A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 32.3 mg produced 87.7 mg of CO2 and 18.0 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.

Answers

Answer 1

Answer:

Empirical formula: C₅H₅O

Molecular formula: C₁₀H₁₀O₂

Explanation:

When a compound containing C, H and O elements is combusted, the general reaction is:

CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O

Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.

Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =

2.0x10⁻³ moles of CO₂ = moles C

Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =

1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H

The mass of the moles of C and H are:

2x10⁻³ moles C ₓ (12g / mol) = 0.024g C

2x10⁻³ moles H ₓ (1g / mol) = 0.002g H

Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O

Moles are:

0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O

Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:

C: 2.0x10⁻³ / 4x10⁻⁴ = 5

H: 2.0x10⁻³ / 4x10⁻⁴ = 5

O:  4x10⁻⁴ / 4x10⁻⁴ = 1

Thus, empirical formula is:

C₅H₅O

The molar mass of the empirical formula is:

12×5 + 1×5 + 16×1 = 81g/mol

As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:

C₁₀H₁₀O₂


Related Questions

1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice? 2. Do you think the results would have been more accurate if a different type of acid or base were used in the standardization? Why, or why not? 3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard?

Answers

Answer:

See explanation

Explanation:

The calculated concentration of the sodium hydroxide is;

Number of moles= mass/molar mass = 1g/40gmol-1 = 0.025 moles

Concentration= number of moles/volume= 0.025×1000/250 = 0.1 M

This calculated concentration will be different from the molarity of NaOH obtained by standardization with acid. The result will not be more accurate if a different acid is used for the standardization this is because sodium hydroxide is deliquescent and absorbs moisture thereby leading to inaccuracy in the calculated molarity.

Any substance that must be used as a primary standard must not absorb moisture, it must be stable and it must be a substance in its pure form.

A student mixes wants to prepare 24.1 mmol of benzamide from benzoyl chloride and NH4OH. If the student uses excess 15 M NH4OH, how many mL of Benzoyl chloride must be used

Answers

Answer:

2.81mL

Explanation:

Based on the reaction:

C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl

Benzoyl chloride + ammonia → Benzamide

1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.

Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.

As molar mass of benzoyl chloride is 141g/mol, mg you require are:

mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.

to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:

3.3981g × (1mL / 1.21g) =

2.81mL

Identify a process that is NOT reversible. A. melting of steel B. freezing water C. melting of ice D. frying an egg E. deposition of carbon dioxide (gas to solid)

Answers

I’m pretty sure it would be D. Frying an egg

A process that is not a reversible reaction is frying an egg.

What are reversible reactions?

Reversible reactions are those reactions in which product will again change into the reactant.

Melting of steel and ice are reversible reaction as after cooling again we get the original state of steel and ice.Freezing of water is also reversible reaction as at normal temperature we get the original state of water.Deposition of carbon dioxide is also a reversible reaction.Frying an egg is a non reversible reaction as after frying an egg we didn't get the original egg again.

Hence option (D) is correct.

To know more about reversible reaction, visit the below link:

https://brainly.com/question/1495850

what is the meaning of the word tetraquark?​

Answers

Answer:

A tetraquark in physics is an exotic meson composed of four valence quarks.

Explanation:

It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.

Hope it helps.

The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g of latic acid is used to make a solution with a volume of 1.00l what is the concentration of hydronium ion in the solution

Answers

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Limiting reagent problem. How many grams of H2O is produced from 40.0 g N2O4 and 25.0 g N2H4. N2O4 (l) + 2 N2H4 (l) → 3 N2 (g) + 4 H2O(g)

Answers

Answer:

28.13 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol

Mass of N2O4 from the balanced equation = 1 x 92 = 92g

Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol

Mass of N2H4 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72 g

Summary:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4.

Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.

From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.

Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.

Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.

In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.

The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:

From the balanced equation above,

64 g of N2H4 reacted to produce 72 g of H2O.

Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.

Therefore, 28.13 g of H2O were obtained from the reaction.

