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Naddika [18.5K]
3 years ago
10

Why do we call them "Neutralization reaction"?This is section 2.6 Chemical Reactions

Chemistry
1 answer:
hram777 [196]3 years ago
7 0
Only some reactions are called neutralization reactions. The most common are acid-base reactions, and the reason is that the consequence of the reaction is the acidity of the acid is reduced, and the basicity of the base is reduced.

A canonical example would be HCl + NaOH = NaCl + H2O. The HCl solution is a very strong acid, e.g. it will react strongly with many things, including your skin if you get it on you. Same for the NaOH solution. However the result, for an equimolar reaction, would just be a solution of table salt (NaCl) in water, which has almost zero reactivity with anything. You can drink it. So we say in this reaction the acid and base have both been "neutralized" -- their potent powers eliminated... hope this helps!
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Oxygen is something that activated Cellar Reapiration that convets nutrients into energy (ATP).
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What were the limition of doberiner classifacation​
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Explanation:

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What is “monomer” mean?
oksano4ka [1.4K]

it is a molecule* that can be joined with other molecules that are identical to form a polymer*

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5 0
3 years ago
If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the
Alina [70]

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

3 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
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