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3 years ago

7. One way to isolate metals from their ores is to react the metal oxide with carbon as shown in the

Chemistry

1 answer:

anyanavicka [17]3 years ago

7 0

Metal oxides react with carbon and are reduced to the metal. The metal in the reaction above is copper.

The equation of the reaction is;

2MO(s) + C(s) -> 2M(s) + CO2(g)

We are told in the question that carbon is in excess so the metal oxide is the limiting reactant.

Number of moles of metal oxide = 34.49 g/M + 16

From PV = nRT

P = 1.50 atm

V = 4.42 L

T = 100°C + 273 = 373 K

n =?

R = 0.082 atmLK-1mol-1

n = PV/RT

n = 1.50 atm × 4.42 L/0.082 atmLK-1mol-1 × 373 K

n = 6.63/30.58

=0.217 moles

If 2 moles of metal oxide yields 1 mole of CO2

34.49 g/M + 16 yields 0.217 moles of CO2

34.49 g/M + 16 = 2 × 0.217

34.49 /M + 16 = 0.434

34.49 = 0.434(M + 16)

34.49 = 0.434M + 6.944

M = 34.49 - 6.944/0.434

M = 63.5

The metal is copper.

Learn more: brainly.com/question/22824409

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Please explain and rank the order of oxidation of methanol, ethanol, 2-propanol, 2-methyl-2propanol reacting with potassium perm

natali 33 [55]

Answer:Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

Explanation:The mechanism of oxidation using Potassium permanganate

involves two steps :

In the first step permanganate ion abstracts a alpha-hydrogen as hydride ion available at alcohol.

Alpha hydrogen is the hydrogen attached to the carbon bearing functional group.

In the second step the permanganate ion subsequently is reduced from +7 oxidation state of Manganese to +5 oxidation state of Manganese.

So for the oxidation of alcohol using potassium permanganate , there must be availability of alpha hydrogens.

we can also relate the order of oxidation with reference to number of alpha hydrogens present on the substrate.

Methanol has 3 available alpha hydrogen

Ethanol has 2 available alpha hydrogen

2-propanol has 1 available alpha hydrogen as it is a secondary alcohol

2-methyl-2propanol has 0 available alpha hydrogens as it is a tertiary alcohol.

Greater the number of available hydrogens easier would be oxidation of alcohols so the order of oxidation would be the following:

Methanol>Ethanol>2-Propanol>2-methyl-2Propanol

please refer the attachment for the structures of following compounds.

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4 years ago

The attraction between a positive metal ion and the electrons surrounding

asambeis [7]

Answer:

metallic bond

Explanation:

Metallic bonds are the attraction between a positive metal ion and the electrons surrounding them.

This bond is found in metals and their alloys. They are interatomic forces.

In this bond, a positive nuclei joins with all other closely packed atoms in the lattice and the electron cloud formed by losing their outermost shell electrons.
Most of the physical properties of metals is due to this bond type.

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3 years ago

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

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The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

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3 years ago

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valkas [14]

Explanation:

KMnO4 + H2S + HCl --> KCl + MnCl2 + H2O + S8

Did Cl change oxidation number?

The oxidation number of Cl in Reactant side is -1

The oxidation number of Cl in product side is -1

Did Oxygen change oxidation number?

The oxidation number of O in Reactant side is -2

The oxidation number of O in product side is -2

Did H change oxidation number?

The oxidation number of H in Reactant side is 1

The oxidation number of H in product side is 1

Did Mn change oxidation number?

The oxidation number of Mn in Reactant side is +7

The oxidation number of Mn in product side is +2

YES

Did S change oxidation number?

The oxidation number of S in Reactant side is -2

The oxidation number of S in product side is 0

YES

Oxidizing agent (element with decrease in oxidation number) = Mn

Reducing agent (element with Increase in oxidation number) = S

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3 years ago

Magnesium oxide does not readily decompose into magnesium and oxygen. The reaction is shown below. MgO(s) + 601.7 kJ mc017-1.jpg

galben [10]

TEMPERATURE IS THE ANSWER

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3 years ago

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