After heating a sample of hydrated CuSO4, the mass of released H2O was found to be 2.0 g. How many moles of H2O were released if the molar mass of H2O is 18.016 g/mol

Answers

Answer:

0.1110 mol

Explanation:

Mass = 2g

Molar mass = 18.016 g/mol

moles = ?

These quantities are realted by the following equation;

Moles = Mass / Molar mass

Substituting the values of the quantities and solving for moles, we have;

Moles = 2 / 18.016 = 0.1110 mol

can I get some urgent help please?​

Answers

Answer:

hi here goes your answer

Explanation:

iv. The lower the PH, the weaker the base

below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (g) ΔH1°= 449.2 kJ R2 O3 (g) + Cl (g) → O2 (g) + ClO (g) ΔH2° = -126 kJ R3 ClO (g) + O (g) → O2 (g) + Cl (g) ΔH3°= -268 kJ

Answers

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

ΔH = -338.8kJ

What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid

Answers

Answer:

2-ethoxy-2-methylpropan-1-ol

Explanation:

On this reaction, we have an "epoxide" (2-methyl-1,2-epoxypropane). Additionally, we have acid medium (due to the sulfuric acid [tex]H_2SO_4[/tex]). The acid medium will produce the hydronium ion ([tex]H^+[/tex]). This ion would be attacked by the oxygen of the epoxide. Then a carbocation would be produced, in this case, the most stable carbocation is the tertiary one. Then an ethanol molecule acts as a nucleophile and will attack the carbocation. Finally, a deprotonation step takes place to produce 2-ethoxy-2-methylpropan-1-ol.

See figure 1

I hope it helps!

Calculate the pH of a buffer solution obtained by dissolving 18.0 g of KH2PO4(s) and 35.0 g of Na2HPO4(s) in water and then diluting to 1.00 L.

Answers

Answer:

pH of the buffer is 7.48

Explanation:

The H₂PO₄⁻/HPO₄²⁻ buffer has a pKa of 7.21. You can find pH of this buffer following H-H equation:

pH = pKa + log [A⁻] / [HA]

pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]

Where [] represents molarity of each specie of the buffer and, as volume is 1.00L, also represents its moles.

Thus, to find pH of the buffer we need to calculate moles of each specie, thus

Moles of 18.0g of KH₂PO₄(Molar mass: 136.086g/mol) = moles of H₂PO₄⁻ are:

18.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.132 moles of KH₂PO₄= H₂PO₄⁻

Moles of 35.0g of Na₂HPO₄(Molar mass: 141.96g/mol) = moles of HPO₄²⁻ are:

35.0g Na₂HPO₄ ₓ (1mol / 141.96g) = 0.2465 moles of Na₂HPO₄= HPO₄²⁻

Replacing in H-H equation:

pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]

pH = 7.21 + log [0.2465] / [0.132]

pH = 7.48

pH of the buffer is 7.48

Select the correct answer. A certain reaction has this form: aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the half-life of this reaction? A. 23.33 seconds B. 0.043 minutes C. 0.0043 seconds D. 23.33 minutes E. 1680 minutes

Answers

Answer:

[tex]\large \boxed{\text{D. 23.34 min}}[/tex]

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the  half-life

[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]

The half life of the reaction is 23.33 minutes.

We know that for a first order reaction;

ln[A]t = ln[A]o  - kt

A plot of ln[A]t  against time (t) will yield a straight line graph with a slope of -k.

From the question, the slope is -2.97 x 10-2 min-1.

So, -2.97 x 10-2 min-1 = - k

k = 2.97 x 10-2 min-1

The half life of a first order reaction is obtained from;

t1/2 = 0.693/k

t1/2 = 0.693/2.97 x 10-2 min-1

t1/2 = 23.33 minutes

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A sample of neon gas at a pressure of 0.609 atm and a temperature of 25.0 °C, occupies a volume of 19.9 liters. If the gas is compressed at constant temperature to a
volume of 12.7 liters, the pressure of the gas sample will be
atm.​

Answers

Answer:

The pressure of the gas sample will be  0.954 atm.​

Explanation:

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

To determine the change in pressure or volume during a transformation at constant temperature, the following is true:

P1 · V1 = P2 · V2

That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.

