Many free radicals combine to form molecules that do not contain any unpaired electrons. The driving force for the radical–radical combination reaction is the formation of a new electron‑pair bond. Consider the formation of hydrogen peroxide. 2OH(g)⟶H2O2(g) Write Lewis formulas for the reactant and product species in the chemical equation. Include nonbonding electrons.

Answers

Answer 1

Answer:

In the attached image the Lewis equation is shown where it is shown how two oxygens react with two hydrogens to meet the octet of the electrons.

Explanation:

Hydrogen peroxide is one of the most named chemicals since it is not only sold as "hydrogen peroxide" in pharmacies but it is also one of the great weapons of immune defense cells to defend ourselves against anaerobic bacteria.

The disadvantage of this compound is that when dividing it forms free oxygen radicals that are considered toxic or aging for our body.

Many Free Radicals Combine To Form Molecules That Do Not Contain Any Unpaired Electrons. The Driving
Answer 2

In the attached image below, you will see the Lewis equation is shown there. There, you will see how two oxygens react with two hydrogens to come about the octet of the electrons.

When two or more atoms bond with each other, they often form a molecule. When two hydrogens and an oxygen share electrons through covalent bonds, a water molecule is formed.

The octet rule is known as when most atoms want to gain stability in their outer most energy level by filling themselves that is the S and P orbitals of the highest energy level with eight electron.

HOOH is the compound  that is form. It is called Hydrogen peroxide. This because it is has reactive oxygen species and the simplest peroxide.

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Many Free Radicals Combine To Form Molecules That Do Not Contain Any Unpaired Electrons. The Driving
Many Free Radicals Combine To Form Molecules That Do Not Contain Any Unpaired Electrons. The Driving
Many Free Radicals Combine To Form Molecules That Do Not Contain Any Unpaired Electrons. The Driving

Related Questions

A volumetric flask contains 25.0 mL of a 14% m/V sugar solution. If 2.5 mL of this solution is added to 22.5 mL of distilled water, what is the % m/V of the new solution.

Answers

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

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Determine the rate of a reaction that follows the rate law:
rate = k[A]”[B]", where:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1

Answers

Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

What is meant by rate of a reaction ?

Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.

Here,

The concentration of A, [A] = 1 M

The concentration of B, [B] = 3 M

The partial order with respect to A, m = 2

The partial order with respect to B, n = 1

The rate constant of the reaction, k = 1.5

The rate of the reaction,

r = k[A]^m [B}^n

r = 1.5 x 1² x 3

r = 4.5 mol L⁻¹s⁻¹

Hence,

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

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Write the following isotope in nuclide notation: oxygen-14

Answers

Answer:

[tex]14\\8[/tex]O

Explanation:

The top number always represents the mass number.

The bottom number always represents the atomic number.

The element always goes after the numbers.

If charge is present, that comes after the element.

Why does a new period start on the periodic table, instead of the row continuing? A. A new period starts when a new energy shell starts. B. A new period starts when a new neutron cycle starts. C. None of these D. It is based on how many protons it has.

Answers

Answer:

B

Explanation:

All the elements in a period have valence electrons in the same shell. The number of valence electrons increases from left to right in the period. When the shell is full, a new row is started and the process repeats.

A new period starts when a new neutron cycle starts. Hence, option B is correct.

What is the period in the periodic table?

A period in the periodic table is a row of chemical elements. All elements in a row have the same number of electron shells.

All the elements in a period have valence electrons in the same shell.

The number of valence electrons increases from left to right in the period.

When the shell is full, a new row is started and the process repeats.

Hence, option B is correct.

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Average Molarity for HCl is .391
Average Molarity for NaOH is .0962

Volume for HCl is:
Trial 1 Your Answer: 14mL
Trial 2 Your Answer: 14mL
Trial 3 Your Answer: 14mL

Volume for NaOH is:
Trial 1: 34.26mL
Trial 2: 33.48mL
Trial 3: 33.84mL

Entry # mass tablet(g) mass antacid(g) Vol HCl(mL) Vol NaOH(mL)
#1: 1.515 0.9010 14.00 34.26
#2: 1.452 0.8370 14.00 33.48
#3: 1.443 0.8280 14.00 33.84

I need help finding the mmoles HCl/mg please.

