How does the solubility of atmospheric carbon dioxide change with variations in pH and salinity?

Answers

Answer 1

Answer:

Decreasing pH decreases the solubility; increasing salinity increases the salinity.

Explanation:

Atmospheric carbon dioxide in seawater is involved in the following equilibria:

(i)   CO₂(g) ⇌ CO₂(aq)

(ii)  CO₂(aq) + H₂O ⇌ H₂CO₃(aq)  

(iii) H₂CO₃(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HCO₃⁻(aq)  

(iv) HCO₃⁻(aq)+ H₂O(l) ⇌ H₃O⁺(aq) + CO₃²⁻(aq)

1. Effect of pH  

Decreasing pH increases [H₃O⁺].

According to Le Châtelier's Principle, the position of equilibrium of (iv) will be pushed to the left.  

This, in turn, pushes the positions of equilibrium of (iii), (ii), and (i) to the left.

The net effect is that CO₂ is pushed out of the solution and back into the atmosphere.

Thus, decreasing the pH decreases the solubility of atmospheric CO₂.

2. Effect of salinity

Salinity includes the measurement of all ions in seawater, not just Na⁺ and Cl⁻.

An important ion that contributes to salinity is Ca²⁺.

Ca²⁺ ions are involved in the equilibrium  

Ca²⁺(aq) + CO₃²⁻(aq) ⇌ CaCO₃(s).

Thus, increasing the salinity ([Ca²⁺]) removes CO₃²⁻ from the solution and pulls the position of equilibrium of (iv) to the right.

The effect goes back through the chain of equilibria.

The net result is that more atmospheric CO₂(g) dissolves in the seawater to replace the CO₃²⁻ that has been removed.

Increasing the salinity increases the solubility of atmospheric CO₂.


Related Questions

Use the balanced chemical equation below. How many grams of the product are formed when 2.34 g of sulfur is completely reacted with fluorine? S8 + 16F2(g) → 8SF4

Answers

Answer:

7.89 g

Explanation:

Step 1: Write the balanced equation

S₈ + 16 F₂(g) → 8 SF₄

Step 2: Calculate the moles corresponding to 2.34 g of S₈

The molar mass of S₈ is 256.52 g/mol.

[tex]2.34g \times \frac{1mol}{256.52g} = 9.12 \times 10^{-3} mol[/tex]

Step 3: Calculate the moles of SF₄ produced from 9.12 × 10⁻³ mol of S₈

The molar ratio of S₈ to SF₄ is 1:8. The moles of SF₄ produced are 8/1 × 9.12 × 10⁻³ mol = 0.0730 mol

Step 4: Calculate the mass corresponding to 0.0730 moles of SF₄

The molar mass of SF₄ is 108.07 g/mol.

[tex]0.0730 mol \times \frac{108.07 g}{mol} = 7.89 g[/tex]



Question 7
How many molecules of water are in 0.042 Lof pure water (ignoring any temperature
effect)?
Express your answerin e-notation using as many decimal places as necessary
Example: 1,000,001 should be expressed as 1.000001e+6
Notes: Avogadro's constant = 6.02214076e+23 mol 1

Answers

Explanation:

Moles of 0.042 L of Pure water = 0.042 / 22.4

[tex]Moles \: of \: 0.042 L \: of \: Pure \: water \: = 1.87 \times {10}^{ - 3} [/tex]

[tex]Molecules \: in \: 0.042 L \: of \: Pure \: water \: = moles \times avogadro \: number[/tex]

Molecules in 0.042 L of Pure Water

= 1.126 * 10^21

Draw the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.Draw the starting amine. Be sure to draw nonbonding electron pairs.
Draw the starting carboxylic acid. Be sure to draw nonbonding electron pairs.

Answers

Answer:

Explanation:

Amidation is a reaction with the formation of an amide. It involves a process where series of reaction between amine and carboxylic acid takes place.

The main objective of this question is to draw with the aid of a diagram.

the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.

But before that,

we are to draw the starting amine and to be sure to include the nonbonding electron pairs.

Also, to draw  starting carboxylic acid and to be sure to include the nonbonding electron pairs.

