Lead (II) sulfide, PbS, reacts with oxygen gas to produce lead (II) oxide and sulfur dioxide. If 0.750 moles of O2 were used during this chemical reaction, how many grams of lead (II) oxide would be produced?

Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We convert mass of phthalate to moles, using its molar mass:

Now we convert mmol of HA to mmol of Ba(OH)₂:

Finally we calculate the molarity of the Ba(OH)₂ solution:

(B) The reaction between Ba(OH)₂ and HCl is:

So the moles of HCl that reacted are:

And the molarity of the HCl solution is:

Answer:

114 K

Explanation:

Given data

We can calculate the temperature at which the student collected the oxygen using the ideal gas equation.

[tex]P \times V = n \times R \times T\\T = \frac{P \times V}{n \times R} = \frac{0.500atm \times 0.629L}{0.0337mol \times 0.0821atm.L/mol.K} = 114 K[/tex]

The oxygen gas was collected at 114 K.

Answer:

The temperature is 114 K or -159ºC

Explanation:

We must use the ideal gas equation to calculate the temperature in K:

P x V = n x R x T

⇒ T = (P x V)/(n x R)

We have to introduce the data expressed in the adequate units:

V = 629 ml x 1 L/1000 ml= 0.629 L

n = 0.0337 moles

P = 0.500 atm

R = 0.082 L.atm/K.mol (it is the gas constant)

T = (P x V)/(n x R) = (0.500 atm x 0.629 L)/(0.0337 mol x 0.082 L.atm/K.mol)

T = 113.8 K ≅ 114 K

Thus, the temperature is 114 K or -159ºC.

Answer:

W = +1624J

Explanation:

V1 = 33.0L

V2 = 61.0L

P = 58 atm

Work = force * distance (∇x)

Work = F.∇x

But force = pressure * area

F = P.A

Work = P.∇x.A

Work = P∇v

Work = P (V₂ - V₁)

Work = 58 * (61 - 33)

Work = 58 * 28

Work = 1624J

The work done is +1624J

Answer:

0.357 g He

Explanation:

1 mol of gas at STP = 22.4 L

2.00 L * 1 mol/22.4 L = 0.08929 mol He

M(He) = 4.00 g/mol

0.08929 mol He * 4.00 g/ 1 mol He = 0.357 g He

Answer:

3:1 M ratio at pH 3-4

Explanation:

Switching one substance for another is called displacement.

Chemical reactions are those reactions in which reactants are react with each other for the formation of product.

Hence, given reaction is the displacement reaction.

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Answer :

Moles = 0.2mol

Explanation:

n= m/M = 15/74.45 = 0.2mol

Answer:

A is 15.0 g

B is 74.45 g/mol

C is 0.201 mol

D is 150.0 mL

E is 1.34 M

Hopes this helps :)

Answer:

The concentration of the solution is 2.86 M

Explanation:

Molarity is a unit of concentration based on the volume of a solution. It is defined as the number of moles of solute that are dissolved in a given volume. In other words, molarity is defined as the number of moles of solute per liter of solution.

The Molarity of a solution is determined by the following expression:

[tex]Molarity (M)=\frac{number of moles of solute}{Volume}[/tex]

Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]).

In this case, you must then know the number of moles of HF, for which you must know the molar mass. Being:

the molar mass of HF is: HF= 1 g/mole + 19 g/mole= 20 g/mole

Then the following rule of three applies: if 20 g of HF are available in 1 mole, 14.3 g in how many moles will they be?

[tex]moles=\frac{14.3 g*1 mole}{20 g}[/tex]

moles= 0.715

So:

Replacing:

[tex]Molarity=\frac{0.715 moles}{0.250 L}[/tex]

Solving:

Molarity= 2.86 [tex]\frac{moler}{liter}[/tex]=2.86 M

The concentration of the solution is 2.86 M

Answer:

Molarity= 1.69M

Explanation:

m= 14.8, Mm= 35, V= 0.25dm3, C= ?

