A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 1.903 g of AgCl. Calculate the percent by mass of each compound in the mixture.

Answers

Answer 1

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess [tex]AgNO_3[/tex] then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles[/tex]

[tex]\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles[/tex]

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = [tex]\frac{x}{58.5}moles[/tex]

Moles of chloride ions in KCl = [tex]\frac{(0.9440-x)}{74.5}moles[/tex]

The total moles of chloride ions = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

[tex]\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles[/tex]

Now we have to determine the value of 'x'.

Moles of AgCl = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

0.0133 mole = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

By solving the term, we get the value of 'x'.

[tex]x=0.171g[/tex]

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

[tex]\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%[/tex]

[tex]\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%[/tex]

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.


Related Questions

What temperature must a gas initially at 10 °c be brought to for the pressure to triple?

Answers

Answer:

I think that depends on the type of gas and the volume of the container.

Calculate Δ Hrxn for the following reaction: CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) given these reactions and their ΔH values: C(s)C(s)H2(g)+++2H2(g)2Cl2(g)Cl2(g)→→→CH4(g),CCl4(g),2HCl(g),ΔH=−74.6 kJΔH=−95.7 kJΔH=−184.6 kJ Express the enthalpy in kilojoules to one decimal place.

Answers

Answer : The enthalpy of the following reaction is, -390.3 KJ

Explanation :

The given balanced chemical reactions are,

(1) [tex]C(s)+2H_2(g)\rightarrow CH_4(g)[/tex]     [tex]\Delta H_1=-74.6KJ/mole[/tex]  

(2) [tex]C(s)+2Cl_2(g)\rightarrow CCl_4(g)[/tex]     [tex]\Delta H_2=-95.7KJ/mole[/tex]

(3) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]     [tex]\Delta H_3=-184.6KJ/mole[/tex]

The final reaction of is,

[tex]CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)[/tex]       [tex]\Delta H_{rxn}=?[/tex]

Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of of the reaction.

The expression for enthalpy for the following reaction will be,

[tex]\Delta H_{rxn}=[2\times \Delta H_3]+[-1\times \Delta H_1]+[1\times \Delta H_2][/tex]

where,

n = number of moles

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

[tex]\Delta H_{rxn}=[2mole\times (-184.6KJ/mole)]+[-1mole\times (-74.6KJ/mole)]+[1\times (-95.7KJ/mole)]=-390.3KJ[/tex]

Therefore, the enthalpy of the following reaction is, -390.3 KJ

Answer:

-390.3 KJ

Explanation:

For Hess's Law, we need to get the corresponding equation below using the sequence of reactions given

By manipulating the reaction, either reversing them or multiplying/dividing them to a certain factor, we can get to the target equation as well as the total enthalpy

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

C(s) + 2H2(g) → CH4(g)     ΔH = −74.6kJ (needs to reverse)

C(s) + 2Cl2(g) → CCl4(g)    ΔH = −95.7kJ (retain)

H2(g) + Cl2(g) → 2HCl(g)   ΔH = −184.6kJ (multiply by 2 to get 4Cl2 and cancel out 4 HCl and 4 H2)

Therefore, it is -390.3 KJ

A container of carbon dioxide (CO2) has an initial temperature of 170 K with a pressure of 50 kPa. When the container is heated the pressure is measured at 283 kPa. The volume is constant at 12 L throughout. What is the final temperature of the carbon dioxide (CO2)?

Answers

Answer: T2= 962.2 K

Explanation:

The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.

So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so

V1=V2  ⇒

[tex]\frac{n1RT1}{P1} = \frac{n2RT2}{P2} \\\\\frac{T1}{P1} = \frac{T2}{P2} \\\\T2= \frac{T1P2}{P1} =\frac{170 K 283KPa}{50 KPa} =962.2 K[/tex]

Identify the missing coefficient in the following equation:

3

2

1

0

Answers

Answer:

2

Explanation:

In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:

   For the mass numbers(superscript):

   235 + 1 = 1 + 139 + 95

       236 = 235

This is not balanced

  For the atomic number:

  92 + 0 = 0 + 53 + 39

      92 = 92

This is balanced.

We simply inspect to see how to balance the mass number.

By putting a coefficient of 2 behind the neutron atom, the equation becomes balanced.