Answers

Answer:

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl)  is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

#1: 0.00144 mmolHCl/mg Sample#2: 0.00155 mmolHCl/mg Sample#3: 0.00153 mmolHCl/mg Sample

Recall that when a reaction is at equilibrium, the forward and reverse reactions occur at the same rate. To illustrate this idea, consider the reaction of A (small, red spheres) and B (large, blue spheres) to form AB.

A+B ⇌ AB

Notice that the reaction never stops. Even after several minutes, there is A and B left unreacted, and the forward and reverse reactions continue to occur. Also note that amounts of each species (i.e., their concentrations) stay the same.

Required:
What is the value of the equilibrium constant for this reaction?
Assume each atom or molecule represents a 1 M concentration of that substance.

Answers

Answer:

Equilibrium constant Kc = [x]² / [A - x] [B - x]

Explanation:

The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium

ie Kc = [products] / [reactants].

The balanced equation of the reaction is given as : A + B ⇄ AB

At the beginning of the reaction,

Initial concentration I = A = 1M

                                       B = 1M

                                      AB = 0M

After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become

                                         C = reactant A = [A - x] M

                                                 rectant B =   [B - x] M

                                              Product AB =  [x] [x] M

At equilibrium, the concentrations are,

                                            E  = rectant A = [A - x] M

                                                   reactant B = [B - x] M

                                                   product AB = [x]² M

therefore , the equilibrium constant, Kc  = [products]/[reactants]

                                                                   = [x]² / [A - x] [B - x]

what is the polarity of black pepper

Answers

Answer:

Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.

Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.

It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.

If an electron has a principal quantum number (n) of 7 and an angular momentum quantum number (l) of 1, the subshell designation is ________

Answers

Answer:

7p

Explanation:

principal quantum number is 7

n=7( principle shell)

angular momentum quantum number gives sub shell

l = 1 means it is p orbital

so answer is 7p orbital

What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12

Answers

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

pH = 3.39

A mixture of krypton and nitrogen gases, at a total pressure of 711 mm Hg, contains 11.7 grams of krypton and 4.10 grams of nitrogen. What is the partial pressure of each gas in the mixture

Answers

Answer:

A. Partial pressure of krypton, Kr is 346.97 mmHg

B. Partial pressure of nitrogen, N2 is 364.03 mmHg.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Total pressure (Pt) = 711 mmHg

Mass of Kr = 11.7 g

Mass of N2 = 4.10 g

Partial pressure of Kr =..?

Partial pressure of N2 =...?

Step 2:

Determination of the number of mole of krypton, Kr and nitrogen, N2. This is illustrated below:

Molar mass of Kr = 84g/mol

Mass of Kr = 11.7g

Mole of Kr =?

Mole = mass /Molar mass

Mole of Kr = 11.7/84 = 0.139 mole

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

Mass of N2 = 4.10g

Mole of N2 =?

Mole = mass /Molar mass

Mole of N2 = 4.1/28 = 0.146 mole

Step 3:

Determination of the mole fraction for each gas. This is illustrated below:

Mole of Kr = 0.139 mole

Mole of N2 = 0.146 mole

Total mole = 0.139 + 0.146 = 0.285 mole

Mole fraction of Kr = mol of Kr/total mol

Mole fraction of Kr = 0.139/0.285

Mole fraction of Kr = 0.488

Mole fraction of N2 = mol of N2/total mol

Mole fraction of N2 = 0.146/0.285

Mole fraction of N2 = 0.512

A. Determination of the partial pressure of krypton, Kr.

This is illustrated below:

Total pressure (Pt) = 711 mmHg

Mole fraction of Kr = 0.488

Partial pressure of Kr =..?

Partial pressure = mole fraction x total pressure

Partial pressure of Kr = 0.488 x 711

Partial pressure of Kr = 346.97 mmHg

B. Determination of the partial pressure of nitrogen, N2

This is illustrated below:

Total pressure (Pt) = 711 mmHg

Mole fraction of N2 = 0.512

Partial pressure of N2 =?

Partial pressure = mole fraction x total pressure

Partial pressure of N2 = 0.512 x 711

Partial pressure of N2 = 364.03 mmHg

Therapeutic drugs generally need to have some hydrophobic and hydrophilic components to be able to effectively reach their target organs and tissues given there are aqueous and nonaqueous parts of the body. The degree to which a compound is hydrophobic and hydrophilic can be determined by measuring its relative solubility in water and octanol, C8H17OH, and water. To do this, a sample of the compound is added to a mixture of water and octanol and mixed well. Water and octanol are immiscible so after the mixture settles, the concentration of the compound in water and the concentration of the compound in octanol is measured. The ratio of the concentrations is called the partition ratio:

Answers

The question is incomplete as some part is missing:

concentration in octanol Partition Ratio = concentration in water

a) What are the intermolecular forces of attraction between octanol molecules? Explain.

b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.

c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.

d) Would nonane (figure 2) be more soluble in water or octanol? Explain.

e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.

f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.

Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C

Answer:

1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.

2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).

3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.

4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.

5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.

What is the mass of 7.68 x 1024 molecules of phosphorus trichloride?

Answers

Answer:

THE MASS OF 7.68 *10^24 MOLECULES OF PHOSPHORUS TRICHLORIDE IS 1746.25 g.

Explanation:

Molar mass of PCl3 = ( 31 + 35.5 *3) = 137.5 g/mol

At 7.68 * 10^24 molecules, how many number of mole is present?

6.03 * 10^23 molecules = 1 mole

7.68*10^24 molecules = x mole

x mole = 7.68 *10^24 molecules/ 6.03 *10^23

x mole = 1.27 *10 moles

x mole = 12.7 moles

Using mole = mass / molar mass

mass = mole * molar mass

mass = 12.7 moles * 137.5 g/mol

mass = 1746.25 g

Hence, the mass of 7.68 *10^24 molecules is 1746.25 g

The NaOH solution is standardized (or its true concentration) is found by reacting it with KHSO4. One of the two products from when NaOH reacts with KHSO4 is H2O. The other product is is a salt consisting of what?

a. NaK (aq)
b. (aq)
c. NaS (aq)
d. None of the above

Answers

the answer to this problem is c
the answer is going to be “C. NaS (aq)” hope you have a good day and hope this helped

Why does the excess of base used in these eliminations favor the E2 over the E1 mechanism for elimination

Answers

Answer:

The base is involved in the rate determining step of an E2 reaction mechanism

Explanation:

Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;

Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.

For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;

Rate = k[alkyl halide] [base]

The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.

Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.

Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light?
1. Gamma rays
2. Infrared radiation
3. Ultraviolet liht
4. X-rays
5. Microwaves
6. Radio waves

Answers

Answer:

3,4,1 and 6,5,2

Explanation:

In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;

Radio waves

Microwaves

Infrared waves

Visible light rays

Ultraviolet rays

X-rays

Gamma rays

(a simple mnemonic is RMIVUXG)

Calculate the work (kJ) done during a reaction in which the internal volume expands from 28 L to 51 L against an outside pressure of 4.9 atm.

Answers

Answer:

W= -11KJ

Explanation:

Given:

volume expands from 28 L to 51 L

pressure =4.9 atm.

We will need to Convert the pressure to Pascal SI

But 1 atm = 101,325 Pa.

Then,

Pressure= (4.9*101323)/1atm = 5*10^5 pa

Then we need to Convert the volumes to cubic meters

But we know that1 m³ = 1,000 L.

V1= 28L * 1m^3/1000L = 0.028m^3

V2=51L × 1m^3 /1000L =0.051m^3

The work done during the expansion of a gas can be calculated as

W= -P(V2-V1)

W= - 5*10^5(0.051m^3 - 0.028m^3)

W= -1.1× 10^4J

Then we can Convert the work to kiloJoule

But1 kJ = 1,000 J.

W= -1.1× 10^4J× 1kj/1000J

= -11KJ

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is

Answers

Answer:

4.26 %

Explanation:

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

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]

A substance used as a cleaner and a fuel is 37.48% C, 49.93% O and 12.58% H by mass. A 0.2804-g sample of the substance occupies a volume of 250.0 mL when it is vaporized at 75o C and 1.00 atm of pressure.
R = 0.0821 L atm/ mol K
a) This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas?
b) If 59.6 kg of the product is actually produced, given the reaction described in (a), what is the percent yield?
c) This compound (the substance you identified in part a) is a potential replacement for gasoline. The products of the complete combustion of this fuel are the same as those for the complete combustion of a hydrocarbon (CO2 and H2O). Calculate the volume of CO2 produced at 27o C and 766 mmHg when 1.00 gallon of this fuel is completely combusted. The density of the fuel is 0.7914 g/mL. 1 gallon = 3.785 liters
d) A claim was made that this fuel is better for the environment because it produces less CO2 per gallon than gasoline, which can be represented by the formula C8H18 (octane). Is this claim true? Octane has a density of 0.6986 g/mL

Answers

Answer:

Explanation:

We shall find out the molecular formula of the substance .

