ANSWER ASAP
Suppose that 10.00 HCl of unknown concentration is neutralized by 20.00 mL of a 1.50 M NaOH solution. Determine the concentration of the HCl solution?


a. 0.0750 M HCl

b. 0.150 M HCl

c. 3.00 M HCl

d. 1.50 M HCl

Answer : The theoretical yield and percent yield for this reaction is, 3.78 grams and 31.7 % respectively.

Explanation : Given,

Mass of [tex]C_6H_5COOH[/tex] = 3.4 g

Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mol

First we have to calculate the moles of [tex]C_6H_5COOH[/tex]

[tex]\text{Moles of }C_6H_5COOH=\frac{\text{Given mass }C_6H_5COOH}{\text{Molar mass }C_6H_5COOH}[/tex]

[tex]\text{Moles of }C_6H_5COOH=\frac{3.4g}{122.12g/mol}=0.0278mol[/tex]

Now we have to calculate the moles of [tex]C_6H_5COOCH_3[/tex]

The balanced chemical equation is:

[tex]C_6H_5COOH+CH_3OH\rightarrow C_6H_5COOCH_3[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]C_6H_5COOH[/tex] react to give 1 mole of [tex]C_6H_5COOCH_3[/tex]

So, 0.0278 mole of [tex]C_6H_5COOH[/tex] react to give 0.0278 mole of [tex]C_6H_5COOCH_3[/tex]

Now we have to calculate the mass of [tex]C_6H_5COOCH_3[/tex]

[tex]\text{ Mass of }C_6H_5COOCH_3=\text{ Moles of }C_6H_5COOCH_3\times \text{ Molar mass of }C_6H_5COOCH_3[/tex]

Molar mass of  = 136.14 g/mole

[tex]\text{ Mass of }C_6H_5COOCH_3=(0.0278moles)\times (136.14g/mole)=3.78g[/tex]

The theoretical yield of [tex]C_6H_5COOCH_3[/tex] produced is, 3.78 grams.

Now we have to calculate the percent yield of the reaction.

Theoretical yield of the reaction = 3.78 g

Experimental yield of the reaction = 1.2 g

The formula used for the percent yield will be :

[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

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

[tex]\text{Percent yield}=\frac{1.2g}{3.78g}\times 100=31.7\%[/tex]

The percent yield of the reaction is, 31.7 %

Answer:

1.04g of iron III carbonate

Explanation:

First, we must put down the equation of reaction because it must guide our work.

2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.

Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate

Mass of iron III nitrate reacted= 1.72g

Molar mass of iron III nitrate= 241.88 g∙mol–1

Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles

From the equation of the reaction;

2 moles of iron III nitrate yields 1 mole of iron III carbonate

7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate

Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass

Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate

The theoretical yield of Fe₂(CO₃)₃ obtained from the reaction is 1.04 g

We'll begin by calculating the mass of Fe(NO₃)₃ that reacted and the mass of Fe₂(CO₃)₃ produced from the balanced equation.

2Fe(NO₃)₃ + 3Na₂CO₃ —> Fe₂(CO₃)₃ + 6NaNO₃

Molar mass of Fe(NO₃)₃ = 241.88 g/mol

Mass of Fe(NO₃)₃ from the balanced equation = 2 × 241.88 = 483.4 g

Molar mass of Fe₂(CO₃)₃ = 291.73 g/mol

Mass of Fe₂(CO₃)₃ from the balanced equation = 1 × 291.73 = 291.73 g

From the balanced equation above,

483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃

From the balanced equation above,

483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃

Therefore,

1.72 g of Fe(NO₃)₃ will react to produce = (1.72 × 291.73) / 483.4 = 1.04 g of Fe₂(CO₃)₃

Thus, the theoretical yield of Fe₂(CO₃)₃ is 1.04 g

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Answer: ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=C\times \Delta T[/tex]

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 13.70 kJ/K

Initial temperature of the calorimeter  = [tex]T_i[/tex] = 298.00 K

Final temperature of the calorimeter  = [tex]T_f[/tex]  = 302.34 K

Change in temperature ,[tex]\Delta T=T_f-T_i=(302.34-298.00)K=4.34K[/tex]

Putting in the values, we get:

[tex]Q=13.70kJ/K\times 4.34K=59.4kJ[/tex]

As heat absorbed by calorimeter is equal to heat released by combustion of propanol

