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3 years ago

__Fe2O3 + _H2O --> ___Fe(OH)3

Chemistry

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

Fe2O3 + __3_H2O --> __2_Fe(OH)3

Explanation:

____Fe2O3 + __3_H2O --> __2_Fe(OH)3

Fe -2 Fe -2

H - 6 H - 6

O - 6 O- 6

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Use the periodic table to find the molar mass of each of the following elements. Sodium (Na): g/mol Oxygen (O): g/mol Carbon (C)

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The Molar mass of an atom corresponds directly with its atomic mass on the periodic table.

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2 years ago

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1.) is H2SO4 a sulfuric acid or a sulfurous acid?

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3 years ago

3. How many atoms in 3.91 moles of sodium?

Gala2k [10]

Answer:

Explanation:

The formula for sodium is Na. It does not form a molecule in some way.

1 mol Na = 6.02*10^23 atoms

3.91 mol = x Cross multiply

x = 3.91 * 6.02 * 10^23

x = 23.65 * 10^23

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Scientific notation is always expressed as a number 1 ≤ x < 10

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3 years ago

In the formation of a solution how does the solvent differ from the solute

dlinn [17]

Solute is a substance that dissolves in a solvent in order to form a solution. Solutes can be in liquid, gaseous or solid phase. Normally, in a solution, solutes are in a lesser amount than the solvents. When a solution has the maximum amount of solutes it can dissolve, then the solution is said to be saturated.

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3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

2 years ago

____Fe2O3 + ___H2O --> ___Fe(OH)3

__Fe2O3 + _H2O --> ___Fe(OH)3