Ration of number of atoms of C , O and H

= [tex]\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}[/tex]

= 3.12 : 3.12 : 12.58

= 1 : 1 : 4

volume of gas at NTP

= 250 x 273 / 350 mL .

= 195 mL .

Molecular weight of the substance = .2804 x 22400 / 195 g

= 32. approx

Let the molecular formula be

(COH₄)n  

n x 32 = 32

n = 1

Molecular formula = COH₄

The compound appears to be CH₃OH

a )

CO + 2H₂ = CH₃OH

28g     4g          32g

59      8

For 8 kg hydrogen , CO required = 56 kg

CO is in excess .  hydrogen is the limiting reagent .

mass of product formed

= 32 x 8 / 4

= 64 kg

b )

percentage yield = product actually formed / product to be formed theoretically  x 100

= 59.6 x 100 / 64

= 93.12 %

c )

2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .

64 g                     2 x 22.4 L

Gram of gas in 1 gallon of fuel

= .7914 x 3785

= 2995.5 g

CO₂ produced at NTP by 2995.5 g CH₃OH

= 2 x 22.4 x 2995.5 / 64 L

= 2096.85 L

At 27° C and 766 mm Hg , this volume is equal to

2096.85 x 300 x 760 / 273 x 766

= 2286.18  L .

d )

C₈H₁₈  =  8CO₂

114g           8 x 22.4 L

gram of fuel per unit gallon

= .6986 x 3785

= 2644.2g

gram of CO₂ produced by 1 gallon of fuel  at NTP

= 8 x 22.4 x 2644.2 / 114

= 4156.5 L

So it produces more CO₂ .

How many water molecules are in a block of ice containing 0.500 mol of water (H2O)?

Answers

Answer:

3.01 × 10²³ molecules

Explanation:

Step 1: Given data

Moles of water (n): 0.500 mol

Step 2: Calculate the molecules of water present in 0.500 moles of water

In order to perform this calculation, we will use the Avogadro's number: in 1 mole of water there are 6.02 × 10²³ molecules of water.

0.500 mol × (6.02 × 10²³ molecules/1 mol) = 3.01 × 10²³ molecules

For a given sample of C3H8(g), the enthalpy change during the reaction is -784kJ. How many grams of CO2 are produced? Group of answer choices g

Answers

Answer:

[tex]m_{CO_2}=46.6gCO_2[/tex]

Explanation:

Hello,

In this case, considering the combustion of propane:

[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)\ \ \ \Delta _CH=-2220.0 kJ/mol[/tex]

We can compute the burnt moles of propane as shown below:

[tex]n=\frac{-784kJ}{-2220.0 kJ/mol} =0.353molC_3H_8[/tex]

Then, by noticing propane and carbon dioxide are in a 1:3 molar ratio, we can compute the grams carbon dioxide by using the shown below stoichiometric procedure:

[tex]m_{CO_2}=0.353molC_3H_8*\frac{3molCO_2}{1molC_3H_8} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=46.6gCO_2[/tex]

Best regards.

Which of the possible compounds has a mass of 163 grams when
6.13 x 1024 molecules of the compound is put on a scale?

Answers

Answer:

CH4

Explanation:

In solving this problem, we must remember that one mole of a compound contains Avogadro's number of elementary entities. These elementary entities include atoms, molecules, ions etc. Recall that one mole of a substance is the amount of substance that contains the same number of elementary entities as 12g of carbon-12. The Avogadro's number is 6.02 × 10^23.

Hence we can now say;

If 163 g of the compound contains 6.13 ×10^24 molecules

x g will contain 6.02 × 10^23 molecules

x= 163 × 6.02 × 10^23 / 6.13 × 10^24

x= 981.26 × 10^23/ 6.13 ×10^24

x= 160.1 × 10^-1 g

x= 16.01 g

x= 16 g(approximately)

16 g is the molecular mass of methane hence x must be methane (CH4)

g A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC

Answers

Answer:

The total photons required = 5.19 × 10²⁸ photons

Explanation:

