Consider a compound that is 31.17% C, 6.54% H, and 62.29% O by mass. Assume that we have a 100 g sample of this compound. What are the subscripts in the empirical formula for this compound?

Answers

Answer 1

Answer: the empirical formula is CH3O and the subscripts are 1, 3, 1

Explanation:Please see attachment for explanation

Consider A Compound That Is 31.17% C, 6.54% H, And 62.29% O By Mass. Assume That We Have A 100 G Sample

Related Questions

Nitrogen fixation is the
conversion of gaseous nitrogen into an organism friendly form (ammonia (NH4 ). preplanned setting of atmospheric levels of nitrogen at 78% stabilizing elemental nitrogen into a gaseous two-atom molecule (N2) the reduction of nitrate (NO3-) to gaseous nitrogen. the repair of D.N.A. by adding nitrogen to the organism's diet

Answers

Answer:

Nitrogen fixation is the  conversion of gaseous nitrogen into an organism friendly form (ammonia) .

Explanation:

Nitrogen fixation -

It is very important and necessary process by which the nitrogen gas present in the atmosphere gets converted to nitrogen derivatives like ammonia  in the soil , is referred to as the process of nitrogen fixation.  

This occurs due to the reason , that the derivatives of nitrogen are of much more importance than the molecular nitrogen.  

Biologically the process is done by the rhizobium bacteria .

Hence ,

From the question,

The correct option is a.

At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.
1. At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm?

Answers

Answer:

0.000273 M

Explanation:

Henry's states that at constant temperature the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure in of that gas in equilibrium with that liquid.

Pressure of Oxygen = mole fraction of Oxygen × 1.00 atm

Mole fraction Oxygen = 21/100 × 1.00atm = 0.21 atm

Molar solubility of Oxygen = KH × PO2 = 0.0013 × 0.21 = 0.000273 M

The amount of a gas that dissolves in a liquid is proportional to the partial pressure of the gas above the liquid. Solubility of oxygen in water exposed to air at 1.00 atm is 0.000273 M.  

Henry's Law:

It states that at constant temperature the amount of a gas that dissolves in a liquid is proportional to the partial pressure of the gas above the liquid.

C = k P

Where,

C = concentration of a dissolved gas

k = Henry's Law constant = 0.0013 M/atm.

P = partial pressure of the gas = [tex]\bold {\dfrac {21}{100} \times 1.00\ atm = 0.21\ atm}[/tex]

 

Put the values in the formula,

[tex]\bold {C = 0.0013 \times 0.21 = 0.000273\ M }[/tex]

Therefore, solubility of oxygen in water exposed to air at 1.00 atm is 0.000273 M.

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An ideal gas, initially at 30°C and 100 kPa, undergoes the following cyclic processes in a closed system: (a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30°C, and finally expanded isothermally to its original state.

Answers

Answer:

a) compressed adiabatically:

T2 = 576.8 K

ΔU = 3422 J/mol

b) cooled at P contant to 30°C:

Q = - 5965.04 J/mol

c) expanded isotermally to P=100 KPa:

W = - 4054.403 J

Explanation:

ideal gas in a mechanically reversible process:

∴ T1 = 30°C = 303 K

∴ P1 = 100 KPa

a) compressed adiabatically to 500 KPa:

ΔU = Q + W      ∴ Q = 0

⇒ ΔU = W = CvΔT.....(1)

∴ Cv = (3/2)R = 12.5 J/K.mol

∴ W = - PδV........(2)

(1) = (2):

⇒ [(R+Cv)/R] Ln (T2/T1) = Ln (P2/P1)

∴ R+Cv/R = 5/2

⇒ (5/2) Ln(T2/T1) = 1.6094

⇒ LnT2 - LnT1 = 0.64376

⇒ LnT2 = 0.64376 + 5.7137 = 6.3575

⇒ T2 = 576.8 K

⇒ ΔU = W = (12.5 J/K.mol)(576.8 - 303 ) = 3422 J/mol

b)n cooled at constant P = 500KPa to 30°C:

∴ T2 = 303 K

∴ T1 = 576.8 K

∴ ΔU = Q + W

⇒ Q = CpΔT

∴ Cp = (5/2)R = 20.8 J/K.mol

⇒ Q = (20.8 J/K.mol)(303 - 576.8)

⇒ Q = - 5695.04 J/mol

c) expanded isothermally a P=100 KPa

∴ ΔU = 0

∴ T = 303 K

∴ P1 = 500 KPa

∴ P2 = 100 KPa

∴ W = nRT Ln(P2/P1)......assuming n = 1 mol

⇒ W = (1 mol)(8.314 J/K,mol)(303 K) Ln(100/500)

⇒ W = - 4054.403 J

Final answer:

This student's question concerns physics and thermodynamic cycles. It describes an ideal gas's behavior when it undergoes adiabatic compression, isobaric cooling, and isothermal expansion in a closed system.