Use dimensional analysis to convert 14.5mi/hr to km/s

Answers

Answer:

0.006 48 km/s

Explanation:

1. Convert miles to kilometres

14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

1 h × (60 min/1h) × (60 s/1 min) = 3600 s

3. Divide the distance by the time

14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s

Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation of two new nuclei and neutrons. The following equation is an example of one such fission process: 235 92U+10n→AZBa+9436Kr+310n
Enter the isotope symbol for the barium (Ba) nucleus in this reaction.In another process in which 235 92U undergoes neutron bombardment, the following reaction occurs:235 92U+10n --> AZSr+143 54Xe+310nEnter the isotope symbol for the strontium (Sr) nucleus in this reaction.

Answers

Final answer:

The isotope symbol for the barium (Ba) nucleus in the given reaction is 13756Ba. The isotope symbol for the strontium (Sr) nucleus in the second reaction is 8838Sr.

Explanation:

In the given nuclear fission reaction, the isotope symbol for the barium (Ba) nucleus is 13756Ba. This can be determined by balancing the atomic numbers and mass numbers on both sides of the reaction.

In the second reaction, the isotope symbol for the strontium (Sr) nucleus is 8838Sr. Again, this can be obtained by balancing the atomic numbers and mass numbers.

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If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu

Answers

First you need to calculate the number of moles of aluminium and copper chloride.

number of moles = mass / molecular weight

moles of Al = 512 / 27 = 19 moles

moles of CuCl = 1147 / 99 = 11.6 moles

From the reaction you see that:

if        2 moles of Al will react with 3 moles of CuCl

then  19 moles of Al will react with X moles of CuCl

X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.

Place the following in order of increasing dipole moment.

I. BCl3 II. BIF2 III. BClF2

A) I < II < III
B) II < I < III
C) II < III < I
D) I < II = III
E) I < III < II

Answers

Final answer:

The order of increasing dipole moment is II < III < I. BCl3 has no dipole moment, BIF2 has a dipole moment, and BClF2 has a greater dipole moment compared to BIF2.

Explanation:

The order of increasing dipole moment is II < III < I.

The dipole moment depends on the electronegativity difference of the bonded atoms and the bond length. In this case, BCl3 has no dipole moment since the molecule is symmetrical and the individual dipole moments cancel out. BIF2 has a dipole moment due to the electronegativity difference between B and I atoms, and BClF2 has a greater dipole moment compared to BIF2 due to the additional electronegativity difference between B and Cl atoms.

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The correct answer is Option A. The order of increasing dipole moment is I < II < III.

BCl₃ is a symmetrical molecule with no net dipole moment because its individual B-Cl bonds cancel out. Hence, BCl₃ has the smallest dipole moment.

For BIF₂, the asymmetry is introduced as iodine and fluorine have different electronegativities, creating a net dipole moment. However, with only one I atom, the effect is moderate.

BClF₂ is more polar than BIF₂ because the different electronegativities of Cl and F combined in an asymmetrical structure increase the net dipole moment.

The octet rule pertains to:
a. only the noble gases.
b. only groups 1A and 2A.
c. elements in Groups 1A - 7A.
d. all elements in the periodic table.

Answers

I think the answer to c

The starting molecule for the krebs cycle is

Answers

Answer:

acetyl CoA

Explanation:

The starting molecule for the krebs cycle is acetyl CoA.

How do the interactions that are broken in water when it is boiled compare with those broken when water is electrolyzed? Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds. Boiling water breaks covalent bonds and electrolysis breaks intermolecular attractions. Boiling water and electrolysis of water break covalent bonds. Boiling water and electrolysis of water break intermolecular forces.

Answers

Answer:

Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds.

Explanation:

When water boils, hydrogen bonds are broken between adjacent water molecules. The hydrogen bond is an intermolecular bond between adjacent oxygen and hydrogen atoms of water molecules.  

During electrolysis, water dissociates in the presence of electric current. Here, ions are formed in the process. Therefore, covalent bonds are broken here.

Final answer:

Boiling water breaks the intermolecular hydrogen bonds between water molecules, causing it to change from liquid to gas. Electrolysis on the other hand, breaks the covalent bonds within individual water molecules, splitting them into hydrogen and oxygen.