Given that:

the radiation wavelength λ= 12.5 cm = 0.125 m

Volume of the container = 0.250 L = 250 mL

The density of water = 1 g/mL

Density = mass /volume

Mass =  Volume ×  Density

Thus; the mass of the water =  250 mL ×  1 g/mL

the mass of the water = 250 g

the specific heat of water s = 4.18 J/g° C

the initial temperature [tex]T_1[/tex] = 20.0° C

the final temperature [tex]T_2[/tex] = 99° C

Change in temperature [tex]\Delta T[/tex] = (99-20)° C = 79 ° C

The heat q absorbed during the process = ms [tex]\Delta T[/tex]

The heat q absorbed during the process = 250 g × 4.18 J/g° C × 79° C

The heat q absorbed during the process = 82555 J

The energy of a photon can be represented by the equation :

= hc/λ

where;

h = planck's constant = [tex]6.626 \times 10^{-34} \ J.s[/tex]

c = velocity of light = [tex]3.0 \times 10^8 \ m/s[/tex]

=  [tex]\dfrac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{0.125}[/tex]

= [tex]1.59024 \times 10^{-24}[/tex] J

The total photons required = Total heat energy/ Energy of a photon

The total photons required = [tex]\dfrac{82555 J}{1.59024 \times 10^{-24}J}[/tex]

The total photons required = 5.19 × 10²⁸ photons

Which aqueous solution will have the highest boiling point temperature? A. 0.100 molal NiBr2(aq) B. 0.250 molal CH3OH(aq) C. 0.100 molal MgSO4(aq) D. 0.150 molal Na2SO4(aq) E. 0.150 molal NH4NO3(aq)

Answers

Answer: 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.

Explanation:

Formula used for Elevation in boiling point :

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

where

[tex]\Delta T_b=T_b-T^o_b[/tex]= elevation in boiling point

[tex[k_b[/tex]  = boiling point constant  

m = molality

i = Van't Hoff factor

A) 0.100 m [tex]NiBr_2[/tex]

i = 3 as [tex]NiBr_2\rightarrrow Ni^{2+}+2Br^-[/tex]

concentration will be [tex]3\times 0.100=0.300[/tex]

B) 0.250 m [tex]CH_3OH[/tex]

i = 1 as [tex]CH_3OH[/tex] is a non electrolyte

concentration will be [tex]1\times 0.250=0.250[/tex]

C) 0.100 molal [tex]MgSO_4(aq)[/tex]

i = 2 as [tex]MgSO_4\rightarrrow Mg^{2+}+SO_4^{2-}[/tex]

concentration will be [tex]2\times 0.100=0.200[/tex]

D. 0.150 molal [tex]Na_2SO_4(aq)[/tex]

i = 3 as [tex]Na_2SO_4\rightarrrow 2Na^{+}+SO_4^{2-}[/tex]

concentration will be [tex]3\times 0.150=0.450[/tex]

E. 0.150 molal [tex]NH_4NO_3(aq)[/tex]

i = 2 as [tex]NH_4NO_3\rightarrrow NH_4^{+}+NO_3^{-}[/tex]

concentration will be [tex]2\times 0.150=0.300[/tex]

The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.

The aqueous solution that would have the highest temperature at boiling point would be:

D). 0.150 molal Na2SO4(aq)

What is a boiling point?

The boiling point is described as the temperature at which the solution starts boiling or the vapor pressure becomes equivalent to the provided external/outer pressure.

To determine the elevation in boiling point, we will use:

Δ[tex]T_{b}[/tex] [tex]= i[/tex] × [tex]k_{b}[/tex] × [tex]m[/tex]

with

[tex]T_{b}[/tex] [tex]= T_{b} - T^{0}_{b}[/tex]

[tex]k_b[/tex] [tex]=[/tex] constant of boiling point

Using this formula,

0.150 molal Na2SO4(aq)

Given,

[tex]i = 3[/tex]

[tex]Na2So4[/tex] will have

[tex]2Na^{+}[/tex] [tex]+[/tex] [tex]SO^{2-}_{4}[/tex]

So,

Concentration [tex]= 3[/tex] × [tex]0.15[/tex][tex]0[/tex]

[tex]= 0.45[/tex][tex]0[/tex]

∵ 0.150 molal [tex]Na2SO4[/tex]Na2SO4(aq) has the maximum concentration.

Thus, option D is the correct answer.

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Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng

Answers

Answer:

143pm is the radius of an Al atom

Explanation:

In a face centered cubic structure, FCC, there are 4 atoms per unit cell.

First, you need to obtain the mass of an unit cell using molar mass of Aluminium  and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.