Explanation:

Let's break these down:

Adiabatic process: In this method, the gas is compressed in such a way that no heat is exchanged with its surroundings. The gas is compressed, raising its pressure and temperature. Isobaric process: During this constant pressure cooling process, as the gas cools, its volume decreases but the pressure stays the same.Isothermal process: In this part, the gas expands and returns to its original state while maintaining a constant temperature.

In the entire cycle, the work done by the gas or on the gas depends on the path the gas follows during each process.

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The solubility of sodium chloride in water is 35.7 g per 100 ml at 0. What is the maximum amount of sodium chloride that will dissolve in 250 ml of water

Answers

Answer:

89.52 gm

Explanation:

35.7 gm of NaCl , dissolves in 100ml of water.

Then amount of NaCl dissolved in 1 ml water = 35.7/100= 3.75 gm

therefore, sodium chloride dissolved in 250 ml of water

= 3.75×250= 89.25 gm

Final answer:

The maximum amount of sodium chloride that will dissolve in 250 ml of water is 89.25 g.

Explanation:

The solubility of sodium chloride in water is 35.7 g per 100 ml at 0°C. To find the maximum amount of sodium chloride that will dissolve in 250 ml of water, we can set up a proportion using the solubility values:

35.7 g / 100 ml = x g / 250 ml

Cross-multiplying and solving for x, we get:

x = (35.7 g * 250 ml) / 100 ml = 89.25 g

Therefore, the maximum amount of sodium chloride that will dissolve in 250 ml of water is 89.25 g.

State whether the following statements are true or false (with justification). (a) 1 mol of N2 has more molecules than 1 mol of Ar. (b) 1 mol of N2 has more mass than than 1 mol of Ar. (c) The molar mass of N2 is greater than the molar mass of Ar

Answers

Answer:

A. False.

Every substance contains the same number of molecules i.e 6.02x10^23 molecules

B. False.

Mass conc. = number mole x molar Mass

Mass conc. of 1mole of N2 = 1 x 28 = 28g

Mass conc. of 1mol of Ar = 1 x 40 = 40g

The mass of 1mole of Ar is greater than the mass of 1mole of N2

C. False.

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of Ar = 40g/mol

The molar mass of Ar is greater than that of N2.

Explanation:

What is the name of the structures that help move substances across a tract surface?

Answers

Answer:

cilia and flagella

Explanation:

In prokaryotic species , cilia are present , and in eukaryotic species , flagella is present .

Cilia and flagella both have same function , i.e. , to enable the movement of the cell , along with the movement of some substance and direct the flow of these substance along the tracts.

Cilia and flagella are composed of basal bodies.

Hence , from the given statement of the question,

The correct term is cilia and flagella .

Final answer:

Cilia are the structures that help move substances across the surface of a tract. They are primarily found in certain types of cells such as those in the respiratory tract and the oviducts.

Explanation:

The structures that assist in moving substances across a tract surface are primarily called cilia. In biology, cilia are tiny hair-like structures that line certain types of cells, especially those in the respiratory tract and the oviducts. They work much like oars on a boat, moving in coordinated waves to propel substances (like mucus or egg cells) along the surface of the tract they occupy. They are responsible in movement of substances across tract.

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A chemist prepares a solution of aluminum sulfate (Al2(SO4) by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filing the flask to the mark with water Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Round your answer to 3 significant digits

Answers

Final answer:

The concentration of the aluminum sulfate solution is calculated to be 25.8 g/dL by dividing the mass of the aluminum sulfate (116.0 g) by the volume of the solution (4.5 dL).

Explanation:

The concentration of a solution is given by the ratio of the mass of the solute to the volume of the solution. In this case, we have 116.0 g of aluminum sulfate dissolved in a solution whose total volume is 450.0 mL. To convert this volume to deciliters (dL), we remember that 1 L = 10 dL and 1 L = 1000 mL. Therefore, the volume of the solution is 450.0 mL * (1 L / 1000 mL) * (10 dL / 1 L) = 4.5 dL. The concentration of aluminum sulfate in the solution is thus 116.0 g / 4.5 dL = 25.8 g/dL.

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What mass (in g) of solute is contained in 764.9 mL of a 0.137 M solution of glucose, C6H12O6? Enter your answer with 3 significant figures and no units.

Answers

Answer:

18.9

Explanation:

Given data

Volume (V): 764.9 mL = 0.7649 LMolar concentration (C): 0.137 M = 0.137 mol/L

We can find the moles (n) of glucose using the following expression.

n = C × V

n = 0.137 mol/L × 0.7649 L

n = 0.105 mol

The molar mass of glucose is 180.16 g/mol. The mass corresponding to 0.105 moles is:

m = 0.105 mol × 180.16 g/mol = 18.9 g

Calculate the energy required to heat 406.0mg of cyclohexane from 33.5°C to 38.9°C .