In this case:

P1= 0.609 atmV1= 19.9 LP2=?V2= 12.7 L

Replacing:

0.609 atm* 19.9 L= P2* 12.7 L

Solving:

[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]

P2= 0.954 atm

The pressure of the gas sample will be  0.954 atm.​

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 

Answers

Answer:

ΔH = [tex]q_{p}[/tex]

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

[tex]q_{com}[/tex] = [tex]q_{p}[/tex]

where

[tex]q_{p}[/tex] = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

[tex]q_{com}[/tex]  = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is [tex]q_{p}[/tex]

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = [tex]q_{p}[/tex] + 0

==> ΔH = [tex]q_{p}[/tex]

what is the difference between acidic and basic protein​

Answers

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10 genes.

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10

Explanation:

A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be

Answers

Answer:

25.99mL is the volume internal volume of the flask

Explanation:

To complete the question:

The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask

The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.

To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:

Mass water = Mass filled flask - Mass of clean flask

Mass water = 60.167g - 34.232g

Mass water = 25.935g of water.

To convert this mass to volume:

25.935g × (1mL / 0.997992g) =

25.99mL is the volume internal volume of the flask

cetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce of water.

Answers

Answer:

0.60 mol

Explanation:

There is some info missing. I think this is the original question.

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.

Step 1: Given data

Moles of water required: 1.5 mol

Step 2: Write the balanced equation

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water

The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol

Need help finding major products

Answers

Answer:

Explanation:

RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻

C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅

C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻

In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺

CH₃CH₂CH₂CH₂CH₂⁺ ⇒  CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

Hence option D is correct .

b )

In the second case carbocation produced is

CH₃CH₂CH₂CH⁺CH₃

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

The product formed is same as in case of first

Option B is correct

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:
P2O5 (s) + H2O (l) =H3PO4 (aq)
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.

Answers

Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

2. We'll begin by writing the balanced equation for the reaction. This is given below:

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

After the reaction between sodium borohydride and the ketone is complete, the reaction mixture is treated with water and H2SO4 to produce the desired alcohol. Explain the reaction by clearly indicating the source of the hydrogen atom that ends up on the oxygen

Answers

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Identify a homogeneous catalyst:

a. SO2 over vanadium (V) oxide
b. H2SO4 with concentrated HCl
c. Pd in H2 gas
d. N2 and H2 catalyzed by Fe
e. Pt with methane

Answers

Answer:

b, H2SO4 with HCl, as they are both liquid acids

Which, if any, of the two molecular geometries is likely to be polar if the outer atoms are identical and there is a difference in electronegativity between the central atom and the outer atoms

Answers

Answer:

Polar molecules are not symmetrical

Explanation:

Even though the structures of the molecules involved were not shown in the question, but I will proceed to give a general explanation of the conditions that describe a polar molecule.

First of all, symmetrical molecules are non-polar and asymmetrical molecules are polar. This is the reason why CF4 will be a nonpolar molecule but H2O will be a polar molecule. Some symmetrical molecules may may posses polar bonds or dipoles but these dipoles eventually cancel out since the molecule is symmetrical in nature.

Summarily, if a molecule possess the same type of atoms attached to the central atom with some symmetry axes, like C3, C4 etc., we will end up with a non polar molecule but if we have a nonplanar molecule, then we will end up finding it to be polar.

Why was it important to establish the Clean Air Act?

Answers

Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.

Explanation:

Answer:

Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.

What mass of Si, in grams, can be produced from the reaction below starting with 225 g SiCl4 and 101 g Mg? SiCl4 + Mg  Si + MgCl2 Given: 1 mol SiCl4 = 169.8963 g SiCl4 1 mol Mg = 24.3050 g Mg 1 mol Si = 28.0855 g Si

Answers

Answer:

[tex]m_{Si}=37.2gSi[/tex]

Explanation:

Hello,

In this case, for the undergoing balanced chemical reaction:

[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]

We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:

[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]

Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:

[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]

Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:

[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]

Best regards.