[tex]Q=q[/tex]

[tex]\text{Moles of propanol}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{1.771g}{60g/mol}=0.030mol[/tex]  

Heat released by 0.030 moles of propanol = 59.4 kJ

Heat released by 1 mole of propanol = [tex]\frac{59.4}{0.030}\times 1=1980kJ[/tex]

ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ/mol

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:

Kb =  7.1 x 10⁻¹³

Explanation:

Ka x Kb = Kw => Kb = 1 x 10⁻¹⁴/1.4 x 10⁻² = 7.1 x 10⁻¹³

The Kb for SO32- at 25ºC is approximately 1.5 x 10^-7.

therefore correct option (b).

To find the Kb for SO32-, we can use the relationship between Ka and Kb. Since the diprotic acid, H2SO3, forms SO32- in the second dissociation step, we can use the formula Kb = Kw/Ka. Kb is the equilibrium constant for the reaction of a base with water to form the hydroxide ion. Kw is the ion product of water, which is 1.0 x 10^-14 at 25ºC. So, using the given Ka2 value of 6.7 x 10^-8, we can calculate the Kb for SO32-:

Kb = Kw/Ka = (1.0 x 10^-14) / (6.7 x 10^-8)

Kb ≈ 1.5 x 10^-7

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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

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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:

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:

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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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: The half-cell reaction occurring at cathode is [tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]

Explanation:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Reduction reaction is defined as the reaction in which a chemical specie accepts electrons. The oxidation state of the chemical specie reduces.

The given balanced chemical equation is:

[tex]PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)[/tex]

The half cell reaction occurring at cathode follows:

[tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]

Hence, the half-cell reaction occurring at cathode is given above.

For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.

The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.

Answer: a) The reaction consumes 0.365 moles of barium chloride.

b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles[/tex]

[tex]\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles[/tex]

[tex]BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]

According to stoichiometry :

1 mole of [tex]BaCl_2[/tex] require 1 mole of [tex]K_2SO_4[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.365=0.365moles[/tex]  of [tex]K_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]K_2SO_4[/tex] is the excess reagent.

As 1 moles of [tex]BaCl_2[/tex] give = 1 moles of [tex]BaSO_4[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{1}{1}\times 0.365=0.365moles[/tex]  of [tex]BaSO_4[/tex]

As 1 moles of [tex]BaCl_2[/tex] give = 2 moles of [tex]KCl[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{2}{1}\times 0.365=0.730moles[/tex]  of [tex]KCl[/tex]

Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Answer:

calories burned = 202 calories

Explanation:

22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.

Answer:

202 calories is the estimate of calories burned

Explanation:

22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.

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:

W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.

Explanation:

heat engine:

∴ Th = 485°C

∴ Tc = 42°C

∴ Qh = 9.75 KJ......heat from hot source

first law:

∴ ΔU = Q + W = 0 .........cyclic process

⇒ Q = - W

∴ Q = Qh + Qc = Qh - (- Qc)

∴ Qc: heat from the cold source ( - )

⇒ Qh - ( - Qc) = - W..............(1)

⇒ Qc/Qh = - Tc/Th...........(2)

from (2):

⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)

⇒ Qc = - 0.8443 KJ

replacing in (1):

⇒ - W = 9.75 KJ - ( - 0.8443 KJ)

⇒ - W = 10.5943 KJ

The formation of 98.7 grams of Fe from the reaction Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with an enthalpy change of -852 kJ per 112 grams of Fe formed, would result in approximately -753 kJ of energy being released. The reaction is exothermic, meaning it releases energy as heat.

The question asks about the energy evolved during the formation of 98.7 grams of iron (Fe) from the reaction given: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with a enthalpy change (ΔH°rxn) of -852 kJ. The reaction is exothermic, meaning it releases energy as heat. The ΔH for the reaction is -852 kJ per 112 grams of Fe formed (the molar mass of Fe is 56 g/mol, so 2 mol of Fe is 112 g).

To find out how much energy is released in forming 98.7 g of Fe, you can use simple proportion: (98.7 g Fe / 112 g Fe) x -852 kJ = approx. -753 kJ. So, the formation of 98.7 grams of Fe would result in approximately -753 kJ of energy being released, pointing to answer 753 kJ among the options given.

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The energy evolved during the formation of 98.7 g of Fe is 753 kJ.