Assume the specific heat capacity of cyclohexane under these conditions is 1.85·J·g−1K−1 . x

Be sure your answer has the correct number of significant digits.

Answers

Answer:

We need 4.06 J to heat of 406.0 mg of cyclohexane

Explanation:

Step 1: Data given

Mass of cyclohexane = 406.0 mg = 0.406 grams

The initial temperature = 33.5 °C

The final temperature = 38.9 °C

The specific heat of cyclohexane = 1.85 J/g*K

Step 2: Calculate the energy required to heat

Q = m*c*ΔT

⇒ with m = the mass of cyclohexane = 0.406 grams

⇒ with c = the specific heat of cyclohexane = 1.85J/g*K

⇒ with ΔT = The change of temperature = T2 - T1 = 38.9 - 33.5 = 5.4

Q = 0.406 g * 1.85J/g*K * 5.40

Q = 4.06 J

We need 4.06 J to heat of 406.0 mg of cyclohexane

The combination of water and CO2 molecules in the atmosphere account for approximately 155 W/m^2 in greenhouse heating.
1. If we start with a black body at 255 K and add 150 W/m^2 in energy flux out of the body, what is the new predicted temperature?

Answers

Final answer:

The problem cannot be solved as presented because the initial energy flux of the black body at 255 K is not given. Without it, we cannot apply the Stefan-Boltzmann law to find the new temperature after adding 150 W/m^2.

Explanation:

The student is asking how to find the new temperature of a black body that was initially at 255 K after adding an energy flux of 150 W/m2. To solve this, we can apply the Stefan-Boltzmann law, which states that the energy radiated by a black body per unit area is directly proportional to the fourth power of the black body's temperature (E = σT4). The constant of proportionality, σ, is known as the Stefan-Boltzmann constant (σ = 5.67 x 10-8 W/m2K4).

where

q= heat flux = 155 W/m²+150 W/m² = 255 W/m²

σ= Stephan-Boltzmann constant = 5.67*10⁻⁸ W/m²K⁴

T= absolute temperature

T₀= absolute initial temperature = 255 K

solving for T

q = σ*(T⁴-T₀⁴)

T = (q/σ + T₀⁴)^(1/4)

replacing values

T = (q/σ + T₀⁴)^(1/4) = (255 W/m²/(5.67*10⁻⁸ W/m²K⁴) + (255 K)⁴)^(1/4) = 305.63 K

T=305.63 K

thus the final temperature is T=305.63 K

Predict the major product obtained upon radical bromination of t-butylcyclohexane.

3-bromo-1-tert-butylcyclohexane
4-bromo-1-tert-butylcyclohexane
2-bromo-1-tert-butylcyclohexane
1-bromo-1-tert-butylcyclohexane
1-bromo-1,1-dimethylethylcyclohexane

Answers

Answer:

1-bromo-1-tert-butylcyclohexane

Explanation:

The parent compound comprises of a cyclohexane to with a tertiary butyl carbon attached. We have been told that the reaction occurs by radical mechanism hence we must recall the order of stability of radicals: tertiary>a secondary> a primary. This implies that the reaction will occur at carbon 1 of the cyclohexane which is a tertiary carbon atom. This leads to the formation of a radical at the 1-position and bromination at that position hence the answer chosen above.

The radical bromination of t-butylcyclohexane favors the formation of 1-bromo-1-tert-butylcyclohexane due to the stability of the tertiary radical intermediate formed during the reaction.

The major product obtained upon radical bromination of t-butylcyclohexane will most likely be 1-bromo-1-tert-butylcyclohexane. This is because radical bromination is highly selective for the most stable radical intermediate, which forms at the position with the greatest number of accessible hydrogens. In the case of t-butylcyclohexane, the tertiary carbon adjacent to the t-butyl group will form the most stable tertiary radical upon abstraction of a hydrogen, which leads to substitution by a bromine atom to form the final product.

Which of the following would be expected to form hydrogen bonds with water? Choose all that apply.

(A) methylamine
(B) N-methylpropanamide
(C) acetaldehyde
(D) cyclopentane
(E) None of the Above

Answers

Answer: C

Explanation:

N-methylpropanamide and methylamine are the compounds that posses hydrogen bonds.

Hydrogen bonds are formed when hydrogen is covalently bonded to a highly electronegative atom. Hydrogen bonds are weaker than covalent bonds but they significantly impact on the chemistry of the molecules in which they occur.