The whole series of the diagrammatic expression for the amine formation can be found in the diagram attached below.

what mass of TiCl4 must react with an excess of water to produce 50.0g of TiO2 if the reaction has a 78.9% yield

Answers

Answer:

\large \boxed{\text{150 g TiCl}_{4}}  

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     189.68                  79.87

           TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl

m/g:                                 50.0

To solve this stoichiometry problem, you must

Convert the actual yield to the theoretical yield  Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂ Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄ Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄

1. Theoretical yield of TiO₂

[tex]\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}[/tex]

2.  Moles of TiO₂

[tex]\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}[/tex]

3,  Moles of TiCl₄

The molar ratio is 1 mol TiO₂:1 mol TiCl₄.

[tex]\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}[/tex]

4.  Mass of TiCl₄

[tex]\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}[/tex]

What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? Give your answer in nm.

Answers

Answer:

the wavelength of radiation emitted  is [tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

[tex]\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}[/tex]

For n = 7:

[tex]\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}[/tex]

[tex]\mathtt{E_7 =-0.27755 \ eV}[/tex]

For n = 4

[tex]\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}[/tex]

[tex]\mathtt{E_4 =- 0.85\ eV}[/tex]

The  electron goes from the n = 7 to the n = 4, then :

[tex]\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}[/tex]

[tex]\mathtt{= 0.57245\ eV}[/tex]

Wavelength of the radiation emitted:

[tex]\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}[/tex]

where;

hc  = 1242 eV.nm

[tex]\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}[/tex]

[tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

There is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+.
A. Write an equation for the half-reaction occurring at each electrode. Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.
B. What minimum voltage is necessary to drive the reaction? Vmin =

Answers

Answer:

See explanation

Explanation:

Anode;

Sn(s) ------> Sn^2+(aq) + 2e

Cathode;

Mn^2+(aq) + 2e ------> Mn(s)

The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;

E°cell= E°cathode - E°anode

E°cell= -1.19 - (-0.14)

E°cell= -1.05 V

2. Which of these is an extensive property?
a. Density
b. Melting point
Temperature
d. Volume

Answers

The answer is D volume

Volume ..........is the answer

2 points
How many significant figures are there in the measurement 2.30 x 10^5
mm ? *
2
3
4.
5

Answers

Answer:

3.

Explanation:

Hello,

In this case, considering the given number:

[tex]2.30x10^5[/tex]

We can see that it does not have zeros to left of the first nonzero digit which is 2, therefore, we just have to consider the 2, 3 and 0 as significant figures because the right-handed zeros are actually considered. Hence, such measurement has 3 significant figures.

Mass and energy are conserved Question 13 options: A) only in chemical changes. B) always in physical changes and sometimes in chemical changes. C) only in physical changes. D) in chemical changes and physical changes.

Answers

Answer:

D) in chemical changes and physical changes.

Explanation:

Mass and energy are conserved in chemical changes and physical changes. Therefore, option D is correct.

What is physical change ?

Physical modifications are those that affect a chemical substance's form but not its chemical content. Physical changes may normally be used to separate compounds into chemical elements or simpler compounds, but they cannot be used to separate mixtures into their component compounds.

A change in physical attributes accompanies a physical change. Melting, turning into a gas, changing strength or durability, changing crystal structure or texture, and changing shape, size, color, volume, or density are a few examples of physical qualities.

In a physical change, the substance's shape or appearance changes, but the type of matter it contains stays the same. In contrast, when matter undergoes a chemical transformation, at least one new substance with novel features is created.

Thus, option D is correct.

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Of the following solutions, which has the greatest buffering capacity?

a. 0.365M HC2H3O2 and 0.497 M NaC2H3O2
b. 0.521 M HC2H3O2 and 0.217 M NaC2H3O2
c. 0.821 M HC2H3O2 and 0.713 M NaC2H3O2
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Answers

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )[/tex]

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. [tex]\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36[/tex]

b. [tex]\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417[/tex]

c. [tex]\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868[/tex]

d. [tex]\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959[/tex]

Therefore, the d. solution has the best buffering capacity.

