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tangare [24]
2 years ago
6

State which substance is undergoing oxidation and which substance is undergoing reduction in the following reaction:

Chemistry
1 answer:
yKpoI14uk [10]2 years ago
8 0

Answer:

Mg(s) = oxidized

H₂SO₄(aq) (hydrogen) = reduced

Explanation:

Oxidized substances lose electrons and see an increase in oxidation number. Reduced substances gain electrons and see a decrease in oxidation number.

Lone elements always have an oxidation number of 0. H₂SO₄ and MgSO₄ are ionic compounds that can be broken up into a cation and polyatomic anion. The anion, SO₄²⁻, has an overall oxidation number of -2. Therefore, to make the overall charge of the ionic compound neutral, the hydrogen in H₂SO₄ has an oxidation state of +1 (bc there are two hydrogens) and the magnesium in MgSO₄ has an oxidation state of +2.

Since Mg(s) is losing electrons (goes from 0 to +2), it is being oxidized. Since the hydrogen in H₂SO₄(aq) is gaining electrons (goes from +1 to 0), it is being reduced.

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Calculate [h3o+] and [s2−] in a 0.10 m solution of the diprotic acid hydrosulfuric acid. (for hydrosulfuric acid ka1 = 9.0 × 10−
Rudiy27
The equation for the first dissociation is:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸ 
Construct ICE table and obtain their equilibrium concentrations:
                  H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M):             0.1                            0              0
C (M):            -x                            +x            +x
E (M):        0.1 -x                            x              x
So:
9.0 x 10⁻⁸ = \frac{X^{2} }{0.1-x}
x = 9.4 x 10⁻⁵ 
From the equilibrium table:
[H₃O⁺] = x = 9.4 X 10⁻⁵ M
[HS⁻] = x = 9.4 X 10⁻⁵ M
The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
                    HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M):          9.4 x 10⁻⁵                   0              9.4 x 10⁻⁵
C (M):            -x                            +x            +x
E (M):     9.4 x 10⁻⁵ -x                    x              9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ = \frac{(x)(9.4 x 10^{-5} + x) }{(9.4 x 10^{-5} - x) }
x = 1.0 x 10⁻¹⁷ M
Therefore the equilibrium concentrations are as follows:
[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M
6 0
3 years ago
How can you predict if a bond will be ionic or covalent by using the periodic table​
babunello [35]

Answer: There is a couple different ways to determine if a bond is ionic or covalent. By definition, an ionic bond is between a metal and a nonmetal, and a covalent bond is between 2 nonmetals. So you usually just look at the periodic table and determine whether your compound is made of a metal/nonmetal or is just 2 nonmetals.

3 0
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Which stage of cell division rusults in the formation of four new haploid cells
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5 0
3 years ago
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stiv31 [10]

Explanation:

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8 0
3 years ago
The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?
juin [17]

The formula for mole fraction is:

mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}    -(1)

The solubility of oxygen gas = 1.0 mmol/L  (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

Density = \frac{mass}{volume}

Mass = 1.0 g/ml\times 1000 mL = 1000 g

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = \frac{1000 g}{18 g/mol} = 55.55 mol

Substituting the values in formula (1):

mole fraction = \frac{0.001}{55.55+0.001}

mole fraction = 1.8\times 10^{-5}

Hence, the mole fraction is 1.8\times 10^{-5}.

7 0
3 years ago
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