Primary, secondary and tertiary amines all form hydrogen bonds. The compounds that has hydrogen bonds are;

methylamine N-methylpropanamide

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An electron charge e mass m and a positron charge e mass m revolve around their common center of mass under the influence of their attractive coulomb force Find the speed v of each particle in terms of e m k and their separation L

Answers

Answer: v = 2π2 Kme2 Z / nh

Explanation:

The formula for velocity of an electron in the nth orbit is given as,

v = 2π2 Kme2 Z / nh

v = velocity

K = 1/(4πε0)

m= mass of an electron

e = Charge on an electron

Z= atomic number

h= Planck’s constant

n is a positive integer.

Characterize the following alkene as having the E or Z configuration. Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any).
Use wedge-and-dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one.

Answers

Answer:

The alkene has Z configuration as shown in the figure.

The products of bromination of this compound are shown in the figure too with they stereochemistry.

Explanation:

The cis/trans nomenclature system in alkenes is insufficient when there are three or more different substituents in the double bond. In these cases, the Z/E nomenclature system, adopted by IUPAC, is used for all alkenes. Z comes from the German word zusammen which means together and E from the German word entgegen which means opposite. They would be equivalent to the terms cis and trans respectively.

If a molecular configuration is Z or E is determined by the priority rules of Cahn, Ingold and Prelog. For each of the two carbon atoms of the double bond, it is determined individually which of the two substituents has the highest priority. If both substituents of higher priority are on the same side, the arrangement is Z. On the other hand, if they are on opposite sides, the arrangement is E.

Halogenation of alkenes

Takes place with the addition of halogen atoms to the double bond to give a neighborhood dihaloalkane.

Halogenation mechanism

Halogenations are carried out at room temperature and in inert solvents such as carbon tetrachloride. In the mechanism it is observed that the opening of the bromonium ion occurs on the opposite side to the positive bromine that is the leaving group, this causes the halogens to be anti in the final product.

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite.
Equation: Al2O3(s) + NaOH(l) + HF(g) --------> Na3AlF6 + H2O(g)
1. If 15.8 kg of Al2O3(s), 55.4 kg of NaOH(l), and 55.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?

Answers

Answer: The mass of cryolite produced is 65.06 kg

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

For aluminium oxide:

Given mass of aluminium oxide = 15.8 kg = 15800 g     (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 102 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of aluminium oxide}=\frac{15800g}{102g/mol}=154.9mol[/tex]

For the given chemical reaction:

[tex]Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)[/tex]

As all the reactants are getting completely utilized. So, the amount of product can be determined by any 1 of the reactant.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 154.9 moles of aluminium oxide produces = [tex]\frac{2}{1}\times 154.9=309.8mol[/tex] of cryolite

Now, calculating the mass of cryolite by using equation 1, we get:

Moles of cryolite = 309.8 moles

Molar mass of cryolite = 210 g/mol

Putting values in equation 1, we get:

[tex]309.8mol=\frac{\text{Mass of cryolite}}{210g/mol}\\\\\text{Mass of cryolite}=(309.8mol\times 210g/mol)=65058g=65.06kg[/tex]

Hence, the mass of cryolite produced is 65.06 kg

1. In a 100 mL volumetric flask, 30.0 mL of 0.150 M NaOH is combined with 25.0 mL of 0.125 M red dye solution and brought up to volume using DI water. Calculate the final concentrations of both the NaOH and the red dye. Show all your calculations.

Answers

Answer:

[NaOH] = 0.045 M

[red dye] = 0.031 M

Explanation:

The NaOH and the red dye will not react between them, so, the process that it's occurring is only a dilution. The final volume of the solution will be the volume of the flask, which is 100 mL.

The number of moles (n) of each substance will not vary, and it's calculated by the multiplication of the concentration (C) by the volume (V). If 1 is the initial solution, and 2 the diluted:

n1 = n2

C1*V1 = C2*V2

For NaOH:

C1 = 0.150 M

V1 = 30.0 mL

V2 = 100 mL

0.150*30 = C2*100

100C2 = 4.5

C2 = 0.045 M

For the red dye:

C1 = 0.125 M

V1 = 25.0 mL

V2 = 100 mL

0.125*25 = C2*100

100C2 = 3.125

C2 = 0.031 M

Final answer:

The final concentration of NaOH is 0.045 M and the final concentration of the red dye is 0.03125 M.

Explanation:

To calculate the final concentrations of NaOH and the red dye, we can use the formula:

11 = 22

Let's calculate the final concentration of NaOH first:

1 (concentration of NaOH) = 0.150 M

1 (volume of NaOH) = 30.0 mL = 0.030 L

2 (final concentration of NaOH) = ?

