular formula, we use the molar mass of the empirical formula and divide the given molar mass of the compound by this value. The empirical formula mass of CS2 is (12.01 g/mol × 2) + (32.07 g/mol × 1) = 76.09 g/mol. The molecular formula is then found by dividing the compound's molar mass by the empirical formula's molar mass:

448.70 g/mol / 76.09 g/mol = 5.9 ≈ 6

Therefore, the molecular formula is C6S12, as the empirical formula must be multiplied by 6 to get the molecular formula.

The density of the cylinder would be 3.652 gram/ cm³

.

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem, you just measured a metal cylinder and obtained the following information: mass - 3.543 grams, diameter 0.53 cm, height = 4.40 cm , and we have to calculate the density of the cylinder,

mass of the cylinder = 3.543 grams

the volume of the cylinder = πr²h

                                            = 3.14 ×.265²×4.4

                                            =0.97 cm³

By using the above formula for density

ρ = mass of the cylinder/volume of the cylinder

  = 3.543 grams/0.97 cm³

  =3.652 grams/ cm³

Thus,the density of the cylinder would be 3.652 grams/ cm³.

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To find the volume and density of a metal cylinder with a mass of 3.543 g, diameter of 0.53 cm, and height of 4.40 cm, we first calculate the volume as 3.08 cm³ using the formula V = πr²h. Then, we determine the density to be 1.15 g/cm³ by dividing the mass by the volume.

You just measured a metal cylinder and obtained the following information: mass - 3.543 g, diameter - 0.53 cm, height = 4.40 cm. To determine the volume and density of the cylinder, follow these steps:

Determining the Volume (V)

First, understand that the volume (V) of a cylinder can be calculated using the formula V = πr²h, where π (pi) is approximately 3.14, r is the radius of the cylinder (half of the diameter), and h is the height of the cylinder. Given a diameter of 0.53 cm, the radius is 0.53 cm / 2 = 0.265 cm. Substituting these values into the formula gives:

V = 3.14 * (0.265 cm)² * 4.40 cm = 3.08 cm³

Determining the Density (D)

Density can be found using the formula D = m / V, where m is the mass and V is the volume. Using the mass of 3.543 g and the volume of 3.08 cm³:

D = 3.543 g / 3.08 cm³ = 1.15 g/cm³

Answer:

[tex]\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]

The only other data you need is the density of water ρw.

Explanation:

We can start by the volume balance

[tex]V_s=V_1+V_2[/tex]

We can replace the volumes with V=M/ρ

[tex]\frac{M_s}{\rho_s}=\frac{M_1}{\rho_1}+\frac{M_2}{\rho_2}[/tex]

If we divide every term by Ms

[tex]\frac{M_s/M_s}{\rho_s}=\frac{M_1/M_s}{\rho_1}+\frac{M_2/M_s}{\rho_2}[/tex]

By definition, wt=Mi/Msol, so we can replace that in the expression

[tex]\frac{1}{\rho_s}=\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}[/tex]

Then we have the expression of the density of the solution

[tex]\rho_s=\frac{1}{\frac{wt1}{\rho_1}+\frac{wt2}{\rho_2}}[/tex]

To replace ρ1 and ρ2, you have to multiply the specific gravity of the components and the density of water.

[tex]\rho_s=\frac{1}{\frac{wt1}{SpGr1\rho_w}+\frac{wt2}{SpGr2\rho_w}}\\\\\rho_s=\frac{\rho_w}{\frac{wt1}{SpGr1}+\frac{wt2}{SpGr2}}[/tex]

Answer:

A) 577.6 years

B) 1580.9 years

Explanation:

You can know the order of a reaction given the units of the rate constant:

Then we know that this is a first order reaction because years is a unit of time as well as seconds. The half life of a first order reaction is given by:

[tex]t_{1/2}=\frac{ln(2)}{k}[/tex]

Here you can solve for the half life:

[tex]t_{1/2}=\frac{ln(2)}{0.0012 years^-1}=577.6 years[/tex]

Now for the rate law for a first order reaction is:

[tex][A]=[A_{o} ]*e^{-kt}[/tex]

Then, if you want to know how long does it take to reach a certain value you solve for time:

[tex]t=-ln([A]/[A]_{o}) *1/k[/tex]

The 15% of its original value is [tex]0.15*[A_{o}][/tex]

You solve for time:

[tex]t=-ln(0.15) *1/0.0012 years^-1=1580.9 years[/tex]

Hope it helps!

