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2 years ago

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a chemical reaction happens in three steps. step 1:step 2:step 3:a→b cc→d ed a→f(fast)(slow)(fast)step 1:a→b c(fast)step 2:c→d e

(slow)step 3:d a→f(fast) what is the rate-determining step

1 answer:

Nezavi [6.7K]2 years ago

8 0

Answer:

Step 2 only

Explanation:

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3 years ago

Read 2 more answers

I need some help with a long chemistry problem. Anything is appreciated!

Mkey [24]

Answer:

The CSI is wrong.

Explanation:

1. Find the volume of the pool

The formula for the volume of a cylinder is V = πr²h .

D = 12 m; h = 10 m

r = D/2 = (12 m/2) = 6.0 m

V = πr²h = π × (6.0 m)² × 10 m = π × 36 m²× 10 m = 360π m³ = 1100 m³

= 1.1. × 10⁶ L

2. Calculate the moles of OH⁻

n = cV = 1.0 × 10⁻² mol·L⁻¹ × 1.3 × 10⁶ L = 11 000 mol of OH⁻

3. Calculate the moles of acetic acid needed for neutralization

HA + OH⁻ ⟶ A⁻ + H₂O

The molar ratio of is 1 mol HA:1 mol OH⁻, so you need 11 000 moL of acetic acid.

4. Calculate the actual moles of acetic acid

You have four 5 L jugs of acetic acid pH 2 .

Volume = 20 L

[H⁺] = 10⁻² mol·L⁻¹ = 0.01 mol·L⁻¹

(a) Set up an ICE table

HA + H₂O ⇌ A⁻ + H₃O⁺

I/mol·L⁻¹: c 0 0

I/mol·L⁻¹: - 0.01 +0.01 +0.01

I/mol·L⁻¹: c - 0.01 0.01 0.01

$K_{\text{a}} = \dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}} = 1.76 \times 10^{-5}$

(b) Calculate the concentration of acetic acid

$\begin{array}{rcl}\dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}}& = & 1.76 \times 10^{-5}\\\\\dfrac{0.01\times 0.01}{c}& = & 1.76 \times 10^{-5}\\\\1 \times 10^{-4} & = & 1.76 \times 10^{-5}c\\c & = & \dfrac{1 \times 10^{-4}}{1.76 \times 10^{-5}}\\\\ & = & \text{6 mol/L}\\\end{array}$

The concentration of the acetic acid is 6 mol·L⁻¹

(c) Calculate the moles of acetic acid

$n = \text{20 L} \times \dfrac{\text{6 mol}}{\text{1 L}} = \textbf{100 mol}$

You have 100 mol of acetic acid.

The CSI is wrong.

You don't have enough acetic acid to neutralize the pool.

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4 years ago