Moles = m/M= C×V

Substitute and Simplify

m/M= C×V

14.8/35= C×0.25

C= 1.69M

Final answer:

The molarity of the ammonium hydroxide solution is 1.684 M.

Explanation:

The molarity of a solution can be calculated using the formula:

Molarity (M) = moles of solute / liters of solution

First, we need to convert the mass of ammonium hydroxide (NH4OH) to moles:

1. Calculate the molar mass of NH4OH:

Total molar mass of NH4OH = (14.01 x 1) + (1.01 x 4) + 16.00 = 35.05 g/mol

2. Calculate the number of moles:

moles = mass / molar mass = 14.8 g / 35.05 g/mol = 0.421 mol

Next, we need to convert the volume of the solution from milliliters to liters:

250.0 mL = 0.250 L

Finally, we can calculate the molarity:

Molarity (M) = 0.421 mol / 0.250 L = 1.684 M

Answer:

option C. Ca(HCO3)2

Explanation:

did it on edg

Answer:

answer c

Explanation:

Answer:

132.93 L

Explanation:

Step 1:

Data obtained from the question:

Volume (V) =?

Number of mole (n) = 2.79 moles

Pressure (P) = 439 mmHg

Temperature (T) = 64°C

Step 2:

Conversion to appropriate unit

For pressure:

760mmHg = 1atm

Therefore, 439 mmHg = 439/760 = 0.58 atm

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

temperature (celsius) = 64°C

Temperature (Kelvin) = 64°C + 273 = 337 K

Step 3:

Determination of the volume.

The volume occupied by N2 can be obtained by using the ideal gas equation as follow:

Volume (V) =?

Number of mole (n) = 2.79 moles

Pressure (P) = 0.58 atm

Temperature (T) = 337 K

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

0.58 x V = 2.79 x 0.082 x 337

Divide both side by 0.58

V = (2.79 x 0.082 x 337)/0.58

V = 132.93 L

Therefore, the volume occupied by N2 is 132.93 L

Answer:

[tex]\large \boxed{1.85:1}[/tex]

Explanation:

The density of a substance is directly proportional to the molar mass and the number of atoms per unit cell, and inversely proportional to the volume of the unit cell.

The BCC unit cell of CsCl contains one K⁺ and one Cl⁻ ion, while the SC unit cell contains four Na⁺ and four Cl⁻ ions.

[tex]\dfrac{\text{CsCl density}}{\text{NaCl density}} = \dfrac{\text{Atoms of Cs}}{\text{Atoms of Ca}} \times \dfrac{\text{MM of CsCl}}{\text{MM of NaCl}} \times \dfrac{\text{Vol. of NaCl unit cell}}{\text{Vol. of CsCl unit cell}}\\\\= \dfrac{1}{4} \times \dfrac{\text{2.88}}{\text{1}} \times \dfrac{1.37^{3}}{1^{3}} = \dfrac{1.85}{1}\\\\\text{The ratio of the density of CsCl to that of NaCl is $\large \boxed{\mathbf{1.85:1}}$}[/tex]

Final answer:

To determine the ratio of the density of CsCl to NaCl, one must consider their molar masses, unit cell structures, and edge lengths. CsCl's density is calculated based on its simple cubic unit cell structure, while NaCl's is based on its FCC unit cell.

Explanation:

The ratio of the density of CsCl to the density of NaCl can be calculated by considering their molar masses, unit cell structures, and the edge lengths of their unit cells.

Sodium chloride (NaCl) crystallizes in a face-centered cubic (FCC) lattice, whereas cesium chloride (CsCl) forms a simple cubic unit cell. Each FCC cell for NaCl contains four formula units, while each simple cubic cell for CsCl contains one formula unit.

Given that the molar mass of CsCl is 2.88 times that of NaCl, and the edge length of NaCl's unit cell is 1.37 times the edge length of CsCl's unit cell, we can calculate the densities by using the formula: density = mass/volume.

The volumes of the unit cells can be found by cubing the respective edge lengths.To find the mass of the unit cells, multiply the molar masses by Avogadro's number and divide by the number of formula units per unit cell.