Regards.

what is 31/50000 If the toxic quantity is 1.5 g of ethylene glycol per 1000 g of body mass, what percentage of ethylene glycol is fatal

Answers

Answer:

[tex]\%m=0.15\%[/tex]

Explanation:

Hello,

In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed with the shown below formula:

[tex]\%m=\frac{m_{ethylene \ glycol}}{m_{ethylene \ glycol}+m_{body\ mass}}*100\%[/tex]

We can use the given masses to obtain:

[tex]\%m=\frac{1.5g}{1.5g+1000g}*100\%\\ \\\%m=0.15\%[/tex]

Best regards.

In an electrolytic cell, the magnitude of the standard cell potential is the:______
a. minimum voltage that must be supplied for a redox reaction to occur
b. maximum voltage that willallow the redox reaction to occur
c. always equal to Eanode - Ecathode
d. none of the above

Answers

Answer:

a. minimum voltage that must be supplied for a redox reaction to occur

c. always equal to Eanode - Ecathode

Explanation:

In an electrolytic cell;  The electromotive force(the maximum standard potential difference) of the cell formed by the system is defined as the standard electrode potential of the right handed electrode minus the standard electrode potential of the left hand electrode. (i.e [tex]\mathbf{E^{\theta}_{cell}=E_{anode} - E_{cathode}}[/tex] )

As we all known that the process by which chemical energy is being converted to electrical energy is called the Electrochemical cell. It consists of two half cells , an oxidation half cell reaction and a reduction half-cell reaction.The overall redox reaction results in a flow of electrons in an electric current which is produced by a minimum voltage.

Therefore, option a and c are both correct.

A patient is prescribed 100mg/day of antibiotic for two weeks. The antibiotic is available in vials that contain 20mg/vial of the drug. How many vials are necessary for the entire treatment?

Answers

Answer:

70

Explanation:

100/20 =5

5 x each day over 2 weeks (14 days) = 70

The vials are necessary for the entire treatment is 70 vials

We have to first calculate the total number of antibiotics taken by the patient in two weeks, then find the total vials neccessary for the entire treatment.

The total antibiotics for a period of two weeks = (100 mg/day)(2 weeks)

Total = 100 mg/day×14 days

Total = 1400 mg for two weeks.

If the antibiotics contain vials in 20 mg/vials of the drugs,

Then the vials necessary for the entire treatment is

vial for the entire treatment  = 1400/20

total vials = 70 vials

Hence the vials are necessary for the entire treatment is 70 vials

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Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the C-side of L-Arg and L-Lys, but not on the C-side of D-Arg and D-Lys. Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the -side of L- and L-, but not on the -side of D- and D-. All the naturally occurring amino acids are D-amino acids, while active centres of serine proteases are nearly all L. All the naturally occurring amino acids are L-amino acids, while active centres of serine proteases are nearly all D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site. None of the above.

Answers

Answer:

D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site

Explanation:

Stereospecificity is the ability to distinguish between stereoisomers of of a particular compound. L- and D- structures of compounds in living organisms are usually present in only one form due to stereospecificity. For example, naturally occuring amino acids in proteins are usually present as L-isomers.

Since enzyme are proteins, their active sites are composed of L-amino acid and they show stereospecificity in the reactions they catalyze. In their binding sites, only substrates complementary in structure can bind in order for catalysis to proceed. Therefore, only amino acids in the L- configuration are complementary to the active site of enzymes.

In the case of serine proteases, The side chains of -Arg and D-Lys will not be positioned properly for binding at the binding site of serine proteases, therefore, no catalysis will occur. On the other hand, L-Arg and L-Lys can bind to the catalytic site of serine proteases since they are complementary fits to the active site of the enzymes.

Magnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.

Magnesium ions produce no emission of absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame

Answers

Answer:

Traces of sodium impurity

Explanation:

Metal ions are identified by a characteristic colour imparted to flame by the metal ion solution. Various metals have various colours which they impart to a flame.

The energy of a flame is not sufficient for the excitation of electrons of Mg to a higher energy level. As a result of this, Mg do not give any color in Bunsen flame.

However, a few tinges of yellow-orange colour which is characteristic of sodium metal do appear when magnesium ions are exposed to a flame as a consequence of traces of sodium impurity in the magnesium ion solution.

What are 1A, 3B, and 7A examples of on the periodic table?

Answers

Answer:

Groups

Explanation:

1 A, 3 B, and 7 A are examples of  group number on the periodic table

Group 1A are  the alkali metals includes lithium sodium and potassium.

