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Mumz [18]
3 years ago
8

Identify the molecules or ions below as Lewis acids, Lewis bases, or neither. If there is more than one possible site in the mol

ecule/ion, focus on the central or the charged atom. H3C-CH2-CH3 __________
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
8 0

Answer: Neither

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

CH_3CH_2CH_3 is a hydrocarbon containing carbon and hydrogen, which have almost same electronegativity values. Thus is a non polar molecule.

It does not donate electrons as it does not contain any lone pair of electrons. It can not accept electrons as it's octet is already complete. Thus is neither a lewis acid nor a lewis base.

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2 years ago
Which are examples of habitat destruction? Check all that apply
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Why is argon and chlorine both gases at room temperature?
vredina [299]
<span>Argon and chlorine are both gases at room temperature because they are non-metals.</span>
7 0
2 years ago
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
2 years ago
1 gallon is how many mm
leva [86]

Answer:

0.133681 (cubic foot)

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