The final step is to compare the calculated densities by taking the ratio of CsCl's density to NaCl's density. This will yield the required density ratio.

Answer:

D

Explanation:

You should wear goggles whether you are the person performing the experiment or the person watching. Analogy: If someone is using fireworks and they say anyone around should wear ear protection would you wear it if you are next to them?

Answer:

Reaction is spontaneous at high temperature and nonspontaneous at low temperature

Explanation:

Given:

Enthalpy change [tex]\Delta H= 545 \frac{KJ}{mol}[/tex]

Entropy change [tex]\Delta S = 1.55[/tex] [tex]\frac{KJ }{K. mol}[/tex]

From the formula of change in free energy,

  [tex]\Delta G = \Delta H - T\Delta S[/tex]

But for spontaneous process the values of quantities are given below

For spontaneous process value of [tex]\Delta G[/tex] is negative

For nonspontaneous process value of [tex]\Delta G[/tex] is positive

Here values of [tex]\Delta H[/tex] and [tex]\Delta S[/tex] are positive, so reaction is spontaneous at

high temperature and nonspontaneous at low temperature

Answer:

The half-life is  [tex]t_h = 3.856*10^{3} minute[/tex]

Explanation:

From the question we are told that

     The sample is  90 Y

      The first  activity is  [tex]A_1 = 2.2 *10^5[/tex]  per minute

       The second  activity is  [tex]A_2 = 5.8 *10^3[/tex]  per minute

        The duration from 1:00 p.m. on December 3, 2006 to 2:15 p.m. on December 17, 2006   is

               [tex]t = 14 \ days \ 1 hr \ 15 min[/tex]

Converting to minutes we have  

               [tex]t = (14 * 24 * 60) + (1* 60) + 15[/tex]

               [tex]t = 20235 \ minutes[/tex]

The first order rate constant for this disintegrations can be mathematically represented  as

               [tex]ln \frac{A_2}{A_1} = - \lambda t[/tex]

 Where [tex]\lambda[/tex] is the rate constant

    Substituting values

                     [tex]ln [\frac{5.8 * 10^{3}}{2.2 *10^{5}} ] = - \lambda * 20235[/tex]

                    [tex]-3.6358 = - \lambda * 20235[/tex]

So

                [tex]\lambda = \frac{3.6358}{20235}[/tex]

                    [tex]\lambda = 1.7968 *10^{-4} minute^{-1}[/tex]

The half life is mathematically represented as

               [tex]t_{h} = \frac{0.693}{\lambda }[/tex]

So           [tex]t_h = \frac{0.693}{1.7968 *10^{-4}}[/tex]

              [tex]t_h = 3.856*10^{3} minute[/tex]

The half-life of the isolated sample of 90Y is calculated using the formula for exponential decay and the given information. The decay constant (lambda) is found and then used to find the half-life, which is approximately 2.67 days.

The subject of this question is the half-life of a isolated sample of 90Y. Half-life is the rate at which radioactive substances decay. The half-life can be calculated using the formula for exponential decay, N = N0e-lambda*t. We are given N0 (the initial amount of material, 2.2 x 10^5 disintegrations per minute), N (the remaining amount of material, 5.8 x 10^3 disintegrations per minute), and t (the time elapsed, 14.25 days). We can use these values to find lambda (the decay constant). Lambda is equal to the natural log of N0/N divided by t. To find the half-life, we use the formula, T = ln(2)/lambda.

Performing the appropriate calculations, we find that the half-life of 90Y is approximately 2.67 days (or 2.67 x 10^0 days in scientific notation).

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There is 1 carbon atom in 5K2CO3.

To determine the number of carbon (C) atoms in 5K2CO3, we need to consider the molecular formula of potassium carbonate (K2CO3). The formula indicates that there are 2 potassium (K) atoms, 1 carbon (C) atom, and 3 oxygen (O) atoms per molecule.

