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3 years ago

What is the chemical formula for iron(II) hydroxide? Fe, OH Fe(OH) FeOH 0 FeOH

Chemistry

2 answers:

Radda [10]3 years ago

5 0

Answer: None of the given options .

Explanation: The chemical formula of Iron (II) hydroxide can be written as

Fe (OH)2

As iron is present in its +2 state and hyrdoxide ion has charge of one negative unit .

Thus iron hydroxide can be written as the above and none of the given options is correct.

denpristay [2]3 years ago

3 0

None of the above, it is Fe(OH)2

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Serjik [45]

The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.

<h3>What is Ideal law ?</h3>

According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.

PV = nRT.

Where,

p = pressure
V = volume (1.75 L = 1.75 x 10⁻³ m³)
T = absolute temperature
n = number of moles
R = gas constant, 8.314 J*(mol-K)

Therefore, the number of moles is

n = PV / RT

State 1 :

T₁ = (25⁰ C = 25+273 = 298 K)
p₁ = 225 kPa = 225 x 10³ N/m²

State 2 :

T₂ = 10 C = 283 K
p₂ = 185 kPa = 185 x 10³ N/m²

The loss in moles of gas from state 1 to state 2 is

Δn = V/R (P₁/T₁ - P₂/T₂ )

V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N

p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)

p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)

Therefore,

Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))

= 0.0213 mol

Hence, The number of moles of gas lost is 0.0213 mol.

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2 years ago

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frosja888 [35]

Answer:

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Explanation:

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6 0

3 years ago

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alexandr1967 [171]

Acids are corrosive, which means they 'eat away' at other materials.

5 0

2 years ago

g Select the correct statements. I. Reduction is the loss of electrons, and oxidation is the gain of electrons II. Reduction is

Karolina [17]

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

Reduction reaction is defined as the reaction in which a substance gains electrons. Here, the oxidation state of the substance decreases.

$X^{n+}+ne^-\rightarrow X$

Oxidizing agents are the agents that helps in the oxidation of other substance and itself gets reduced. These agents undergoes reduction reactions.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. Here, oxidation state of the substance increases.

$X\rightarrow X^{n+}+ne^-$

Reducing agents are the agents that helps in the reduction of the other substance and itself gets oxidized. These agents undergoes reduction reactions.

Oxidation state is the number which is given to an atom when it looses or gains electron. It is written as a superscript. In a compound, the total charge is equal to the sum of the charges of all atoms in that compound. <u>For Example:</u> In $MnO_4^-$ , manganese has +7 oxidation number and oxygen has -2.

So, the charge on the compound = $[=7+(4\times (-2))]=-1$

Hence, the correct answer is Option D.

6 0

3 years ago

3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O

AysviL [449]

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

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:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 mole
NO: 2 moles
H₂O: 4 moles

The molar mass of the compounds is:

Cu: 63.55 g/mole
HNO₃: 63 g/mole
Cu(NO₃)₂: 187.55 g/mole
NO: 30 g/mole
H₂O: 18 g/mole

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

Cu: 3 moles ×63.55 g/mole= 190.65 grams
HNO₃: 8 moles ×63 g/mole= 504 grams
Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
NO: 2 moles ×30 g/mole= 60 grams
H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

$mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3} x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }$

<u><em>mass of Cu(NO₃)₂= 5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

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2 years ago