The given reaction is:

Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)

The enthalpy change for the reaction is -852 kJ.

To calculate the energy evolved during the formation of 98.7 g of Fe, we need to use the molar mass of Fe.

The molar mass of Fe is 55.845 g/mol.

By converting the given mass of Fe to moles and using the stoichiometry of the balanced equation, we can calculate the energy evolved.

98.7 g Fe * (1 mol Fe / 55.845 g Fe) * (-852 kJ / 2 mol Fe) = -753 kJ

Therefore, the energy evolved during the formation of 98.7 g of Fe is 753 kJ.

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Question:

A) 12

B) 29

C) 2.1 × 10⁻²

D) 8.7 × 10⁻²

E) 47

Answer:

The correct option is;

E) 47

Explanation:

Kc, which is the equilibrium constant of a chemical reaction is derived by finding the ratio between the product of the equilibrium concentration of the product raised to their respective coefficients to the product of the equilibrium concentration of the reactants also raised to their respective coefficients.

Here we have;

[H₂] = 0.14 M

[Cl₂] = 0.39 M

[HCl] = 1.6

The reaction is given as follows;

H₂ (g) + Cl₂ (g) ⇄ 2HCl (g)

The formula for Kc is given as follows;

[tex]Kc = \frac{[HCl]^2}{[H_2][Cl_2]} = \frac{1.6^2}{0.14 \times 0.39} = 46.886[/tex]

Therefore, the Kc for the reaction is approximately equal to 47.

Answer:

The total pressure in the container is 0.389 atm

Explanation:

Step 1: Data given

Kp = 0.739

The initial pressure of H2S = 0.215 atm

Step 2: The balanced equation

H2S(g) ⇆ H2(g) + S(g)

Step 3: The initial pressures

pH2S = 0.215 atm

pH2 = 0 atm

pS = 0 atm

Step 4: The pressures at the equilibrium

pH2S = 0.215 - X atm

pH2 = X atm

pS = X atm

Step 5:

Kp = 0.739 = (pS)*(pH2) / (pH2S)

0.739 = X*X / (0.215 - X)

0.739 = X² / (0.215 - X)

X² = 0.739*(0.215-X)

X² = 0.1589 - 0.739X

X² +0.739X - 0.1589 = 0

X = 0.174

pH2S = 0.215 - 0.174 atm = 0.041 atm

pH2 = 0.174 atm

pS = 0.174 atm

Step 6: Calculate the total pressure in the container

Total pressure = 0.041 atm + 0.174 atm + 0.174 atm

Total pressure = 0.389 atm

The total pressure in the container is 0.389 atm

Answer:

44.63g

Explanation:

First, let us calculate the number of mole of KBr in 1.50M KBr solution.

This is illustrated below:

Data obtained from the question include:

Volume of solution = 250mL = 250/1000 = 0.25L

Molarity of solution = 1.50M

Mole of solute (KBr) =.?

Molarity is simply mole of solute per unit litre of solution

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of solute (KBr) = 1.50 x 0.25

Mole of solute (KBr) = 0.375 mole

Now, we calculate the mass of KBr needed to make the solution as follow:

Molar Mass of KBr = 39 + 80 = 119g/mol

Mole of KBr = 0.375 mole

Mass of KBr =?

Mass = number of mole x molar Mass

Mass of KBr = 0.375 x 119

Mass of KBr = 44.63g

Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution

Final answer:

To make a 1.50 M potassium bromide solution with a volume of 250.0 mL, you would need approximately 44.625 grams of potassium bromide.

Explanation:

To calculate the mass of potassium bromide required, we can use the formula:

mass (g) = concentration (M) x volume (L) x molar mass (g/mol)

The concentration is given as 1.50 M, the volume is 250.0 mL (which can be converted to 0.250 L), and the molar mass of potassium bromide (KBr) is 119.0 g/mol. Plugging in these values, we get:

mass (g) = 1.50 M x 0.250 L x 119.0 g/mol = 44.625 g

Therefore, you would need approximately 44.625 grams of potassium bromide to make 250.0 mL of a 1.50 M potassium bromide solution.

Answer:

3.88 × 10²⁴ molecules

Explanation:

In order to solve this question, we need to consider the Avogadro's number. know that 1 mole of particles contains 6.02 × 10²³ particles. This applies to different kinds of particles: atoms, molecules, electrons.

