CHEMISTRY COLLEGE

What is the mass, in grams, of 1.20×1021 molecules of aspirin, c9h8o4?

Answers

Answer 1

Answer:

Answer: The mass of given number of molecules of aspirin is 0.359 grams.

Explanation:

We are given:

Number of molecules of aspirin = $1.20* 10^(21)$

We know that:

Molar mass of aspirin $(C_9H_8O_4)$ = 180.16 g/mol

According to mole concept:

$6.022* 10^(23)$ number of molecules are contained in 1 mole of a compound

Also, $6.022* 10^(23)$ number of molecules of aspirin has a mass of 180.16 grams

So, $1.20* 10^(21)$ number of molecules will have a mass of $(180.16)/(6.022* 10^(23))* 1.20* 10^(21)=0.359g$

Hence, the mass of given number of molecules of aspirin is 0.359 grams.

Answer 2

Answer:

The mass in grams of 1.20 x10^21 molecules of asprin is 0.359 grams

calculation

find the number of moles of aspirin by use of Avogadro's law that is 1 mole =6.02 x10^23 molecules

what of 1 .20 x10^21 molecules

= (1 mole x 1.20 x10 ^21 molecules)/6.02 x10^23 molecules)= 1.993 x10^-3 moles

mass of aspirin= moles x molar mass

molar mass of aspirin = (12 x9)+(1 x8) +(16x4)=180 g/mol

mass= 1.993 x10^-3 moles x180 g/mol = 0.359 grams

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1.81 g H2 is allowed to react with 10.2 g N2, producing 2.19 g NH3.What is the theoretical yield in grams for this reaction under the given conditions?3H2(g)+N2(g)→2NH3(g)

Answers

The theoretical yield : = 10.251 g

Further explanation

Given

Reaction

3H₂(g)+N₂(g)→2NH₃(g)

1.81 g H₂

10.2 g N₂

2.19 g NH₃

Required

The theoretical yield

Solution

Find limiting reactant :

H₂ : 1.81 g : 2 g/mol = 0.905 mol

N₂ : 10.2 g : 28 g/mol = 0.364 mol

mol : coefficient

H₂ = 0.905 : 3 = 0.302

N₂ = 0.364 : 1 = 0.364

H₂ as a limiting reactant(smaller ratio)

Moles NH₃ based on H₂, so mol NH₃ :

= 2/3 x mol H₂

= 2/3 x 0.905

=0.603

Mass NH₃ :

= mol x MW

=0.603 x 17 g/mol

= 10.251 g

Draw the mechanism of the reaction of N,N-dimethylaniline and the diazonium salt of sulfanilic acid to form the azo dye.

Answers

Answer:

See attachment

Explanation:

The amine group of N,N-dimethylaniline is an electron-donating group and will cause the benzene ring to act as a nucleophile in an electrophilic aromatic substitution reaction with the diazonium salt of sulfanilic acid.

Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the smallest value of K at 298 K? A + B → 2 C; E°cell = -0.030 V A + 2 B → C; E°cell = +0.98 V A + B → C; E°cell = +1.22 V A + B → 3 C; E°cell = +0.15 V More information is needed to determine.

Answers

The reactions that would have the smallest value of K is

A + B → 2 C; E°cell = -0.030 V

Option A

Generally the equation for the number of electrons transferred is mathematically given as

$nFE^o_(cell)=RT\ln K$

where

T= Temperature

F=25C(298K)

R = Gas constant

R= 8.314 J/K.mol

F = Faraday's constant

F= 96500 C

We see from the equation that the E-cell is directly proportional to K(equilibrium constant of the reaction)

Hence, The reactions that would have the smallest value of K is

A + B → 2 C; E°cell = -0.030 V

For more information on Reaction

brainly.com/question/11231920

Answer:

The reaction with smallest value of K is :

A + B → 2 C; E°cell = -0.030 V

Explanation:

$nFE^o_(cell)=RT\ln K$

where :

n = number of electrons transferred

F = Faraday's constant = 96500 C

$E^o_(cell)$ = standard electrode potential of the cell

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K$ = equilibrium constant of the reaction

As we cans see, that standard electrode potential of the cell is directly linked to the equilibrium constant of the reaction.

Higher $E^o_(cell)$ higher will be the value of K.
Lower $E^o_(cell)$ lower will be the value of K.

So, the reaction with smallest value of electrode potential will have smallest value of equilibrium constant. And that reaction is:

A + B → 2 C; $E^o_(cell) =-0.030 V$

The periodic table displaysOA. all of the known elements that exist in the world today.
OB. only the important elements that exist in the world.
OC. only the important compounds that exist in the world.

Answers

OA. all the known elements that exist in the world today.

A chemist designs a galvanic cell that uses these two half-reactions:O2 (g) + 4H+(aq) + 4e− → 2H2O (l) Eo =+1.23V
Zn+2 (aq) + 2e− → Zn(s) Eo=−0.763V

Answer the following questions about this cell.

Write a balanced equation for the half-reaction that happens at the cathode.
Write a balanced equation for the half-reaction that happens at the anode.
Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions

Answers

Answer: The reaction is spontaneous and there is not enough information to calculate the cell voltage.

Explanation:

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

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

For a:

The half reactions for the cell occurring at cathode follows:

$O_2(g)+4H^+(aq)+4e^-\rightarrow H_2O(l);E^o_(cathode)=+1.23V$

For b:

The half reactions for the cell occurring at anode follows:

$Zn(s)\rightarrow Zn^(2+)+2e^-;E^o_(anode)=-0.763V$ ( × 2)

For c:

The balanced equation for the overall reaction of the cell follows:

$O_2(g)+4H^+(aq)+2Zn(s)\rightarrow H_2O(l)+2Zn^(2+)(aq)$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_(cell)$

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_(cell)$ of the reaction, we use the equation:

$E^o_(cell)=E^o_(cathode)-E^o_(anode)$

Putting values in above equation, we get:

$E^o_(cell)=1.23-(-0.763)=1.993V$

As, the standard electrode potential of the cell is coming out to be positive, the reaction is spontaneous in nature.

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Zn^(2+)]^2)/([H^(+)]^4* p_(O_2))$

As, the concentrations and partial pressures are not given. So, there is not enough information to calculate the cell voltage.

Hence, the reaction is spontaneous and there is not enough information to calculate the cell voltage.

You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution of NaClNaCl, and there isn’t any NaClNaCl to be found in the lab. You do, however, have some Na2SO4Na2SO4. Can you substitute the same number of grams of Na2SO4Na2SO4 for the NaClNaCl in a solution? Why or why not?

Answers

Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.

Final answer:

You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.

Explanation:

No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.

For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.

Learn more about Chemical Substitution here:

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