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3 years ago

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

1 answer:

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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g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol

Ronch [10]

Answer:

$M=0.0637M$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)$

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

$n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}$

Finally, the resulting molarity in 30.8 mL (0.0308 L):

$M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M$

Regards.

3 0

3 years ago

In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond?

masha68 [24]

Answer:

A . 6.3 In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond

7 0

3 years ago

Light bending as it passes through a rain drop is an example of Transmission Diffraction Refraction Reflection

Yuliya22 [10]

The answer would be the third one listed, Refraction

4 0

3 years ago

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

5 0

3 years ago

1. The diagram above shows the repeating groups of atoms that make up two samples. Will the

serious [3.7K]

Answer:

Will likely be the same

Explanation:

We can see in both pictures there is a black molecule and a red molecule. However, we also have a purple molecule in one image and a yellow in the other. It would LIKELY be the same because we have more of the same molecules then more different molecules. Hope this helps

4 0

3 years ago