In the Decomposition reaction, a reactant gets split into two or more products. The half-life of the reaction is 577.6 years and the time taken to reach 15% of the original is 1580.9 years.

Half-life is the time taken by the radioactive isotope to decay one-half of its total amount.

From the units of the rate of the reaction, the order can be estimated as, for a first-order reaction unit is 1/s.

The half-life of the reaction can be given as,

[tex]\rm t\dfrac{1}{2} = \dfrac{\rm ln (2)}{k}[/tex]

The rate constant (k) for the decomposition reaction is 0.0012 per year.

Substituting values in the above equation:

[tex]\begin{aligned} \rm t\dfrac{1}{2} &= \dfrac{\rm ln (2)}{0.0012}\\\\&= 577.6\;\rm years\end{aligned}[/tex]

The rate law of the first-order reaction is given as,

[tex]\begin{aligned}\rm [A] = [A_{o}] \times e^{-kt}\\\\\rm t = -ln ([A][A_{o}]) \times \dfrac{1}{k}\end{aligned}[/tex]

Solving for time (t):

[tex]\begin{aligned}\rm t &= \rm -ln (0.15) \times \dfrac{1}{0.0012}\\\\& = 1580.9\;\rm years\end{aligned}[/tex]

Therefore, the half-life is 577.6 years and the time taken is 1580.9 years.

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Answer:

179.4306 g

Explanation:

Given that:

Mass of [tex]Bi_2O_3[/tex] = 198 g

Molar mass of [tex]Bi_2O_3[/tex] = 465.96 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{198\ g}{465.96\ g/mol}[/tex]

[tex]Moles\ of\ Bi_2O_3= 0.4293\ mol[/tex]

From the balanced reaction,

[tex]Bi_2O_3+3C\rightarrow 2Bi+3CO[/tex]

1 mole of [tex]Bi_2O_3[/tex] on reaction produces 2 moles of bismuth

So,

0.4293 mole of [tex]Bi_2O_3[/tex] on reaction produces 2 × 0.4293 moles of bismuth

Moles of bismuth = 0.8586 moles

Molar mass of bismuth = 208.9804 g/mol

So, mass of bismuth = Moles × Molar mass = 0.8586 × 208.9804 g = 179.4306 g

Final answer:

The theoretical yield of bismuth in the given reaction is 0.848 moles or 177.2 grams.

Explanation:

To determine the theoretical yield of bismuth (Bi) in the reaction between 198 g of Bi2O3 and excess carbon, we need to balance the equation first. The balanced equation for the reaction is:

Bi2O3 + 3C → 2Bi + 3CO

From the balanced equation, we can see that for every mole of Bi2O3, we get 2 moles of Bi. To calculate the theoretical yield, we need to convert the given mass of Bi2O3 to moles using its molar mass (465.96 g/mol) and then use the stoichiometry to find the moles of Bi.

Mass of Bi2O3 = 198 g

Molar mass of Bi2O3 = 465.96 g/mol

Moles of Bi2O3 = (198 g) / (465.96 g/mol) = 0.424 mol

Moles of Bi = 2 * (0.424 mol) = 0.848 mol

The theoretical yield of bismuth in this reaction is 0.848 moles or you can convert it to grams using the molar mass of bismuth (208.98 g/mol) to get the theoretical yield in grams.

Answer: The force that must be applied is 15 N.

Explanation:

Force exerted on the object is defined as the product of mass of the object and the acceleration of the object.

Mathematically,

[tex]F=m\times a[/tex]

where,

F = force exerted = ?

m = mass of the object = 3 kg

a = acceleration of the object = [tex]5m/s^2[/tex]

Putting values in above equation, we get:

[tex]F=3kg\times 5m/s^2=15N[/tex]

Hence, the force that must be applied is 15 N.

