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3 years ago

Which element is oxidized in the reaction below? fe(co)5 (l) + 2hi (g) fe(co)4i2 (s) + co (g) + h2 (g)?

1 answer:

5 0

Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂ (s) is x. For whole compound, the charge is zero.

Sum of oxidation numbers in all elements = Charge of the compound.

Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
x + 4*0 + 2*(-1) = 0
x + 0 - 2 = 0
x = +2
Hence the oxidation state of Fe in product Fe(CO)₄I₂ (s) is +2.

Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
y = 0

Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.

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Why must sample spots be above the solvent in paper chromatography

marishachu [46]

Because if they are submerged in the solvent, they would dissolve! This would prevent them from seperating and not allow you to actually record anything

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3 years ago

If you add enzyme to a solution containing only the product(s) of a reaction, would you expect any substrate to form?

Masteriza [31]

Answer:

Explanation:

Provided the reaction that leads to the formation of the products can proceed in both forward and backward directions, the correct answer would be yes because the reaction will proceed backward until equilibrium is reached.

For a reaction that can proceed both forward and backward, the addition of a catalyst increases the rate of reaction in both directions based on the fact that a catalyst cannot alter the equilibrium of a reaction.

Hence, if an enzyme is added to the product of a reaction that has the potential to proceed in both forward and reverse reactions, a substrate would be expected to form because the reaction will proceed backward until an equilibrium is reached.

The correct option is e.

8 0

3 years ago

The iron content of foods can be determined by dissolving them in acid (forming Fe3+), reducing the iron(III) to iron(II), and t

____ [38]

Answer:

Oxidation half-reaction:

Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻

Reduction half-reaction:

Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)

Explanation:

The reaction that takes place is:

Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)

The oxidation half-reaction is:

Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻

It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.

The reduction half-reaction is:

Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)

It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.

4 0

3 years ago

1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH

6 0

3 years ago

Which of the following energy does NOT travel as a wave?

frosja888 [35]

Answer:

Wind

Explanation:

It's impossible for them to travel in a wave.

3 0

3 years ago