Group 3B in most periodic tables, lanthanum and actinium are considered to be a part of Group3B.

Group 7A are the halogens includes chlorine ,bromine and iodine.

What are  some characteristics of Group1A elements ?

They are all soft, silver metals. Due to their low ionization energy, these metals have low melting points and are highly reactive. The group 1A elements with their ns1 valence electron configurations are very active metals. They lose their valence electrons very readily.

All the Group 3B elements are rather soft, silvery-white metals, although their hardness increases with atomic number. They have higher ionization energies than the Group 1A and 2A elements, and are ionized to form a 3+ charges.

They have very high electronegativity. They have seven valence electrons . They are highly reactive, especially with alkali metals and alkaline earths metals.

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Which of the following conversions involves an oxidation of the underlined element?
A. H3PO3 → H2P03
B. 03 → 02
C. S03 → S042-
D. BrO2 → BrO-
E. CIO3 → ClO4

Answers

A.H3PO - H2P03 should be but good luck

How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Answers

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

[tex]23.6 g \times \frac{1mol}{266.69g} = 0.0885 mol[/tex]

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

[tex]0.0885molAlBr_3 \times \frac{2molAl}{2molAlBr_3} = 0.0885molAl[/tex]

0.085 moles of Al are required to form 23.6 g of AlBr₃.

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Which of the following has the longest half-life?
a. 0-19
b. H-3
c. Zn-71
d. Rb-85

Answers

Answer:

b. H-3

Explanation:

The half life of the following isotopes is given as;

a. 0-19

26.88 seconds

b. H-3

12 years

c. Zn-71

2.4 minutes

d. Rb-85

This is a stable isotope. It does not decomposes.

Comparing the half lives, the isotope with the longest half life is Hydrogen - 3. With half life of 12 years.


Identify if the following are chemical or physical properties: "CP" for
chemical property and "PP" for physical property
1. Oxygen is odorless and colorless
2. Copper turns green when exposed to the environment
3. The piece of metal is magnetic
4. The density of water is 1.0 g/cm3
5. Diamonds are a very hard substance
6. The tree is 8 meters high

Answers

Answer:

1.pp

2.cp

3.pp

4.pp

5.pp

6.pp

9.57: How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4(s) at standard state conditions?

Answers

Answer:

459.6kj

Explanation:

the chemical equation =

3mn(s) + 2CO2(g) --- Mn3O4(s)

we have change in H to be -1378.83

we are required to find the amount of heat that is liberated for a mole of manganese.

3 moles of manganese = -1378.83

1 mole of manganese = ?

when we cross multiply, we will have a ratio that we will use to find this amount

1378.83/3 = 459.6kj

The complete combustion of copper(I) sulphide is according to the following equation:
2Cu2S(s) + 302(g) →→2Cu2O(s) + 2SO2(g)
If the mass of Cu2S in the mixture is 14.0 g, calculate
(a) the number of molecules of oxygen gas reacted.
(b) the mass of SO2 gas produced.
(c)the volume of SO2 gas at STP

Answers

Answer:

a) 7.94 x 10²³ molecules, b) 5.62 g SO2, c) 1.97 L

Explanation:

a) We need to first convert Cu2S to moles. Since molar mass of Cu2S is 159.14 g/mol, 14.0 g = 0.0880 mol. Using molar ratios (3 mol O2/2 mol Cu2S, 0.0880 mol of Cu2S = 0.132 mol O2. Since 1 mol contains 6.02 x 10²³ molecules, 0.132 mol O2 = 7.94 x 10²² molecules O2.

b) Since the molar ratios of Cu2S to SO2 is 1:1, 0.0880 mol of Cu2S produces 0.0880 mol SO2. To convert mol to grams, we use the molar mass of SO2 (64.06 g/mol) to figure out that 0.0880 mol SO2 = 5.63 g SO2.

c) At STP, 1 mol occupies 22.4 liters. 0.0880 mol SO2 x 22.4 L/1 mol = 1.97 L

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What volume of a 0.124 M KOH solution neutralizes 23.4 mL of 0.206 M HCl solution?
A) 15.9 mL
B) 38.9 mL
C) 31.8 mL
D) 1.00 × 104 mL
E) 5.00 × 102 mL

Answers

A) 15.9 mL I hope it is if not I’m sorry

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

Neutralization reaction has resulted in the formation of salt and water with the reaction of acid and base.