Since there is only 1 carbon atom in the formula, regardless of the coefficient 5 in front, there will always be 1 carbon atom in 5K2CO3.

Therefore, there is 1 carbon atom in 5K2CO3.

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Answer: The theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2[/tex]

Explanation : Given,

Volume of acetophenone = 135  microliters = 135 × 10⁻⁶ L = 0.135 mL

conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)

Density of acetophenone = 1.03 g/mL

Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g

Mass of 4-nitrobenzaldehyde = 127 mg  = 0.127 g

Conversion used : (1 mg = 0.001 g)

First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde

[tex]\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}[/tex]

[tex]\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol[/tex]

and,

[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}[/tex]

[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

[tex]C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3[/tex]

From the balanced reaction we conclude that

As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone

So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone

From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of 4-nitrochalcone

From the reaction, we conclude that

As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone

So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone

Now we have to calculate the mass of 4-nitrochalcone

[tex]\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}[/tex]

Molar mass of 4-nitrochalcone = 253.25 g/mole

[tex]\text{ Mass of 4-nitrochalcone}=(0.000840moles)\times (253.25g/mole)=0.21273g=212.73mg=2.13\times 10^2mg[/tex]

(1 g = 1000 g)

Therefore, the theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2mg[/tex]

Answer:

True:

Explanation:

Hybrid orbitals do not exist in isolated atoms. They form only in covalently bonded atoms.

Hybridization happens when several atomic orbitals combine to form other orbitals with the same energy and greater stability.

A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that combined to produce the set.

Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds, and both can appear in an atom at the same time.

The statements about hybridization that are true include:

B. Hybrid orbitals within the same atom have the same energy and shape.

C. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals.

D. An atom can have both hybridized and unhybridized orbitals at the same time.

A sublevel is an energy level that is typically associated with the valence electrons found outside an atomic nucleus.

In Chemistry, there are four (4) types of sublevel and these includes:

I. s orbital (sublevel): it has one (1) orbital i.e 1s.

II. p orbital (sublevel): it has three (3) orbitals.

III. d orbital (sublevel): it has five (5) orbitals.

IV. f orbital (sublevel): it has seven (7) orbitals.

Hybridization can be defined as a phenomenon which involves the linear combination of two or more atomic orbitals of a molecule, so as to form the same number of hybrid orbitals, with each of the orbital having the same energy and shape.

Generally, the two types of hybridization an atom can have include:

1. Hybridized orbitals.

2. Unhybridized orbitals

Hence, we can deduce the following points from the above:

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Answer: The mass of [tex]NH_3[/tex] produced is, 3.03 grams.

Explanation : Given,

Mass of [tex]N_2[/tex] = 2.5 g

Mass of [tex]H_2[/tex] = 2.5 g

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

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

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

[tex]\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=\frac{2.5g}{28g/mol}=0.089mol[/tex]

and,

[tex]\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=\frac{2.5g}{2g/mol}=1.25mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

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

From the balanced reaction we conclude that

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

So, 0.089 moles of [tex]N_2[/tex] react with [tex]0.089\times 3=0.267[/tex] moles of [tex]H_2[/tex]

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

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

From the reaction, we conclude that

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

So, 0.089 mole of [tex]N_2[/tex] react to give [tex]0.089\times 2=0.178[/tex] mole of [tex]NH_3[/tex]

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

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

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

[tex]\text{ Mass of }NH_3=(0.178moles)\times (17g/mole)=3.03g[/tex]

Therefore, the mass of [tex]NH_3[/tex] produced is, 3.03 grams.

Answer:

16.36A

Explanation:

We'll begin by writing a balanced dissociation equation of aqueous Cr2(SO4)3. This is illustrated below:

Cr2(SO4)3 —> 2Cr^3+ 3(SO4)^2-

From the above, we can see that Cr is trivalent.

Next, let us determine the number of faraday needed to deposit metallic Cr. This is illustrated below:

Cr^3+ 3e- —> Cr

From the above equation, 3 faradays are needed to deposit metallic Cr

1 faraday = 96500C

Therefore, 3 faraday = 3 x 96500C = 289500C.