2 (final volume of solution) = 100 mL = 0.100 L

Plugging in the values into the formula gives us:

(0.150 M)(0.030 L) = 2 (0.100 L)

Solving for 2, we find that the final concentration of NaOH is 0.045 M.

Now, let's calculate the final concentration of the red dye:

1 (concentration of red dye) = 0.125 M

1 (volume of red dye) = 25.0 mL = 0.025 L

2 (final concentration of red dye) = ?

2 (final volume of solution) = 100 mL = 0.100 L

Using the formula:

(0.125 M)(0.025 L) = 2 (0.100 L)

We find that the final concentration of the red dye is 0.03125 M.

A particular 3.0 L solution contains 2.5 mmol (millimoles) of a substance. What is the concentration of this solution in number of molecules per cubic meter? g

Answers

Answer: 0.83mol/m3

Explanation:

Number of mole = 2.5 mmol = 0.0025mol

Recall

1L = 0.001m3

Therefore 3L = 3x0.001 = 0.003m3

0.003m3 contains 0.0025mol

Therefore, 1m3 will contain = 0.0025/0.003 = 0.83mol

The concentration in mol/m3 is 0.83mol/m3

The concentration of this solution in number of molecules per cubic meter is equal to 5 x 10^23 molecules.

Calculation of molecules

To calculate the amount of molecules present in a given substance, one must use Avogadro's constant, which corresponds to:

                                     [tex]6 \times 10 ^{23}molecules =1mol[/tex]

For this question, we must first calculate the molarity, so that:

                                               [tex]\frac{2.5 \times 10^{-3}mol}{xmol} =\frac{3L}{1L}[/tex]

                                            [tex]x = 0.83 \times 10^{-3} mol/L[/tex]

After that, you must convert the unit of volume to cubic centimeters, that is, milliliters, so that:

                                   [tex]0.83 \times 10^{-3} mol/L = 0.83 mol/ml[/tex]

Finally, just multiply the value obtained by the avogadro constant:

                                [tex]0.83 \times 6\times 10^{23}= 5 \times 10^{23} molecules[/tex]

So, the concentration of this solution in number of molecules per cubic meter is equal to 5 x 10^23 molecules.

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Draw a structure for a compound that meets the following description: An optically active compound, C5H10O with an IR absorption at 1730 cm-1.

Answers

Question: A optically active compound, C5H10O, exhibits IR absorption at 1730 cm-1.

Its carbon NMR shifts are given below. The number of hydrogen's at each carbon, determined by DEPT, is given in parentheses after the chemical shift.

13C NMR: δ 22.6 (3), 23.6 (1), 52.8 (2), 202.4 (1)

Draw the structure of this compound in the window below

Explanation:

3-methylbutanal is a butanal substituted by the methyl group at the 3rd position. It is a volatile constituent in the olive. Also, it is used as a flavoring agent and a plant metabolite, it is also a Saccharomyces cerevisiae metabolite. It is also called as the Isovaleraldehyde organic compound. The liquid is colorless at STP, and also found in very low concentrations. It is also seen to be produced commercially for different use. Mostly used compound reagent in the preparation of pharmaceuticals and pesticides.

Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfred Nobel (1833-1896) founded the Nobel Prizes with a fortune he made by inventing dynamite, a mixture of nitroglycerin and inert ingredients that was safe to handle.

1. Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin ( C 3 H 5 NO 3 3 ) into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide.

2. Suppose 69.0 L of carbon dioxide gas are produced by this reaction, at a temperature of − 5.0 °C and pressure of exactly 1 atm . Calculate the mass of nitroglycerin that must have reacted. Be sure your answer has the correct number of significant digits.

Answers

Answer:

1. 4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)

2. 238 g of C₃H₅(NO₃)₃ has been reacted.

Explanation:

This is the chemical reaction:

4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)

For the second part, let's apply the Ideal Gases Law to find out the moles of CO₂ that were produced.

P . V = n . R . T

1 atm . 69L = n . 0.082 L.atm/mol.K . 268K

(1 atm . 69L) / (0.082 L.atm/mol.K . 268K) = n  → 3.14 moles

In the equation, ratio between nitroglycerin and CO₂ is 12:4.

12 moles of CO₂ were produced by 4 moles of C₃H₅(NO₃)₃

Then, 3.14 moles of CO₂ would have been produced by (3.14  .4) / 12 = 1.04 moles of C₃H₅(NO₃)₃

Let's convert the moles to mass, to find out the mass of nitroglycerin that must have reacted (mol . molar mass)

1.04 mol . 227.08 g/mol = 238 g

What are the strengths, weaknesses, and implications of the precautionary principle as a method for deciding whether a technology should be used?

Answers

Answer:

The precautionary principle would prevent the implementation of technologies that possess a risk to humans, animals, and the environment. The strengths are that it will definately save the population and planet from a new technology that could cause long-term harm.