A sample of N2(g) was collected over water at 25 oC and 730 torr in a container with a volume of 340 mL. The vapor pressure of water at 25 oC is 23.76 torr. What mass of N2 was collected

Answers

Answer:

0.36 g of N2.

Explanation:

The following data were obtained from the question:

Temperature (T) = 25 °C

Volume (V) = 340 mL

Measured pressure = 730 torr

Vapour pressure = 23.76 torr

Mass of N2 =..?

First, we shall determine the true pressure of N2. This can be obtained as follow:

Measured pressure = 730 torr

Vapor pressure = 23.76 torr

True pressure =..?

True pressure = measured pressure – vapor pressure

True pressure = 730 – 23.76

True pressure = 706.24 torr.

Converting 706.24 torr to atm, we have:

760 torr = 1 atm

Therefore,

706.24 torr = 706.24 / 760 = 0.929 atm

Next, we shall convert 340 mL to L. This is illustrated below:

1000 mL = 1 L

Therefore,

340 mL = 340/1000 = 0.34 L

Next, we shall convert 25 °C to Kelvin temperature. This is illustrated below:

Temperature (K) = Temperature (°C) + 273

T(K) = T (°C) + 273

T (°C) = 25 °C

T(K) = 25 °C + 273

T (K) = 298 K

Next, we shall determine the number of mole of N2. This can be obtained as follow:

Pressure (P) = 0.929 atm

Volume (V) = 0.34 L

Temperature (T) = 298 K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =...?

PV = nRT

0.929 x 0.34 = n x 0.0821 x 298

Divide both side by 0.0821 x 298

n = (0.929 x 0.34 ) /(0.0821 x 298)

n = 0.0129 mole

Finally, we shall determine the mass of N2 as shown below:

Mole of N2 = 0.0129 mole

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

Mass of N2 =.?

Mole = mass /Molar mass

0.0129 = mass of N2/ 28

Cross multiply

Mass of N2 = 0.0129 x 28

Mass of N2 = 0.36 g

Therefore, 0.36 g of N2 was collected.

An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.010 mol of Acetic acid. What molarity of sodium acetate will you need to add to the solution

Answers

Answer:

Molarity of sodium acetate you will need to add is 0.0324M

Explanation:

Assuming volume of the buffer is 1L.

The pH of a buffer can be determined using Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is pKa of the weak acid,  [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid

Replacing for the acetic buffer (pKa = 4.76):

pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]

As you have 0.010 moles of acetic acid in 1L:

[Acetic Acid] = 0.010mol / 1L = 0.010M

And you require a pH of 5.27:

5.27 = 4.76 + log [Sodium Acetate] / [0.010M]

0.51 = log [Sodium Acetate] / [0.010M]

10^0.51 = [Sodium Acetate] / [0.010M]

3.236 =  [Sodium Acetate] / [0.010M]

3.236 [0.010M] = [Sodium Acetate]

0.0324M = [Sodium Acetate]

Molarity of sodium acetate you will need to add is 0.0324M

2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations

Answers

Answer:

a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)

workdone (w) = -8442.6 J  ≈ -8.443 KJ

heat transferred (q) of the ideal gas = - w

q = 8.443 KJ

b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0

the workdone(w) in the ideal gas= - 4567.5 J  ≈ - 4.57 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

Explanation:

given

mole of an ideal gas(n) = 2.5 mol

Temperature (T) = 20°C

= (20°C + 273) K  = 293 K

Initial pressure of the ideal gas(P₁) = 20 atm

Final pressure of the ideal gas(P₂) = 5 atm.

2) (a)for adiabatic reversible process,

note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.

Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]

= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]

= 6090.01 J × [-1.3863]

= -8442.6 J  ≈ -8.443 KJ

So, the work done (w) of ideal gas = -8.443 KJ

For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-8.443 KJ)

q = 8.443 KJ

heat transfer (q) of the ideal gas = 8.443 KJ

(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.

Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )

= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]

= - 6090.01 J × 0.75

= - 4567.5 J  ≈ - 4.57 KJ

∴work done(w) of an ideal gas = - 4.57 KJ

For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0

From first law of thermodynamics:-

U = q + w

0 = q + w

q = - w

q = - (-4.5675 KJ)

q = 4.5675 KJ

the heat transfer (q) of an ideal gas = 4.5675 KJ

11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KCl solution? Are these results what you would expect? Why?