Explanation:

There is a distinct difference between the type of interactions broken when water is boiled and when it's electrolyzed. Boiling water primarily breaks the intermolecular attractions, specifically hydrogen bonds that exist between water molecules, thereby allowing water molecules to escape into the air as steam or water vapor. On the other hand, electrolysis of water involves breaking of the covalent bonds within individual water molecules, thereby splitting water into its constituent hydrogen and oxygen atoms. This process requires more energy compared to boiling water due to the strength of covalent bonds when compared to hydrogen bonds.

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Which formula equation shows a reversible reaction?
a. 2Na+F2–> 2NaF
b. CaCO3–> CaO+CO2
c. NH4CI(s)—–NH3(
g.+HCI(
g. 2H2O2(aq)–pt—> 2H2O(I)+O2(g)

Answers

Answer:

NH₄Cl(s) ⇄ NH₃ (g) + HCl(g)

Explanation:

A reversible reaction is indicated by using a double arrow ⇄

The upper arrow (→), from left to right,  indicates the direct or forward reaction, which goes from left to right.

In the direct reaction, the reactants are the substances shown on the left side of the equation, and the products are the substances shown on the right side.

The lower arrow (←), from right to left, indicates the reverse reaction, which goes from right to left.

In the reverse reaction, the reactants are the substances shown of the right side and the products are the substances shown of the left side.

Summarizing:

Forward reaction: NH₄Cl(s) → NH₃ (g) + HCl(g)

Reverse reaction NH₄Cl(s) ← NH₃ (g) + HCl(g)

Reversible reaction NH₄Cl(s) ⇄ NH₃ (g) + HCl(g)

In the moment that the rates of both forward and reverse reactions are equal it is said that the equilibrium has been reached.

Answer:

just did edge test:

c. NH4CI(s)<—–>NH3( g)+HCI(g)

How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NaCl(aq) Report your answer to 3 decimal places.

Answers

Answer: 2.796 grams

Explanation:

[tex]Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4[/tex]

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

[tex]\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles[/tex]

[tex]\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles[/tex]

According to stoichiometry:

1 mole of [tex]BaCl_2[/tex] reacts with 1 mole of [tex]Na_2SO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will react with=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]Na_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product. [tex]Na_2SO_4[/tex]  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of [tex]BaCl_2[/tex] produces 1 mole of [tex]BaSO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will produce=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]BaSO_4[/tex]

Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g[/tex]

Thus 2.796 grams of [tex]BaSO_4[/tex]  are produced.

Which atom is most likely to attract electrons in a bond?

A) K
B) Na
C) S
D) Se

Answers

C) S

Because of the trends of electronegativity. Increases across a period and decreases down a group.
Final answer:

In a bond, Sulfur (S) is most likely to attract electrons due to its higher electronegativity. This is based on the trend in the periodic table where electronegativity increases across a group from left to right and decreases down a group.

Explanation:

In a bond, elements that have high electronegativity are more likely to attract electrons. Electronegativity is a measure of the ability of an atom to attract the electrons in a bond. It increases across a period from left to right, and decreases down a group in the periodic table. Based on this premise, among the atoms listed, Sulfur (S) and Selenium (Se) have higher electronegativities than Potassium (K) and Sodium (Na).

However, between Sulfur and Selenium, Sulfur (S) is more likely to attract electrons in a bond because it is higher up in the table, given that electronegativity decreases down a group. Therefore, the answer to your question is C) Sulfur (S).

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Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of H2O(g). C2H4(g) + H2O(g) ⇌ C2H5OH(g) Kc = 9.0 × 103 [C2H4]eq = 0.015 M [C2H5OH]eq = 1.69 M

Answers

Final answer:

The equilibrium concentration of H2O(g) in the given reaction is approximately 12.91 × 10^-6 M.

Explanation:

To determine the equilibrium concentration of H2O(g) in the given reaction, we need to use the equilibrium constant expression and the equilibrium concentrations of the other species. The equilibrium constant expression for the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g) is Kc = [C2H5OH]/([C2H4][H2O]). Given that [C2H4]eq = 0.015 M and [C2H5OH]eq = 1.69 M, we can substitute these values into the expression to solve for the equilibrium concentration of H2O(g).