The weakness is that this principle may inhibit new technologies that are needed to help under-developed countries from preventing diseases. The precautionary principle states that technologies should entirely risk-free.

Explanation:

Coal, which is primarily carbon, can be converted to natural gas, primarily CH4, by the following exothermic reaction: C(s)+2H2(g)⇌CH4(g) A. What is the effect of adding more C to the reaction mixture? B. What is the effect of adding more H2 to the reaction mixture? C. What is the effect of raising the temperature of the reaction mixture? D. What is the effect of lowering the volume of the reaction mixture? E. What is the effect of adding a catalyst to the reaction mixture? F. Which of the following will favor CH4 at equilibrium?
catalyst to the reaction mixture", "raising the temperature of the reaction mixture", "both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture", "adding more C to the reaction mixture"]

Answers

Final answer:

The effects on the equilibrium of converting carbon to methane through the reaction C(s) + 2H2(g) ⇌ CH4(g) depend on changes in the system. Adding more carbon or hydrogen, or lowering volume, favors the formation of methane, while raising the temperature favors reactants. Adding a catalyst speeds up equilibrium but doesn't affect its position.

Explanation:

Effects of Changes on a Chemical Equilibrium

The conversion of carbon to methane in the presence of hydrogen gas is described by the exothermic reaction C(s) + 2H2(g) ⇌ CH4(g). This process is subjected to changes that can affect its equilibrium according to Le Châtelier's principle:

Adding more C to the reaction would shift the equilibrium to the right, favoring the formation of more CH4, as the system tries to counteract the change.Adding more H2 would also shift the equilibrium to the right for the same reason, resulting in more CH4 production.Raising the temperature of an exothermic reaction will shift the equilibrium to the left, favoring the reactants and thus producing less CH4.Lowering the volume of the reaction mixture increases pressure, shifting the equilibrium toward the side with fewer gas molecules, which in this case is the right side, favoring CH4.Adding a catalyst to the reaction mixture does not shift the equilibrium; it only speeds up the rate at which equilibrium is reached.

Regarding which options favor CH4 at equilibrium:

A catalyst does not affect the position of the equilibrium.Raising the temperature favors the reactants (not CH4) in an exothermic reaction.Adding more H2 or lowering the volume favors CH4 formation.Adding more C would also favor CH4 production.

Thus, both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture would favor the formation of CH4 at equilibrium.

Adding more H₂ or lowering the volume of the reaction mixture will shift the equilibrium to favor CH₄ formation. Raising the temperature will decrease CH₄ production and adding a catalyst or more C will not change the equilibrium position of the reaction.

The reaction given, C(s) + 2H₂(g) ⇌ CH₄(g), is governed by the principles of chemical equilibrium. Let’s analyze each part:

Effect of adding more C: Adding more solid carbon does not affect the equilibrium position because solids are not included in the expression for the equilibrium constant. Therefore, the amount of CH₄ will remain unchanged.Effect of adding more H₂: Adding more hydrogen gas will shift the equilibrium to the right, increasing the production of CH₄ due to Le Chatelier’s principle.Effect of raising the temperature: Since the reaction is exothermic, raising the temperature will shift the equilibrium to the left, decreasing the amount of CH₄ formed.Effect of lowering the volume: Lowering the volume will increase the pressure. Since there are fewer moles of gas on the right (1 mole) than on the left (2 moles), the equilibrium will shift to the right, increasing CH₄ production.Effect of adding a catalyst: Adding a catalyst will increase the rate at which equilibrium is reached but will not affect the equilibrium position itself, so the amount of CH₄ remains the same.Which will favor CH₄ at equilibrium: The options “both adding more H₂ to the reaction mixture and lowering the volume of the reaction mixture” will favor CH₄ formation

Many drugs are sold as their hydrochloric salts (R2NH2+Cl−), formed by reaction of an amine (R2NH) with HCl. Part 1 out of 4 Draw the major organic product formed from the formation of acebutolol with HCl. Acebutolol is a β blocker used to treat high blood pressure. Omit any inorganic counterions.Figure:a structure of Acebutolol is shown in the figure

Answers

Answer:

Hi

Acebutolol hydrochloride is the form of the hydrochloride salt of acebutolol, a synthetic derivative of butyranide with a hypotensive and antiarrhythmic activity. Acebutolol acts as a cardioselective beta-adrenergic antagonist with very little effect on bronchial receptors, having intrinsic sympathomimetic properties. Acebutolol is used in ventricular arrhythmias. Other indications may include hypertension, alone or in combination with other drugs. The salt scheme is found in the attached file.

Explanation:

How many simple distillation columns are required to purify a stream containing five components into five 'pure"products? Sketch all possible sequences.