Answer: The given number in scientific notation is [tex]2.367\times 10^{7}[/tex]

Explanation:

Scientific notation is the notation where a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

We are given:

A number having value = 23,665,700

Converting this into scientific notation, we get:

As, the decimal is shifting to left side, the power of 10 will be positive.

[tex]\Rightarrow 23,665,700=2.367\times 10^{7}[/tex]

Hence, the given number in scientific notation is [tex]2.367\times 10^{7}[/tex]

Answer: The number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]

Explanation:

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

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

Given mass of bromine = 1.37 g

Molar mass of bromine = 79.904 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of bromine}=\frac{1.37g}{79.904g/mol}=0.0171mol[/tex]

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, 0.0171 moles of bromine will contain = [tex]0.0171\times 6.022\times 10^{23}=1.03\times 10^{22}[/tex] number of bromine atoms.

Hence, the number of atoms of bromine present in given number of mass is [tex]1.03\times 10^{22}[/tex]

Final answer:

To find the number of bromine atoms in 1.37 g of bromine, calculate the atomic mass using isotope information, convert grams to moles, and then use Avogadro's number, yielding approximately 1.03 × 10²² bromine atoms.

Explanation:

To calculate the number of bromine atoms in 1.37 g of bromine, we first need to determine the molar mass of bromine. The atomic mass of bromine is calculated based on the isotopic composition, considering Bromine has two isotopes, 79Br and 81Br with masses of 78.9183 amu and 80.9163 amu respectively, and relative abundances of 50.69% and 49.31%. Using these values, we calculate the average atomic mass of bromine:

Atomic mass = (0.5069 × 78.9183 amu) + (0.4931 × 80.9163 amu) = 79.904 amu.

We then convert grams of bromine to moles by dividing by the molar mass:

Number of moles = 1.37 g ÷ 79.904 g/mol = 0.01715 moles.

Using Avogadro's number, 6.022 × 10²³ atoms/mol, we multiply the number of moles of bromine by this constant to find the number of atoms:

Number of atoms = 0.01715 moles × 6.022 × 10²³ atoms/mol = 1.03 × 10²² atoms.

Thus, there are 1.03 × 10²² atoms of bromine in 1.37 g of bromine. Remember, when giving your answer in scientific notation, it is important to express it in the form of a number between 1 and 10 multiplied by a power of 10.

The Clausius-Clapeyron equation is used to calculate the vapor pressure of water at 40°C given its vapor pressure at 20°C and heat of vaporisation. By rearranging the equation and inserting the known values, the vapor pressure at 40°C can be found, which will be higher than that at 20°C.

The Clausius-Clapeyron equation relates the vapor pressure and temperature of a substance to its heat of vaporisation. To calculate the vapor pressure of water at 40°C, given the vapor pressure at 20°C (2.34 kPa) and the heat of vaporisation (2537.4 kJ/kg), we can rearrange the equation to solve for the new vapor pressure:

Clausius-Clapeyron Equation: ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

Where P1 is the initial vapor pressure, P2 is the final vapor pressure, ΔHvap is the molar heat of vaporisation, R is the universal gas constant in J/(mol·K), T1 is the initial temperature in Kelvin, and T2 is the final temperature in Kelvin.

The calculation involves converting both temperatures from Celsius to Kelvin and plugging the values into the equation to solve for P2. After using this equation, we find that the vapor pressure at 40°C is significantly higher than at 20°C.

Answer:

It is necessary in order to let air in.

Explanation:

When the stopper is not removed, a vacuum is formed inside the separatory funnel as the liquid drains out. The lower air pressure inside the separatory funnel then causes the liquid to drain improperly.