Kc = [C2H5OH]/([C2H4][H2O])

9.0 × 10^3 = 1.69/((0.015)([H2O]))

[H2O] = 1.69/((0.015)(9.0 × 10^3))

[H2O] ≈ 12.91 × 10^-6 M

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The equilibrium concentration of H2O(g) is approximately 0.0125 M.

The equilibrium concentration of H2O(g) given the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g), the equilibrium constant (Kc), and the equilibrium concentrations of C2H4 and C2H5OH at a particular temperature. We start by writing the expression for the equilibrium constant:

Kc = [C2H5OH]/[C2H4][H2O]

Then, we plug in the known values:

9.0 × 10^3 = 1.69/[0.015][H2O]

Now, we solve for [H2O]:

[H2O] = 1.69/(9.0 × 10^3 × 0.015)

[H2O] ≈ 1.69/(135) ≈ 0.0125 M

A solution that is 20% acid and 80% water is mixed with a solution that is 50% acid and 50% water. If twice as much 50% acid solution is used as 20% solution, then what is the ratio of acid to water in the mixture of the solutions?

Answers

Answer:

The ratio acid to water in the mixture is 2:3

Explanation:

Let the volume of 20% acid solution used to make the mixture = x units

So, the volume of 50% acid solution used to make the mixture = 2x units

Total volume of the mixture = x + 2x = 3x units

For 20% acid solution:

C₁ = 20% , V₁ = x

For 50% acid solution :

C₂ = 50% , V₂ = 2x

For the resultant solution of sulfuric acid:

C₃ = ? , V₃ = 3x

Using

C₁V₁ + C₂V₂ = C₃V₃

20×x + 50×2x = C₃×3x

So,

20 + 50×2 = C₃×3

Solving

120 = C₃×3

C₃ = 40 %

Thus, for the resultant mixture,

Acid percentage = 40%

Water percentage = (100 - 40)% = 60%

Ratio acid to water in the mixture = 40:60 = 2:3

Answer:

2:3

Explanation:

one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40 C and a pressure of 1.05 atm can be produced by the reaction of 6.23 g KMnO4 with 45.0 ml of 6.00 m HCl?

Answers

Answer: The volume of chlorine gas produced in the reaction is 2.06 L.

Explanation:

For potassium permanganate:

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

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

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol[/tex]

For hydrochloric acid:

To calculate the moles of hydrochloric acid, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

[tex]6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol[/tex]

For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

[tex]2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O[/tex]

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{2}{16}\times 0.27=0.033moles[/tex] of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{5}{16}\times 0.27=0.0843moles[/tex] of chlorine gas.

To calculate the volume of gas, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]40^oC=[40+273]K=313K[/tex]

Putting values in above equation, we get:

[tex]1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L[/tex]

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

Final answer:

The question involves a stoichiometry problem in chemistry, where we calculate the volume of chlorine gas from the reaction of potassium permanganate with hydrochloric acid. We identify the limiting reactant and then use the ideal gas law to find the volume under the given conditions of temperature and pressure.

Explanation:

The student's question concerns a chemical reaction between potassium permanganate (KMnO4) and hydrochloric acid (HCl) to produce chlorine gas (Cl2). This is a stoichiometry problem where we will calculate the volume of chlorine gas generated at specific conditions using the ideal gas law. To find the number of moles of chlorine gas that can be produced, we first determine the limiting reactant. Then, using the ideal gas law (PV=nRT), we can calculate the volume at the given temperature and pressure.

First, we must find the reaction equation:
KMnO4 + 16HCl → 2KCl + 5Cl2 + 2MnCl2 + 8H2O

Next, we calculate the moles of KMnO4 and HCl. Following this, we determine the limiting reactant and use it to calculate moles of Cl2 produced. Lastly, we use the ideal gas law (PV=nRT, with R=0.0821 L·atm/(mol·K)) to find the volume of Cl2 at 40°C and 1.05 atm.

What is the percent-by-mass concentration of acetic acid (CH3COOH) in a vinegar solution that contains 51.80 g acetic acid in a 1.000−L solution? The density of this solution is 1.005 g/mL.

Answers

Final answer:

The percent-by-mass concentration of acetic acid in the vinegar solution is approximately 5.15%, calculated by dividing the acetic acid mass by the total solution mass and then multiplying by 100.