Answers

Answer: one simple distillation column is required to separate the stream into five pure products. With four different flat bottom flask, for collection of the distilled products

Explanation: simple distillation works with the difference in boiling points of the liquid to be separated. For the separation of five different constituent to be possible, we have to know the boiling points of the constituents.

For your understanding, let's define constituents in the liquid to be A, B, C, D, E. And the boiling points increases respectively. Start by heating the liquid to the boiling point of A to extract A. After a while check if the constituents A is still dropping in the flat bottom flask, if it has stopped dropping, it simply means that we have extracted all A constituents in the liquid, label the Flask A. Get another flask to extract constituent B.

Heat the mixture to the boiling point of B, after a while check if constituent B is still dropping in the flat bottom flask, if it has stopped dropping,it means that we have extracted all B constituent in the liquid, label the Flask B. Get another flask for C.

Repeat the same process for C and D.

After Extracting D we don't need to distillate E because we already have a pure form of E inside to the conical flask.

SEE PICTURE TO UNDERSTAND WHAT A SIMPLE DISTILLATION LOOKS LIKE

Assuming the boiling point increased by 2 °C, what is the approximate molality of NaCl when the video ends? Kbp(water) = 0.512 °C/(mol/kg)

Answers

Answer : The molality of NaCl is, 1.95 mol/kg

Explanation :

Formula used for Elevation in boiling point :

[tex]\Delta T_b=i\times k_b\times m[/tex]

where,

[tex]\Delta T_b[/tex] = change in boiling point = [tex]2^oC[/tex]

[tex]k_b[/tex] = boiling point constant  for water = [tex]0.512^oC/m[/tex]

m = molality

i = Van't Hoff factor = 2 (for electrolyte)

The dissociation [tex]NaCl[/tex] will be,

[tex]NaCl\rightarrow Na^++Cl^{-}[/tex]

So, Van't Hoff factor = Number of solute particles = [tex]Na^++Cl^{-}[/tex] = 1 + 1 = 2

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

[tex]2^oC=2\times (0.512^oC/m)\times m[/tex]

[tex]m=1.95mol/kg[/tex]

Therefore, the molality of NaCl is, 1.95 mol/kg

The molality of NaCl is, 1.95 mol/kg

The calculation is as follows;

[tex]\Delta T _ b = i \times k_b \times m\\\\2^{\circ}C = 2\times (0.512^{\circ} C/m) \times m[/tex]

m = 1.95 mol/kg

Here m represent the molarity

[tex]\Delta T_b[/tex] represents a change in boiling point i.e. [tex]2^{\circ}C[/tex]

[tex]k_b[/tex] represents the constant boiling point for water  

i = Van't Hoff factor = 2 (for electrolyte)

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Element "Z" has 2 naturally occurring isotopes with the following masses and natural abundances:

a. Z-45 44.8776 amu 32.88%
b. Z-47 46.9443 amu 67.12%
c. Calculate the atomic mass of Z.

Answers

Answer: the atomic Mass of Z is 46.26

Explanation:Please see attachment for explanation

Consider a bathtub holding a system of water (where the boundaries of the tub are the boundaries of the system) under the following conditions:
(a) The bathtub has a closed drain and is already full of water at the start of analysis. No steam rises from the water. You want to examine heat transfer loss from the water contained in the tub over time.
(b) The bath tub drain is partially open and hot water is continually added to keep the tub filled and the temperature constant. You want to examine the flow rates required to maintain the temperature of the water in the tub.
For each system (a) and (b), determine if it is characterized as open or closed and state your reasoning.

Answers

Answer:

a)  the system is closed

b) the system is open

Explanation:

a) the system can be characterised as closed system. If we define the boundaries of this system as an envelope that contains the bathtub, then there will not be any mass transfer to the environment through the boundary → then the system is closed ( the mass of the system remains constant)

b) on the other hand , taking the same system as in the first case, if the hot water is continually added to the tub , then there is water flow through the boundary (from the environment to the system) → then the system is open ( the mass of the system can be altered)

18. Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.

Answers

Answer:

              Ethanol is completely miscible due to presence of Hydrogen bonding.

              Ethanethiol is partially miscible due to absence of Hydrogen Bonding.

Explanation:

                     The miscibility of liquids depend upon the intermolecular interactions between the two liquids. The stronger the intermolecular interactions the more miscible will be the liquids.

Among the two given examples, Ethanol is more miscible in water because it exhibits hydrogen bonding which is considered the strongest intermolecular interaction. Hydrogen bonding occurs when the hydrogen atom is bonded to more electronegative atoms like Fluorine, Oxygen and Nitrogen. In this way the hydrogen atom gets partial positive charge and the electronegative atom gets partial negative charge. Hence, these partial charges results in attracting the opposite charges on other surrounding atoms.

