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3 years ago

How many atoms of each element are in one molecule of this product.

Chemistry

1 answer:

anastassius [24]3 years ago

7 0

I assume about 5 atoms are in one molecule of this product

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The following chemical reaction occurs in a basic solution.

Dafna1 [17]

Answer:

6 moles of electrons

Explanation:

Let us consider the species NO3− and ClO−. The NO3− is oxidized to NO the oxidation number of nitrogen is decreased from +5 to +2.

The oxidation number of chlorine is increased from +1 to +3. This implies that six electrons were transferred in the balanced reaction equation shown in the question. Hence the answer.

3 0

3 years ago

How many protons are present in one copper(ii) ion?

AlladinOne [14]

Copper has 29 protons, when dealing with Cu^2+ all that means is it lost two electrons. so now the element has 29 protons and 27 electrons. Protons are positive and electrons are negative and neutrons are neutral. So say you had an element X^2- then you have gained two more electrons so the element has an overall negative charge. hope that helps

6 0

3 years ago

The electron configuration for aluminum (Al) is shown below:

Serjik [45]

Answer:

It is the third choice

Explanation:

Just did assignment

3 0

3 years ago

Which chloride is a coloured solid rtp<br>

lianna [129]

Answer:

sodium chloride

hope that helped :)

7 0

2 years ago

Read 2 more answers

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