Explanation:

To find the percent-by-mass concentration of acetic acid in vinegar, the mass of acetic acid is divided by the total mass of the solution and then multiplied by 100. First, convert the solution volume to mass using the given density. The density of the solution is 1.005 g/mL, which means 1.000 L (or 1000 mL) of solution has a mass of 1005 g (1000 mL × 1.005 g/mL). The mass of acetic acid is given as 51.80 g. Thus, the percent-by-mass concentration is calculated as (51.80 g / 1005 g) × 100%.

The calculation gives a percent-by-mass concentration of approximately 5.15%. This value represents the mass of acetic acid expressed as a percentage of the total mass of the solution.

The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Select one:
a. 52.8
b. 728
c. 0.198
d. 3490

Answers

Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

PLEASE ANSWER FAST
If iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as coal burns. If a furnace burns an amount of coal containing 198.20 g of FeS2, how much SO2 (an air pollutant) is produced?

4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
Select one:
a. 211.7
b. 52.92
c. 590.8
d. 582.1

Answers

First we need to calculate the number of moles of FeS[tex]_{2}[/tex]:

number of moles = mass (grams) / molecular mass (g/mol)

number of moles of FeS[tex]_{2}[/tex] = 198.2/120 = 1.65 moles

From the chemical reaction we deduce that:

if            4 moles of FeS[tex]_{2}[/tex] produces 8 moles of SO[tex]_{2}[/tex]

then  1.65 moles of FeS[tex]_{2}[/tex] produces X moles of SO[tex]_{2}[/tex]

X = (1.65×8)/4 = 3.3 moles of SO[tex]_{2}[/tex]

Now we can calculate the mass of SO[tex]_{2}[/tex]:

mass (grams) = number of moles × molecular mass (grams/mole)

mass of SO[tex]_{2}[/tex] = 3.3×64 = 211.2 g

The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????,k, is 6.1×10−8 s−1.6.1×10−8 s−1. What is the value of the rate constant at 785.0 K?

Answers

Answer : The rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]600.0K[/tex] = [tex]6.1\times 10^{-8}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]785.0K[/tex] = ?

[tex]Ea[/tex] = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]600.0K[/tex]

[tex]T_2[/tex] = final temperature = [tex]785.0K[/tex]

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

[tex]\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}][/tex]

[tex]K_2=1.45\times 10^{-2}s^{-1}[/tex]

Therefore, the rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]

The rate constant at the new temperature is 1.81 ×10^−2 s−1.

We have the reaction written as follows; C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g).

We have to use the formula;

ln(k2/k1) =Ea/R(1/T1 - 1/T2)

Now;

k2 = ?

k1 = 6.1×10−8 s−1

Ea = 262 ×10 ^3 J/mol

T1= 600.0 K

T2 = 785.0 K

R = 8.314 J/K. mol

ln (k2/6.1×10−8) = 262 ×10 ^3 /8.314 (1/600.0 - 1/785.0)

ln (k2/6.1×10−8) = 12.6

k2/6.1×10−8 = e^12.6

k2 = 6.1×10−8 (e^12.6)

k2 = 1.81 ×10^−2 s−1.

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Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined with an excess of O2 and ignited in a bomb calorimeter. After the reaction, it was found that the temperature of the calorimeter had increased from 25.000C to 26.061C. The calorimeter contained 1.000 kg of water. The heat capacity of the calorimeter was 4.643 kJ/C. Determine the heat of reaction, in kJ/mol propane. The reaction was:

Answers

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

[tex]q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ[/tex]

Moles of propane =[tex]\frac{1.860 g}{44 g/mol}=0.0422 mol[/tex]

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

[tex]\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol[/tex]

How can the value of Ksp be related to the molar solubility of a compound? View Available Hint(s) How can the value of Ksp be related to the molar solubility of a compound? When a common ion is present, the solubility of the compound decreases, and this is reflected in a lower value of Ksp. The square of a compound's molar solubility equals the value of Ksp for the compound. The value of Ksp equals the concentration of the compound in a saturated solution, which can be converted to the molar solubility using the molar mass. The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

Answers

Answer:

The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

Explanation:

Consider a slightly soluble solid with formula M₃X₂. Its solubility product expression is

[tex]\begin{array}{rcccc}M_{3}X_{2}(s) & \rightleftharpoons&3M^{2+}(aq) & + & 2X^{3-}(aq)\\& & 3s & &2s\\\\K_\text{sp}& = & [3s]^{3}[2s]^{2}&= & 108s^{5}\\\\\end{array}[/tex]

Thus, the molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

A is wrong. The solubility product constant is a constant. It does not change in the presence of a common ion.