While, in case of Ethanethiol the hydrogen atom is not bonded to any high electronegative atom hence, there will be no hydrogen bonding and therefore, there will be less interactions between the neighbour atoms.

The difference in solubility between ethanol and ethanethiol in water can be explained by the difference in polarity and the ability to form hydrogen bonds with water molecules.

Ethanol (C2H5OH) has a hydroxyl (-OH) group, which is polar and can form hydrogen bonds with water molecules. The oxygen atom in the hydroxyl group has a partial negative charge due to its higher electronegativity, and the hydrogen atoms in water have a partial positive charge. This allows for the formation of hydrogen bonds between the ethanol molecules and water molecules, which overcomes the energy required to break the hydrogen bonding network in water. As a result, ethanol can mix with water in any proportion, making it completely miscible.

On the other hand, ethanethiol (C2H5SH) contains a thiol (-SH) group instead of a hydroxyl group. Although the thiol group can still form hydrogen bonds, the sulfur atom is less electronegative than the oxygen atom in ethanol. This results in a weaker hydrogen bond with water molecules. Additionally, the larger size of the sulfur atom compared to the oxygen atom leads to a greater Van der Waals radius, which can disrupt the hydrogen bonding network in water more than the smaller oxygen atom. As a result, ethanethiol is not as soluble in water as ethanol, and its solubility is limited to 1.5 grams per 100 milliliters of water.

In summary, the greater polarity and hydrogen bonding capability of ethanol with water molecules make it completely miscible, while the weaker hydrogen bonding and disruption of water's hydrogen bonding network by ethanethiol limit its solubility in water.

You recrystallized acetanilide from water and found out that the final mixture contains 90 % of acetanilide by weight while the rest is water.
Calculate the molar percentage of acetanilide in the mixture if the molecular weight of acetanilide is 135 g/mol and the molecular weight of water is 18 g/mol. Round your answer to the nearest tenth, e.g. 1.1%.

Answers

Answer: 54.5%

Explanation:Please see attachment for explanation

Almost all elements and compounds can be a solid liquid and gas under special circumstances. What do you think might change the state of an element or compound?​

Answers

Answer:

The state of the element or compound decided by two forces mainly:

The force of the attraction

The Kinetic(thermal) energy between the molecules of the element.

Explanation:

Kinetic Energy depends upon the temperature , and on increasing the temperature the kinetic energy also increases.

Force of attraction between the particles is increased by applying pressure to the substance.

When the substance is in solid state then , the force of attraction between the particles is much more then the thermal energy of the particles.

The thermal energy or kinetic energy reduces the force of attraction between the particles.Hence if the kinetic energy is increased then the substances goes from solid to liquid state.

Thus on, increasing the temperature the substance changes the state from solid to liquid

[tex]Solid\rightleftharpoons Liquid[/tex]

If you increase further , the kinetic energy then the substance goes from liquid to gaseous state again.

[tex]liquid\rightleftharpoons Gases[/tex]

Role of Kinetic energy : The kinetic energy separate the particles further to large distance.Hence they become far apart result in change in state.

Affect of increasing pressure produce the reverse products.

[tex]Gas\rightleftharpoons liquid\rightleftharpoons Solid[/tex]

How many grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel?

Answers

0.87 grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 2.32 g of 5.5% w/w gel

Let x represent the number of grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a (x + 1.45)g of 5.5% w/w gel

Hence:

(x * 8%) + (1.45 * 4%) = (x + 1.45) * 5.5%

0.08x + 0.058 = 0.055x + 0.07975

0.025x = 0.02175

x = 0.87 g

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Final answer:

To prepare a 5.5% w/w progesterone gel, you should mix 4.383 g of an 8% progesterone gel with 1.45 g of a 4% progesterone gel.

Explanation:

A weight/weight percent (w/w%) refers to the amount of a substance (in this case, progesterone) contained in a total solution. The equation we want to use here is 'mass of solute/mass of solution = concentration'. We know the concentrations of the individual gels and that of the final gel we want to achieve. We need to find out the amount of 8% gel to mix with 1.45 g of 4% gel to get a 5.5% gel.

This would be solved as follows:

Let's denote the weight of the 8% gel as x. The contribution of progesterone from the 8% gel is 0.08x, and from the 4% gel is 0.04*1.45 = 0.058 g. The total weight of the final gel solution is x+1.45 g. The total amount of progesterone is 0.08x + 0.058 g. To get a 5.5% solution, we set up the equation 0.08x + 0.058 / 1.45 + x = 0.055. Solving for x gives us 4.383 g.

Therefore, the answer is that 4.383 g of an 8% progesterone gel should be mixed with 1.45 g of the 4% gel to obtain a 5.5% gel.

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