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3 years ago

How does a chemical equation model a chemical reaction

Chemistry

1 answer:

ale4655 [162]3 years ago

4 0

Answer:

reactant + reactant = product + product

Explanation:

(pretend = is the arrow)

the reaction is shown in the equation by modelling how the reactants react with eachother to form the products on the other side of the arrow

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An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form

ozzi

In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

$C:H:I=n_{C}:n_{H}:n_{I}$

Putting the values,

$C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1$

Thus, molecular formula of compound will be $C_{6}H_{11}I$ .

4 0

3 years ago

Write the acid-base reaction that occurs when an aqueous solution of HCl is added to an aqueous solution of NaOH. (Use the lowes

Sloan [31]

H3O+(aq) + OH-(aq) --> 2H2O (l)

NaHCO3(s) --> NaH 2+ (aq) + CO3 2- (aq)

NaH 2+ (aq) + H2O (l) --> Na+ (aq) + H3O+ (aq)

H2O (l) + CO3 2- (aq) --> OH- (aq) + HCO3- (aq)

(I'm not completely sure if I did the third question right) I'm sorry if I got it wrong

4 0

3 years ago

Read 2 more answers

You have 350 mL of 3.4 M hydrochloric acid (HCl). How many grams of HCl gas are dissolved? Bonus: what is the volume of the HCl

Crank

Answer:

1. 43.44g of HCl

2. 26.67 L of HCl

Explanation:

1) Molarity of a solution = number of moles (n) ÷ Volume (V)

According to the provided information in this question,

V = 350 mL = 350/1000 = 0.350L

Molarity = 3.4 M

Using Molarity = n/V

3.4 = n/0.350

n = 3.4 × 0.350

n = 1.19mol

Using the formula below to calculate the mass of HCl;

mole = mass/molar mass

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

mole = mass/MM

mass = 1.19 mol × 36.5g/mol

mass = 43.44g of HCl

2) At STP, HCl has a pressure of 1atm, a temperature of 273K

V = ?

n = 1.19 mol

R = 0.0821 Latm/molK

Using PV = nRT

V = nRT/P

V = 1.19 × 0.0821 × 273/1

Volume = 26.67L

5 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

Answer: The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

Explanation:

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

Case 1:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 2:

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

Case 3:

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

The kidneys play a major role in the urinary system. Which of these is NOT a function of the kidneys?

Jet001 [13]

Answer:

Explanation:

3 0

3 years ago