Answers

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)?

Answers

Answer:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

Explanation:

Hello,

In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:

- The molar mass of the solute, in order to convert from moles of solute to grams of solute.

- The density of solution, to convert from volume of solution to mass of solution.

For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:

[tex]w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%[/tex]

Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.

Regards.

For the w/w% of the solution, information about the molecular mass of the solute, and density of the solution has been required.

Molarity can be defined as the moles of the solute per liter of the solution. The molarity can be used for the determination of the weight of the solute, by the information about the molecular weight of the compound.

Thus, for the w/w% of the solution, the weight of the solute has been determined with information about the molecular mass of the solute.

The weight of the solvent has been determined with the density of the solution. The density has been defined as the mass per unit volume.

Thus, for the w/w% of the solution, the weight of the solvent has been determined by the density of the solution.

For more information about the w/w% of the solution, refer to the link:

https://brainly.com/question/12369178

How to do this
Q1 and Q2
Only want to know how to find molecular formula

Answers

Answer:

Question 1

A. Empirical formula is C8H8O3

B. Molecular formula is C8H8O3

Question 2.

A. Empirical formula is CH2

B. Molecular formula is C4H8

Explanation:

Question 1:

A. Determination of the empirical formula:

Carbon (C) = 63.2%

Hydrogen (H) = 5.26%

Oxygen (O) = 31.6%

Divide by their molar mass

C = 63.2/12 = 5.27

H = 5.26/1 = 5.26

O = 31.6/16 = 1.975

Divide by the smallest

C = 5.27/1.975 = 2.7

H = 5.26/1.975 = 2.7

O = 1.975/1.975 = 1

Multiply through by 3 to express in whole number

C = 2.7 x 3 = 8

H = 2.7 x 3 = 8

O = 1 x 3 = 3

Therefore, the empirical formula for the compound is C8H8O3

B. Determination of the molecular formula of the compound.

From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules.

Now from the question given, we were told that 1 molecule of the compound has a mass of 2.53×10¯²² g.

Therefore, 6.02×10²³ molecules will have a mass of = 6.02×10²³ x 2.53×10¯²² = 152.306 g

Therefore, 1 mole of the compound = 152.306 g

The molecular formula of the compound can be obtained as follow:

[C8H8O3]n = 152.306

[(12x8) + (1x8) + (16x3)]n = 152.306

[(96 + 8 + 48 ]n = 152.306

152n = 152.306

Divide both side by 152

n = 152.306/152

n = 1

The molecular formula => [C8H8O3]n

=> [C8H8O3]1

=> C8H8O3

Question 2:

A. Determination of the empirical formula of the compound.

Mass sample of compound = 0.648 g

Carbon (C) = 0.556 g

Mass of Hydrogen (H) = mass sample of compound – mass of carbon

Mass of Hydrogen (H) = 0.648 – 0.556

Mass of Hydrogen (H) = 0.092 g

Thus, the empirical formula can be obtained as follow:

C = 0.556 g

H = 0.092 g

Divide by their molar mass

C = 0.556/12 = 0.046

H = 0.092/1 = 0.092

Divide by the smallest

C = 0.046/0.046 = 1

H = 0.092/0.046 = 2

Therefore, the empirical formula of the compound is CH2.

B. Determination of the molecular formula of the compound.

Mole of compound = 0.5 mole

Mass of compound = 28.5 g

Molar mass of compound =.?

Mole = mass /Molar mass

0.5 = 28.5/ Molar mass

Cross multiply

0.5 x molar mass = 28.5

Divide both side by 0.5

Molar mass = 28.5/0.5 = 57 g/mol

Thus, the molecular formula of compound can be obtained as follow:

[CH2]n = 57

[12 + (1x2)]n = 57

14n = 57

Divide both side by 14

n = 57/14

n = 4

Molecular formula => [CH2]n

=> [CH2]4

=> C4H8.

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