In the neutralization reaction, the strength of the acid and base can be given by:

Molarity of acid [tex]\times[/tex] Volume of acid = Molarity of base [tex]\times[/tex] Volume of base

Given, the molarity of KOH base = 0.124 M

The volume of acid (HCl) = 23.4 ml

Molarity of acid (HCl) = 0.206 M.

Substituting the values:

0.206 [tex]\times[/tex] 23.4 = 0.124 × Volume of base (KOH)

Volume of base (KOH) = 38.874 ml.

Volume of KOH = 38.9 ml

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

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At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. What is the composition in mole fractions of a solution that has a vapor pressure of 40. torr at 20°C?

Answers

Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]

where, x = mole fraction in solution

[tex]p^0[/tex] = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_1+p_2\\[/tex][tex]p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0[/tex]

[tex]x_{benzene}=x[/tex],

[tex]x_{toluene}=1-x_{benzene}=1-x[/tex],

[tex]p_{benzene}^0=75torr[/tex]

[tex]p_{toluene}^0=22torr[/tex]

[tex]p_{total}=40torr[/tex]

[tex]40=x\times 75+(1-x)\times 22[/tex]

[tex]x=0.34[/tex]

Thus (1-x0 = (1-0.34)=0.66

Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66

What mass (in grams) of aspirin (C₉H₈O₄) is produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield from the reaction below? C₇H₆O₃ (s) + C₄H₆O₃ (s) → C₉H₈O₄ (s) + HC₂H₃O₂ (aq).

Answers

Answer:

71.3 g

Explanation:

molar mass of C₇H₆O₃ = 138.13 g

molar mass of [tex]C_{9}H_{8}O_{4}[/tex] = 180.17 g

find the moles of reactant: C₇H₆O₃

C₇H₆O₃ that reacted = mass/molar mass = 57.6 g C₇H₆O₃ / 138.13 g C₇H₆O₃

= 0.417 mol [tex]C_{7}H_{6}O_{3}[/tex]

From the reaction equation, 1 mole of C₇H₆O₃ yields one mole of aspirin

find theoretical yield of aspirin:

0.417 mol C7H6O3 x 180.17 g C9H8O4 / 1 mol C9H8O4[tex]\frac{0.417 mol C_{7}H_{6}O_{3}}{} x \frac{180.17 g C_{9}H_{8}O_{4}}{1 mol C_{9}H_{8}O_{4}}[/tex]

= 75.1 g C9H8O4

Actual yield= % yield × theoretical yield/100

Actual yield = 95.0 × 75.1/100

Actual yield = 71.3 g

Taking into account the reaction stoichiometry and actual yield, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.

The balanced reaction is:

22 C₇H₆O₃ (s) + 32 C₄H₆O₃ (s) → 24 C₉H₈O₄ (s) + 33 HC₂H₃O₂ (aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₇H₆O₃: 22 moles C₄H₆O₃: 32 moles C₉H₈O₄: 24 moles HC₂H₃O₂: 33 moles

The mass molar of each compound is:

C₇H₆O₃: 138 g/moleC₄H₆O₃: 102 g/mole C₉H₈O₄: 180 g/moleHC₂H₃O₂: 60 g/mole

Then by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

C₇H₆O₃: 22 moles× 138 g/mole= 3036 gramsC₄H₆O₃: 32 moles× 102 g/mole= 3264 grams C₉H₈O₄: 24 moles× 180 g/mole= 4320 gramsHC₂H₃O₂: 33 moles× 60 g/mole= 1980 grams

Then you can apply the following rule of three: if by stoichiometry 3036 grams of C₇H₆O₃ produce 3264 grams of C₉H₈O₄, 57.6 grams of C₇H₆O₃ will produce how many mass of C₉H₈O₄?

