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Semmy [17]
2 years ago
7

the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin

e gas are needed to react with 30 aluminum atoms?
Chemistry
1 answer:
Vanyuwa [196]2 years ago
3 0

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = \frac{30 * 3}{2}\\ = 45 molecules of Cl₂.

Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

Learn more: brainly.com/question/24918379

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Answer:

is used to break reactant bonds and/or make new product bonds.

Explanation:

<em>The correct answer would be that </em><em>activation energy is used to break reactant bonds and/or make new product bonds</em><em>.</em>

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3 years ago
I NEED HELP PLEASE, THANKS!
Crazy boy [7]

Answer:

Here's what I get.

Explanation:

1. Brønsted-Lowry theory

An acid is a substance that can donate a proton to another substance.

A  base is a substance that can accept a proton from another substance.

2. pH of ammonia

The chemical equation is

\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \text{OH}$^{-}$

For simplicity, let's re-write this as

\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$

(a) Set up an ICE table.

                     B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.335             0        0

C/mol·L⁻¹:       -x                +x       +x

E/mol·L⁻¹:  0.335 + x          x         x

\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335

(b) Solve for [OH⁻]

\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}

(c) Calculate the pOH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39

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Answer:

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