Answer: The volume of argon gas at STP is 57.4 mL

Explanation:

STP conditions are:

Pressure of the gas = 1 atm = 760 torr

Temperature of the gas = 273 K

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas

[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas

We are given:

[tex]P_1=725torr\\V_1=65.0mL\\T_1=22^oC=(22+273)K=295K\\P_2=760torr\\V_2=?\\T_2=273K[/tex]

Putting values in above equation, we get:

[tex]\frac{725torr\times 65.0mL}{295K}=\frac{760torr\times V_2}{273K}\\\\V_2=57.4mL[/tex]

Hence, the volume of argon gas at STP is 57.4 mL

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 65.0 mL = 0.065 L (as 1 ml = 0.001 L),      

     [tex]T_{1}[/tex] = [tex]22.0^{o}C[/tex] = (22 + 273) K = 295 K,

      [tex]P_{1}[/tex] = 725 torr = 0.954 atm           (as 1 torr = 0.00131579 atm),

     [tex]V_{2}[/tex] = ?,   [tex]T_{2}[/tex] = 273 K,

      [tex]P_{2}[/tex] = 1 atm

And, according to ideal gas equation,

               [tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

           [tex]\frac{0.954 atm \times 0.065 L}{295 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]

            [tex]V_{2}[/tex] = 0.0574 L

As, 1 L = 1000 ml. So, 0.0574 L = 57.4 ml.

Thus, we can conclude that the volume of this Ar gas at STP is 57.4 L.

SO3 2- has three possible resonance structures, each with one of the three oxygen atoms having a double bond to the central sulfur atom, while the other two have single bonds. The sulfur atom is surrounded by two bonds and one lone pair of electrons in any given structure.

The student asked to draw the resonance structures for SO3 2-. Resonance structures are a way of illustrating the delocalization of electrons within certain molecules, often containing double bonds. In the case of the SO3 2- ion, we have sulfur connected to three oxygen atoms.

In each of the resonance structures, sulfur has an oxidation number of +6 and each oxygen atom has an oxidation number of -2. There are three different resonance structures possible for this ion. This is because one of the oxygen atoms can have a double bond to the sulfur atom while the other two oxygen atoms have single bonds, with sulfur carrying one lone pair. The double bond can be with any of the three oxygen atoms, leading to three possible structures, hence the term resonance structures.

The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp².

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Answer:

[tex]4.264*10^{6}[/tex]

Explanation:

Hello !

The correct answer is option C)  [tex]4.264*10^{6}[/tex]

To write in scientific notation, numbers between 0 and 9 are used before the point and the rest of the numbers afterwards.

Then we count how many places we ran the point. If you run to the right, the exponent of 10 is positive and if you run to the left, it is negative.

This way it is  [tex]4.264*10^{6}[/tex]

Answer:

14.20

Explanation:

The pH can be calculated as:

KOH is a strong base and it will dissociate completely in the solution as:

[tex]KOH\rightarrow K^++OH^-[/tex]

According to the equation and also mentioned that KOH will dissociate completely. So,

Molarity of the hydroxide ions furnished = Molarity of the KOH = 1.6 M

So,

pOH = -log[OH⁻] = -log(1.6) = -0.20

Also,  

pH + pOH = 14  

So, pH = 14 - ( - 0.20 ) = 14.20

Answer:

0.5455

Explanation:

The moles of benzene in the process stream in 1 sec = 450 moles

The moles of toluene in the process stream in 1 sec = 375 moles

So, according to definition of mole fraction:

[tex]Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]

Applying values as:

[tex]Mole\ fraction\ of\ benzene=\frac {450}{450+375}[/tex]

Mole fraction of benzene in the process stream = 0.5455

Explanation:

MWQ means the minimum weighable quantity.

Mathematically,         MWQ = [tex]\frac{Sensitivity}{1 - \text{fraction of accuracy}}[/tex]

or,        MWQ = [tex]\frac{sensitivity}{\text{fractional error}}[/tex]

It is given that sensitivity is 4.5 mg and maximum permitted error is 3.6%.

Therefore, fraction error = [tex]\frac{3.6}{100}[/tex] = 0.036

Hence, we will calculate MWQ as follows.

                  MWQ = [tex]\frac{sensitivity}{\text{fractional error}}[/tex]

                            = [tex]\frac{4.5 mg}{0.036}[/tex]

                            = 125 mg

Thus, we can conclude that the MWQ of the given balance is 125 mg.