Mass of an unit cell

As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:

4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g

Edge length

As density of aluminium is 2.71g/cm³, the volume of an unit cell is:

1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³

And the length of an edge of the cell is:

∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m

Radius:

As in FCC structure, Edge = √8 R, radius of an atom of Al is:

4.044x10⁻¹⁰m = √8 R

1.430x10⁻¹⁰m = R.

In pm:

1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =

143pm is the radius of an Al atom

The radius of the atom of Al in the FCC structure has been 143 pm.

The FCC lattice has been contributed with atoms at the edge of the cubic structure.

The FCC has consisted of 4 atoms in a lattice.

The mass of the unit cell of Al can be calculated as:

[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole

4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles

The mass of 1 mole Al has been 26.98 g/mol.

The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g

The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.

The volume of the Al cell can be calculated as:

Density = [tex]\rm \dfrac{mass}{volume}[/tex]

Volume = Density × Mass

The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g

The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]

The volume of the cube has been given as:

Volume = [tex]\rm edge\;length^3[/tex]

6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]

Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm

Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm

Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

In an FCC lattice structure, the radius of the atom can be given by:

Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

1 m = [tex]\rm 10^1^2[/tex] pm

1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.

The radius of the atom of Al in the FCC structure has been 143 pm.

For more information about the FCC structure, refer to the link:

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Calcium carbide, CaC2, reacts with water to form calcium hydroxide and the flammable gas ethyne (acetylene) in the reaction: What mass of ethyne can be produced

Answers

Answer:

1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂

Explanation:

Hello,

To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.

Equation of reaction

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Molar mass of calcium carbide (CaC₂) = 64.1g/mol

Molar mass of water (H₂O) = 18g/mol

Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol

Molar mass of ethyne (C₂H₂) = 26g/mol

From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂

1 mole of CaC₂ = mass / molar mass

Mass = 1 × 64.1

Mass = 64.1g

1 mole of C₂H₂ = mass / molar mass

Mass = 1 × 26

Mass = 26g

Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂

Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction

If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of the reaction?

Answers

Answer:

The correct answer is - 0.570 grams

Explanation:

The balanced chemical reaction is given by

Cu(NO3)2(aq)     + 2NaOH(aq)    -------->    Cu(OH)2(s)      + 2NaNO3(aq)

        1.0 mole            2.0 mole                 1.0 mole          2.0 mole

number of mol of Cu(OH)2,

n = Molarity * Volume

= [tex]35.0*0.167 = 5.845[/tex] millimoles

As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2

Mass of Cu(OH)2 = number of mol * molar mass

= [tex]97.5*5.845*10^-3[/tex]

= 0.570 grams

Thus, the correct answer is - 0.570 grams

Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/

Answers

Answer:

10.328 m

Explanation:

normal atmospheric pressure = 101325 Pa

density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3

pressure = pgh

where p = density

g = acceleration due to gravity = 9.81 m/s^2

h = height of column

imputing values, we have

101325 = 1000 x 9.81 x h

height of column h = 101325/9810 = 10.328 m

solution solution solution​

Answers

Answer:

Oxygen present in food items makes then rancid due to the presence of oils and fats. If the food is flushed with nitrogen, it prevents it from being oxidised (the nitrogen acts as an antioxidant).

Hope it helps ! :)

Identify the correctly written chemical reaction
A. Reactant + Reactant = Product
B. Reactant + Reactant → Product + Product
C. Reactant + Product → Reactant + Product
D. Product + Product Reactant + Reactant

Answers

Answer:

B. Reactant + Reactant -> Product + Product

Explanation:

Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction

Products are substances that are produced as a result of the reaction

Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving.  In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards

And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.  

To keep things simple:

Products can't be at the beginning of a reaction since they weren't formed yet.

Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products

As such, only one possible chemical reaction would follow that reasoning:

    Reactant + Reactant ->  Product + Product

Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.

What is a chemical equation?

A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.

In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.

Hence, option B is correct.

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A 32.3-gram sample of gas is found to have a volume of 1.9 liters at 301 K and 1.21 atm. What is the molar mass of this gas? Show all of the work used to solve this problem.

Answers

Answer:

351.1g/mol

Explanation:

you can find the answer using The ideal gas equation

n= PV/RT

n=(1.21*1.9/0.082*301)mol

n=0.092 mol

molar mass=Mass/mole

m=32.3g/0.092mol

m=351.1g/mol

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