Molar Mass of Cr = 52g/mol

Now let us determine the quantity of electricity needed for 2.68g of Cr metal

This is shown below:

52g of Cr required 289500C.

Therefore, 2.68g of Cr will require = (2.68 x 289500)/52 = 14920.38C

Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:

Q (quantity of electricity) = 14920.38C

t (time) = 15.2mins = 15.2 x 60 = 912secs

I (current) =?

Apply the equation Q = It

Q = It

14920.38 = I x 912

Divide both side by 912

I = 14920.38/912

I = 16.36A

Therefore, a current of 16.36A is needed for the process.

The current needed is 16.36 A. The quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process.

[tex]Cr_2(SO_4)_3 ---- > 2Cr^{3+}+ 3SO_4^{2-}[/tex]

The number of faradays needed to deposit metallic Cr. This is illustrated below:

[tex]Cr^{3+}+ 3e^- ---- > Cr[/tex]

Given:

3 faradays are needed to deposit metallic Cr

1 faraday = 96500C

Therefore, 3 faraday = 3 * 96500C = 289500C.

Molar Mass of Cr = 52g/mol

52g of Cr required 289500C.

Therefore, 2.68g of Cr will require = (2.68 * 289500)/52 = 14920.38C

Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:

Q (quantity of electricity) = 14920.38C

t (time) = 15.2mins = 15.2 x 60 = 912secs

To find:

I (current) =?

Apply the equation,

Q = It

14920.38 = I * 912

I = 14920.38/912

I = 16.36A

Therefore, a current of 16.36A is needed for the process.

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Answer:n=1

Explanation:

Final answer:

The energy level n=1 has the least energy.

Explanation:

The energy levels of electrons in an atom are determined by the principal quantum number, denoted as n. The higher the value of n, the higher the energy level. Therefore, the energy level n=7 has the highest energy. On the other hand, the energy level n=1 has the lowest energy, which makes it the answer to your question.

Answer:

21.28 grams solute can be added if the temperature is increased to 30.0°C.

Explanation:

Solubility of solute at 20°C = 32.2 g/100 grams of water

Solute soluble in 1 gram of water = [tex]\frac{32.2}{100}g=0.322 g[/tex]

Mass of solute in soluble in 56.0 grams of water:

[tex]0.322\times 56.0=18.032 g[/tex]

Solubility of solute at 30°C = 70.2g/100 grams of water

Solute soluble in 1 gram of water = [tex]\frac{70.2}{100}g=0.702 g[/tex]

Mass of solute in soluble in 56.0 grams of water:

[tex]0.702 \times 56.0=39.312 g[/tex]

If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C

Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g

Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g

Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:

39.312 g - 18.032 g = 21.28 g

21.28 grams solute can be added if the temperature is increased to 30.0°C.

Final answer:

By comparing the solubility at two different temperatures, an additional 21.280 g of solute can be dissolved in 56.0 g of water when the temperature increases from 20.0 °C to 30.0 °C.

Explanation:

Understanding how temperature influences solubility is crucial when trying to determine the concentration of a solute that can be dissolved in a solvent at different temperatures. According to the data provided, the solubility of a solute in water at 20.0 °C is 32.2 g per 100 g of water and at 30.0 °C it increases to 70.2 g per 100 g of water.

To calculate the additional amount of solute that can be dissolved when the solution temperature is raised from 20.0 °C to 30.0 °C, we should first calculate the solubility in 56.0 g of water at both temperatures:

So, 18.032 g of solute can be dissolved in 56.0 g of water at 20.0 °C.

So, 39.312 g of solute can be dissolved in 56.0 g of water at 30.0 °C.

The difference between these two amounts (39.312 g - 18.032 g) gives us the additional amount of solute that can be added at the higher temperature:

Additional solute = 39.312 g - 18.032 g = 21.280 g

Therefore, an additional 21.280 g of solute can be added to the solution when the temperature is raised from 20.0 °C to 30.0 °C without reaching saturation.