In this case, 1 mole of molecules of oxygen gas contains 6.02 × 10²³ molecules of oxygen gas. We will use this relation to find the number of molecules of oxygen gas in 6.44 moles of oxygen gas.

[tex]6.44mol \times \frac{6.02 \times 10^{23}molecules }{1mol} = 3.88 \times 10^{24}molecules[/tex]

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:

3:1 M ratio at pH 3-4

Explanation:

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:

1. Number of gas particles (atoms or molecules)

2. Number of moles of gas

3. Average kinetic energy

Explanation:

Since the two gas has the same volume and are under the same conditions of temperature and pressure,

Then:

1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole

2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.

3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.

The answer choices which are the SAME for these two gas samples are

This refers to the type  of energy which has to do with the motion of an object.

Hence, we can see that since the two gas has the same volume and are under the same conditions of temperature and pressure,

We would infer that:

Read more about kinetic energy here:

https://brainly.com/question/25959744

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium 1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵    

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium

1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵  

Answer:

The correct answer is B. 734 g

Explanation:

The chemical formulae of lithium sulfate is Li₂SO₄. With this, we can calculate the molecular weight (MM) of lithium sulfate as follows:

MM(Li₂SO₄) = (2 x molar mass Li) + molar mass S + (4 x molar mass O)

                   = (2 x 6.9 g/mol) + 32 g/mol + (4 x 16 g/mol)

                   = 109.9 g/mol

We need to prepare a solution with a molarity of 2.67 M. That means that the solution has to have 2.67 moles of Li₂SO₄ per liter of solution. We can convert from mol to grams with the calculated molecular weight and then we have to multiply by the volume of solution (2500 ml= 2.5 L), as follows:

Mass = 2.67 mol/L x 109.9 g/mol x 2.5 L ≅ 734 g

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]

The standard heat of formation of [tex]\( \text{CuO(s)} \) is \( -149.6 \text{ kJ/mol} \).[/tex]

To find the standard heat of formation of [tex]\( \text{CuO(s)} \)[/tex], we can use the given thermochemical equations and apply Hess's Law.

Given Equations:

1. [tex]\( 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \)[/tex]

2. [tex]\( \text{Cu}_2\text{O(s)} \rightarrow \text{CuO(s)} + \text{Cu(s)} \quad \Delta H^\circ = 11.3 \text{ kJ} \)[/tex]

Goal:

Find the standard heat of formation of [tex]\( \text{CuO(s)} \), \( \Delta H_f^\circ \text{(CuO(s))} \).[/tex]

Steps:

1. Write the formation reaction for [tex]\( \text{CuO(s)} \)[/tex]

[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]

2. Manipulate the given equations to match the formation reaction:

a. Reverse Equation 2 to get [tex]\( \text{CuO(s)} \) and \( \text{Cu(s)} \)[/tex] on the reactant side:  

[tex]\[ \text{CuO(s)} + \text{Cu(s)} \rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \][/tex]

b. Add this to Equation 1 (which needs no change) to eliminate[tex]\( \text{Cu}_2\text{O(s)} \): \[ \begin{align*} \text{CuO(s)} + \text{Cu(s)} &\rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \\ 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} &\rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \\ \end{align*} \][/tex]

Adding these equations:

[tex]\[ \text{CuO(s)} + \text{Cu(s)} + 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)} + 4 \text{CuO(s)} \][/tex]

Simplify:

[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]

3. Combine the enthalpy changes:

[tex]\[ \Delta H^\circ = -287.9 \text{ kJ} + (-11.3 \text{ kJ}) = -299.2 \text{ kJ} \][/tex]

Since the above reaction is for 2 moles of [tex]\( \text{CuO(s)} \)[/tex], the enthalpy change for the formation of 1 mole of [tex]\( \text{CuO(s)} \)[/tex] is:

[tex]\[ \Delta H_f^\circ \text{(CuO(s))} = \frac{-299.2 \text{ kJ}}{2} = -149.6 \text{ kJ/mol} \][/tex]

Answer:

sun

Explanation:

Answer:

The sun is where almost all energy on Earth came from.

Answer:

[NaOH] = 0.107M

Explanation:

25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O

The reaction is 1 to 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)

moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole

moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...

Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.

Answer:

[NaOH] = 0.107M

Explanation:

25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O

The reaction is 1 to ratio 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)

moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole

moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...

Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.