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3 years ago

1.Which of the following are considered pure substances

1 answer:

5 0

1.) homogeneous mixtures

2.) NaCI, because sodium is a metal and chlorine is a nonmetal

3.) I'm thinking the first the option

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What type of energy is carried by electrical charges as they move in a circuit?

Usimov [2.4K]

B. Electrical energy

3 0

3 years ago

When the fish-tank water has a pH of 8.0, the hydronium ion concentration is 1.0 × 10-8 mole per liter. What is the hydronium io

Oduvanchick [21]

You must know and use the formula for pH.

pH = - log [H3O+], where [H3O+] is the molar concentration of hydronium ion.

So, when pH is 8.0 => 8.0 = - log [H3O+] and you can use antilogarithm (the inverse function of logarithm) to find [H3O+], in this way:

[H3O+] = 10^-8 = 1 * 10 ^-8 M

When, pH = 7.0 =>

7.0 = - log [H3O+] => [H3O+] = 1 * 10^ -7 M

Answer: 1*10^-7 mole / liter

5 0

3 years ago

(Exercise 5.2.14 modified) A common method used to produce bleach (sodium hypochlorite) is by the reaction Cl2 + 2NaOH  NaCl +

Oksana_A [137]

Answer:

Chlorine is limiting reactant

Explanation:

Based on the reaction:

Cl₂ + 2NaOH → NaClO + NaCl + H₂O

1 mole of chlorine reacts with 2 moles of NaOH

To find limiting reactant, we need to determine the moles of the reactants:

Moles Cl₂ -Molar mass: 70.9g/mol-:

800lb Cl₂ * (453.6g / 1lb) * (1mol / 70.90g) =

5118 moles Cl₂

Moles NaOH -Molar mass: 40g/mol-:

1200lb NaOH * (453.6g / 1lb) * (1mol / 40g) =

13608 moles NaOH

For a complete reaction of 13608 moles of NaOH you need:

13608 moles NaOH * (1mol Cl₂ / 2 moles NaOH) = 6804 moles of Cl₂

As the solution contains just 5118 moles of chlorine,

<h3>Chlorine is limiting reactant</h3>

8 0

3 years ago

So its not A so its B C or D plz help

Rainbow [258]

Answer:

the answer is B

hope this helped!

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