The rate law is defined as the rate of reaction in which reactants are expressed in their molar concentration raised to the power of their stoichiometric coefficient.

In the given reaction, the slow step determines the rate of reaction.

The chemical reaction is:

2A + B [tex]\rightarrow[/tex] D

The intermediate reaction of the mechanism follows:

Step 1: A + B [tex]\rightleftharpoons[/tex] C (slow)

Step 2: A + C  [tex]\rightleftharpoons[/tex] D (fast)

In the given reaction, the first step is a slow step, which determines the rate of the reaction. The rate for the equation can be given as:

Rate = k [A]² [B]

Therefore, the rate law expression for the overall reaction is Rate = k [A]² [B].

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The rate law for the given overall reaction, considering that the first step is the slow step, would be rate = k[A][B]. The reaction is first order with respect to both A and B, making it an overall second-order reaction.

The question requires determining the rate law for an overall reaction from a given two-step reaction mechanism. In the provided reaction, step 1: A + B ⟶ C is the rate-determining (or slow) step while step 2: A + C ⟶ D is a fast step.

Given this scenario, the rate-determining step dictates the rate law of the overall reaction. The rate law for a reaction where the slow step is the first step can be written as , where

is the rate constant and m and n are the orders of the reaction with respect to reactants A and B respectively.

In this case, since both A and B are involved in the slow step, the rate law would be rate = k[A][B], where the reaction is first-order with respect to both A and B, making it an overall second-order reaction.

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Answer:

The volume of the air breath per day is =12240 L day ^ -1 and he volume of oxygen breathe each day is= 2570.4 L

Explanation:

Complete question: The oxidizing agent our bodies use to obtain energy from food is oxygen (from the air). If you breathe 17 times a minute (at rest), taking in and exhaling 0.50 L of air with each breath, what volume of air do you breathe each day? Air is 21% oxygen by volume. what volume of oxygen do you breathe each day?

Solution

Given that,

The volume of the air with each breath = 0.50L

The frequency of breath is = 17 times per minute

the next step is to calculate the volume of the air breath per day

The volume of the breathed air = volume of air breathed/ breath * 17 breaths/min * 60 minutes / per hour * 24 hours/ per day

= 0.50L/ breath * 17 breath /min * 60 minutes / per hour * 24 hours/ per day

= 12240 L day ^ -1

The next step is to calculate the oxygen breathed per day:

The volume of oxygen breathed = volume of air breathed * percent oxygen in air/100

= 12240 L day ^ -1 * 21 /100

= 2570.4 L day ^ -1

Therefore for each day, the volume of oxygen breathe is 2570.4 L

Answer:

Volume of oxygen we breathe each day = 2570.4 L/day

Explanation:

Given that :

volume of air used per breath = 0.5 L

Frequency of breath = 17 times per minute

We know that the percentage volume of air in the atmosphere is approximately = 21%

Hence; the volume of air breathed = [tex]\frac{volume \ of \ air breathed}{breath } *\frac{17 \ breaths}{ minute } * \frac{60 \ minutes }{hour}*\frac{24 \ hour}{day}[/tex]

the volume of air breathed =  [tex]\frac{0.5 \ L }{breath } *\frac{17 \ breaths}{ minute } * \frac{60 \ minutes }{hour}*\frac{24 \ hour}{day}[/tex]

the volume of air breathed = 12,240 L/day

To calculate the volume of oxygen breathed per day; we have:

Volume of oxygen breathed = [tex]volume \ of \ air \ breathed * \frac{percentage \ of \ air }{100}[/tex]

Volume of oxygen breathed = [tex]12240* \frac{21 }{100}[/tex]

Volume of oxygen breathed = 2570.4 L/day

Answer:

0.0023 moles of H₂A

Explanation:

moles H₂A in soln = moles NaOH used in titration

moles = Molarity x Volume in Liters

moles H₂A = moles NaOH used

Which is = (0.100M)(0.02258L) = 0.0023 mol H₂A

The number of moles in a diprotic acid is 0.0023 moles of H₂A. This can be identified using law of dilution.