[tex]mass of C_{9} H_{8} O_{4} =\frac{57.6 grams of C_{7} H_{6} O_{3} x3264 gramsof C_{9} H_{8} O_{4} }{3036 grams of C_{7} H_{6} O_{3}}[/tex]

mass of C₉H₈O₄= 60.03 grams

On the other side, actual yield is the amount of product actually obtained from a reaction. Assuming 95.0% yield  and considering the previously calculated mass the maximum amount of product that can be produced in the reaction, you can apply the following rule of three:  If 100% equals 60.03 grams of the compound produced, 95% equals how much mass of the compound?

[tex]mass of C_{9} H_{8} O_{4} =\frac{95 percentx60.03gramsof C_{9} H_{8} O_{4} }{100 percent}[/tex]

mass of C₉H₈O₄= 57.0285 grams

Finally, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.

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Calculate the solubility at of AgCl in pure water and in a ).0010 M solution. You'll find data in the ALEKS Data tab. Round both of your answers to significant digits.

Answers

Answer:

1.34x10⁻⁵ mol / L is the solubility of AgCl

Explanation:

Ksp of AgCl is defined as:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:

Ksp = [Ag⁺] [Cl⁻]

Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):

Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:

[Ag⁺] = X

[Cl⁻] = X

Where X is the amount of AgCl that dissolvesin water. X = solubility:

Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]

1.8x10⁻¹⁰ = [X] [X]

1.8x10⁻¹⁰ = X²

X =

1.34x10⁻⁵ mol / L is the solubility of AgCl

Calculate the molality of a 15.0% by mass solution of MgCl 2 in H 2O. The density of this solution is 1.127 g/mL. Group of answer choices

Answers

Answer:

1.86 m

Explanation:

Step 1: Calculate the mass of solute and solvent in 100 grams of solution

We have a 15.0% by mass solution of MgCl₂ (solute) in H₂O (water). In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.

Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂

The molar mass of MgCl₂ is 95.21 g/mol.

[tex]15.0g \times \frac{1mol}{95.21 g} = 0.158mol[/tex]

Step 3: Convert the mass of water to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]85.0g \times \frac{1kg}{1,000g} = 0.0850kg[/tex]

Step 4: Calculate the molality of the solution

[tex]m = \frac{0.158mol}{0.0850kg} = 1.86m[/tex]

The molality is 1.86 m.

Given:

Mass solution of MgCl₂ in H₂O= 15%

Density= 1.127g/mL

To find:

Molality=?

Let's solve this question step by step:

Step 1: Calculate the mass of solute and solvent in 100 grams of solution

As it is given that 15.0% by mass solution of MgCl₂ (solute) in H₂O (water).

In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.

Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂

The molar mass of MgCl₂ is 95.21 g/mol.

Number of moles is given mass divided by the molar mass.

[tex]\text{Number of moles}= \frac{15g}{95.21g/mol} =0.158mol[/tex]

Step 3: Convert the mass of water to kilograms.

As we know, 1 kg = 1,000 g.

[tex]=\frac{85}{1000} =0.085kg[/tex]

Step 4: Calculate the molality of the solution.

Molality is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

[tex]\text{Molality}=\frac{0.158\text{mol}}{0.085\tetx{kg}} =1.86 m[/tex]

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Which of the following can't undergo aldol condensation.

(CH3)3CCHO
CH3CH2CHO
PhCOCH3
CH3COCH3

Answers

Answer:

(CH3)3CCHO

Explanation:

Carbonyl compounds that contain alpha hydrogen atoms are the only ones that can undergo aldol condensation reaction.

Aldol condensation is one of the condensation reactions in organic chemistry that occurs when an enol or an enolate ion reacts with another carbonyl compound to yield a β-hydroxyaldehyde or β-hydroxyketone, this is immediately followed by the dehydration of the product to yield a conjugated enone.

(CH3)3CCHO lacks alpha hydrogen atoms hence it does not undergo the aldol condensation.

A scuba diver goes deeper underwater the diver must be aware that the increased pressure affects the human body be increasing the

Answers

Answer:

The amount of dissolved gases in the body. Have a good day! =)

Explanation:

Which lists metric units, in order, from smallest to largest?

A) milligram, centigram, gram

B) kilogram, gram, centigram

C) kilogram, hectogram, decagram

D) decagram, hectogram, milligram

Answers

Answer:

milligram, centigram, decigram

Hope this answer correct :)

Answer:

C

Explanation:

hope it helps.

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