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Zarrin [17]
3 years ago
5

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

in each of the following equations:
(a)Mg + NiCl2 ->MgCl2 + Ni
(b)PCl3 + Cl2 -> PCl5
(c)C2H4 + 3O2 -> 2CO2 + 2H2O
(d)Zn + H2SO4 -> H2 + ZnSO4
(e)2K2S2O3 + I2- > K2S4O6+2KI
(f)3Cu + 8HNO3 = 3Cu(NO3)2 + 4H2O + 2NO
Chemistry
1 answer:
KATRIN_1 [288]3 years ago
8 0

Answer:

Explanation:

a) Mg + NiCl_2 \rightarrow MgCl_2 + Ni

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and NiCl_2 acts as oxidizing agent.

b) PCl_3 + Cl_2 \rightarrow PCl_5

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

PCl_3 acts as reducing agent and Cl_2 acts as oxidizing agent.

c) C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

C_2H_4 acts as reducing agent and O_2 acts as oxidizing agent.

d) Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and H_2SO_4 acts as oxidizing agent

e) K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

K_2S_2O_3 acts as reducing agent and I_2 acts as oxidizing agent.

f) 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and HNO_3 acts as oxidizing agent

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barxatty [35]
A cartilage doesn’t produce blood cells
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3 years ago
Ethanoic acid, CH3COOH, ionizes to form an ethanoate ion (CH3COO-) in aqueous solution. what else does it form? is it a monoprot
barxatty [35]

Ethanoic acid ionizes in aqueous solutions to form two ions which are CH_3COO^- and H^+

<h3>Ionization of ethanoic acid</h3>

Ethanoic acid goes by the chemical formula CH_3COOH.

In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

CH_3COOH ---- > CH_3COO^- + H^+

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single H^+

More on ethanoic acid can be found here: brainly.com/question/9991017

#SPJ1

8 0
2 years ago
The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
MAVERICK [17]

Answer:

Approximately 2.46\; \rm mol.

Explanation:

Make use of the molar mass data (M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}) to calculate the number of moles of molecules in that 64.0\; \rm g of \rm C_2H_2:

\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}.

Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

Coefficient of \rm H_2O in this equation: 2.

In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

Hence the ratio: \displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1.

Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

3 0
3 years ago
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
The spectrochemical series is I &lt; Br&lt; &lt; Cl^- &lt; F
vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0
3 years ago
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