B is wrong. It is correct only for compounds with formula MX.

C is wrong. Ksp does not equal the concentration of the compound in solution.

The value of Ksp is related to the molar solubility in that molar solubility helps calculate the ion concentrations in a solution, which are used to calculate Ksp. The presence of a common ion affects both solubility and Ksp through the common ion effect.

The value of Ksp, or the solubility product constant, is closely related to the molar solubility of a compound, which is a measure of how much of the compound can dissolve in a given amount of solvent to form a saturated solution. The connection between Ksp and molar solubility can be understood through a series of steps:

First, molar solubility is calculated by converting the solubility of a compound (often given in g/L) to moles per liter using the molar mass of the compound.Next, the dissociation equation of the compound is used to determine the concentration of each ion in the solution.Finally, these ion concentrations are used in the Ksp expression to calculate the solubility product constant.

Furthermore, if a common ion is present in the solution, it affects the solubility of the compound and thus the Ksp value. This is a demonstration of the common ion effect, where the presence of a common ion decreases the solubility and Ksp of the compound according to Le Chatelier's principle.

Therefore, understanding the relationship between Ksp and molar solubility is essential for predicting the solubility of compounds and determining the possible concentrations of ions in solution.

Which of the following would increase the rate of a chemical reaction between hydrochloric acid (HCl) and solid zinc metal (Zn)?

A. Decreasing the concentration of HCl
B. Pulverizing the zinc metal into a fine powder
C. Performing the reaction at a lower temperature
D. Decreasing the amount of Zn

Answers

Answer:

B. Pulverizing the zinc metal into a fine powder

Answer: B. Pulverizing the zinc metal into a fine powder

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

A. Decreasing the concentration of [tex]HCl[/tex] and [tex]Zn[/tex]: Thus rate of the reaction would decrease on decreasing the concentration of hydrochloric acid and zinc.

B.  Pulverizing the zinc metal into a fine powder : If the solute particles is present in smaller size, more it can take part in the chemical reaction due to large surface area, hence increasing the rate of reaction.

C. Performing the reaction at a lower temperature. Decreasing the temperature means the energy of the particles is less and thus lesser reactants would cross the energy barrier and thus lesser will be the rate.

Identify each definition that applies to the compound in red. Check all that apply. HCI + NaOH → H2O + NaCl
Arrhenius acid
Bronsted-Lowry acid
Arrhenius base
Bronsted-Lowry base

Answers

Answer:

A. Arrhenius acid    

B. Bronsted-Lowry acid

Explanation:

An acid base reaction invoves an acid and a base and yields salt and water as products.

What is a neutralization?

A neutralization is a reaction between an acid and a base to yield salt and water only. The highlighted compound is not shown here however we we shall tell what is compound is in this reaction.

HCI - Arrhenius acidNaOH  - Arrhenius baseH20 -  waterNaCl - Salt

Hence, this an acid base reaction in the Arrhenius sense.

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Which region in the IR spectrum could be used to distinguish between butanoic acid and 2-butanone?
a. 1680-1750 cm-1
b. 3200-3600 cm-1
c. 1600 cm1
d. 2500-3300 cm-1

Answers

Answer:2500-3300[tex]cm^{-1}[/tex]

Explanation:We can distinguish between 2-butanone and butanoic acid by analyzing the spectra .                                          

The structure of the two compounds are slightly different as butanoic acid has an OH group whereas for 2-butanone there is no OH group present.

Please refer the attachments for structure of compounds.

Both the compounds will show carbonyl stretching frequency at around 1700-1750cm^{-1}[/tex] due to the presence of carbonyl group in both the compounds.