While performing titrations, the law of dilution is used.

Molarity is defined as the quantity of moles of solute partitioned by the volume of the arrangement in liters.

Moles of acid = Moles of base

n= M /V

Moles H₂A = moles NaOH used

= (0.100M)(0.02258L)

= 0.0023 mol H₂A

Find more information about Molarity here:

brainly.com/question/17138838

Answer:

.

Explanation:

Oil or Lipid + Base lead to Glycerol and Soap

Explanation:

Reactants can be referred to as starting materials for a chemical reaction. They undergo a change to form products. Reactants are usuallyconsumed in the reaction process.

Reactants are written at the left side before the arrow sign in a reaction.

Products are the ending materials of a reaction. They are what is left after the reactants has been consumed.

Prpducts are written on the right side after the arrow sign in a reaction

Basically, a reaction is given as;

Reactants --> Products

A + B --> C + D

A and B are reactants and C and D are products

In the reaction;

“carbon plus oxygen makes carbon dioxide”

C + O2 --> CO2

The reactants are; C and O2.

The product is CO2

Answer:

The answer to your question is given below

Explanation:

A. The difference between reactants and products in a chemical equation is that the reactants are located on the left hand side of the equation while the products are located on the right hand side of the equation.

B. When carbon (C) combine with oxygen (O2) to produce carbon dioxide (CO2), the details of the reaction are given below:

Reactants => C and O2

Product => CO2.

The balanced equation is given below:

C + O2 —> CO2

C and O2 are located on the left side which indicates that they are the reactants while CO2 is located at the right side which indicates that it is the product.

Answer:

39.3%

Explanation:

Our guide in solving the problem must be the reaction equation hence it is pertinent to put down first:

CaF2 + H2SO4 --> CaSO4 + 2HF

We have a very important information in the question, sulphuric acid is present in excess. This implies that calcium fluoride is the limiting reactant.

Number of moles of calcium fluoride reacted= mass of calcium fluoride reacted/ molar mass of calcium fluoride

Molar mass of calcium fluoride= 78.07 g/mol

Number of moles of calcium fluoride= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of 0.28 moles of hydrogen fluoride = number of moles× molar mass

Molar mass of hydrogen flouride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g this is the theoretical yield of HF

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

Answer:

A. The theoretical yield of HF is 5.64g

B. The percentage yield of HF is 39%

Explanation:

Step 1:

The balanced equation for the reaction:

CaF2 + H2SO4 --> CaSO4 + 2HF

Step 2:

Determination of the mass of CaF2 that reacted and the mass of HF produced from the balanced equation. This is illustrated below:

Molar Mass of CaF2 = 40 + (19x2) = 40 + 38 = 78g/mol

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF from the balanced equation = 2 x 20 = 40g.

From the balanced equation above,

78g of CaF2 reacted and 40g of HF were produced.

A. Determination of the theoretical yield of HF.

This is illustrated below:

From the balanced equation above,

78g of CaF2 reacted to produce 40g of HF.

Therefore, 11g of CaF2 will react to produce = (11 x 40)/78 = 5.64g of HF.

The theoretical yield of HF is 5.64g

B. Determination of the percentage yield.

The percentage yield of HF can be obtained as follow:

Actual yield = 2.2g

Theoretical yield = 5.64g

Percentage yield =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 2.2/5.64 x 100

Percentage yield = 39%

The percentage yield of HF is 39%

Answer:

The balanced chemical equation is given as:

[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]

Explanation:

When magnesium metal burns in presence of oxygen it gives white color powdered compound called magnesium oxide.

The balanced chemical equation is given as:

[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]

According to reaction, 2 moles of magnesium metal when reacts with 1 mole of oxygen gas it gives 2 moles of  solid magnesium oxide.

Answer:

False

Explanation:

You have bacterias inside of each one of us, some are good and some are bad.

They can be found in the air ou objects, and objects don't decay.