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cricket20 [7]
3 years ago
13

How many amperes would be needed to produce 60.0 grams of magnesium during the electrolysis of molten mgcl 2 in 2.00 hours?

Chemistry
1 answer:
Shalnov [3]3 years ago
8 0

Answer:

\boxed{\text{66.2 A}}

Explanation:

1. Write the equation for the reaction.

M_r:                 24.30

          MgCl₂ ⟶ Mg + Cl₂

m/g:                   60.0

2. Calculate the moles of Mg

Moles of Mg = 60.0 g Mg × (1 mol Mg/ 24.30 g Mg) = 2.469 mol Mg

3. Calculate the moles of electrons

Moles of electrons = 2.469 mol Mg × (2 mol electrons/1 mol Mg)

= 4.938 mol electrons

4. Calculate the number of coulombs

Q = 4.938 mol electrons × (96 485 C/1 mol electrons) = 476 500 C

5. Calculate the current required

Q  = It

I = Q/t

t = 2.00 h × (60 min/1h) × (60 s/1 min) = 7200 s

I = 476 500 C/7600 s= 66.2 C/s = 66.2 A

You need a current of \boxed{\textbf{66.2 A}}.

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Explanation:

Calcium chloride is <u>deliquescent.</u> If exposed to air, it will absorb sufficient water from the air to allow it to dissolve. After a short while, instead of a white lump, you will have a pool of clear liquid.

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3 years ago
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Balance the chemical equations.<br> 1FeCl3 + KOH → Fe(OH)3 + KC1
wariber [46]

Answer:

FeCl3 + 3KOH → Fe(OH)3 + 3KCl

Explanation:

3 0
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
3 years ago
Help in chem please!!!!!!
IgorLugansk [536]

This is a one-step unit analysis problem.  Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

3 0
3 years ago
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