A broad intense peak at around 3000[tex]cm^{-1}[/tex] will be observed only for butanoic acid due to the presence of OH group and 2-butanone would not show any broad intense peak at around 3000[tex]cm^{-1}[/tex].

So by analyzing the IR spectra and identifying the intense broad peak of OH we can easily distinguish between butanoic acid and 2-butanone.

Due to hydrogen bonding in between two carboxylic acid molecules ,peak broadening for OH group in butanoic acid takes place.

Generally the ketones show IR stretching in around 1700-1730cm^{-1}[/tex] due to the presence of carbonyl group.

Generally carboxylic acids show IR stretching  for carbonyl group at around 1700-1760cm^{-1}[/tex] and a broad intense peak for OH at around 2500-3000cm^{-1}[/tex].

Final answer:

The C=O stretching region in the IR spectrum can be used to distinguish between butanoic acid and 2-butanone.

Explanation:

In the IR spectrum, the region that could be used to distinguish between butanoic acid and 2-butanone is the C=O stretching region. The C=O stretching vibration band of saturated aliphatic ketones, such as 2-butanone, appears at around 1715 cm-1. However, butanoic acid contains a carboxyl group (COOH), which exhibits a broad and intense peak in the 1680-1750 cm-1 range.

The scientific principle which is the basis for balancing chemical equations is

Answers

Answer:

Law of Conservation of Mass

Explanation:

Final answer:

The Law of Conservation of Matter forms the basis for balancing chemical equations, ensuring that the number of each element is equal on both sides of the equations. This principle also serves to describe a reaction's stoichiometry, where amounts of reactants and products in a reaction are considered.

Explanation:

The scientific principle which forms the basis for balancing chemical equations is the Law of Conservation of Matter. According to this principle, the same number of each element must be represented on the reactant (input) and product (output) sides of an equation. This ensures that equations accurately reflect the reality that matter is not created or destroyed in a chemical reaction.

For instance, consider the following chemical equation: PC15 (s) + H₂O(1) →→→ POC13 (1) + 2HCl(aq). In this equation, we balance the equation by ensuring that for each of the elements involved (P, C, and H), the number of atoms of that element is equal on both sides of the equation.

Beyond simply balancing the equation, this principle also helps describe a reaction's stoichiometry, which involve the relationships between amounts of reactants and products. Coefficients from the balanced equation can be used in computations relating to reactant and product masses, molar amounts, and other quantitative properties.

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The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Select one:
a. 52.8
b. 728
c. 0.198
d. 3490

Answers

Answer:

a) 52.8

Explanation:

M1V1 = M2V2

(7.91 M)(x ml) = (2.13 M) (196.1 ml)

(7.91M) (xml) = 417.693 M.ml

x ml = 417.693/ 7.91

x   =  52.8

Answer:

A. 52.8

Explanation:

We make use of the dilution formula to solve this

[tex]M_1\times V_1 =M_2\times V_2[/tex]

Where ([tex]M_1[/tex] and [tex]M_2[/tex] are the initial molarity and final molarity and [tex]V_1[/tex] and [tex]V_2[/tex] are  the initial volume and final volume)

Plugging into the values in the formula  

[tex]M_1\times V_1 =M_2\times V_2[/tex]

[tex]7.91M \times V_1=2.13M \times 196.1mL[/tex]

[tex]V_1=\frac {(2.13M \times 196.1mL)}{7.91M}[/tex]

= 52.8 mL is the Answer

Select the phrase that best describes scientific findings communicated through popular media.
A) generally free of technical jargon
B) peer reviewed
C) commonly used by most scientists
D) extremly reliable

Answers

Answer:

The correct answer is option A) generally free of technical jargon

Explanation:

Hello!

Let's solve this!

When we read popular magazines or newspapers, we see scientific findings. They generally do not use scientific or technical vocabulary so that more people can understand it. They are also not the most reliable sources, since to look for reliable information, we have to look for scientific journals.

After the analysis we conclude that the correct answer is option A) generally free of technical jargon

1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?

Select one:
a. 45600
b. 91200
c. 91200
d. 11400

2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the

Select one:
a. exponent.
b. half-life.
c. isotope.
d. nucleus.

3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.

Select one:
a. 138 days
b. 276 days
c. 414 days
d. 552 days


Answers

1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?

Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.

2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.

3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.

What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.

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