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3 years ago

Determine the number of ions produced in the dissociation of the compound listed. AlF3

1 answer:

6 0

Answer is: the number of ions produced in the dissociation of aluminium fluoride is 4.

Chemical dissociation of aluminium fluoride in water:
AlF₃(aq) → Al³⁺(aq) + 3F⁻(aq).
There are four ions, one aluminium cation and three fluoride anions.
Aluminium has oxidation +3, because it lost three electrons, to have electron configuration as noble gas neon and fluorine has oxidation -1, because it gain one electron to have electron configuration as noble gas neon.

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Find the kinetic energy of a boy of mass 6.5kg running at a velocity of 6.0m/s

ololo11 [35]

Explanation:

Given,

M=6.5kg,velocity(v)=6m/s

K.E=1/2mv^2

=1/2×6.5×6^2

=1/2×6.5×36

=18×6.5

=117 J

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2 years ago

1 Defing polymerization

elena-14-01-66 [18.8K]

Answer:

Polymerization is a process through which a large number of monomer molecules react together to form a polymer. The macromolecules produced from a polymerization may have a linear or a branched structure.

Explanation:

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2 years ago

Read 2 more answers

Which compound contains both ionic and covalent bonds? 1. ki 2. cacl2 3. ch2br2 4. nacn?

Savatey [412]

NaCn contains an Na+ ion and a CN- negative ion, so it has an ionic bond. The CN ion is formed by a covalent bond of the carbon and nitrogen, so it also has a covalent bond. Therefore, the answer is NaCN.

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3 years ago

Read 2 more answers

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

mestny [16]

Answer:

The answer is:

(a) $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of $AgNO_3$ ,

= 30.0 ml

or,

= 0.030 L

Molarity of $AgNO_3$ ,

= 0.050 M

Moles of NaCl will be:

= $\frac{Given \ mass}{Molar \ mass}$

= $\frac{0.0860}{58.44}$

= $0.00147 \ mol$

now,

Moles of $AgNO_3$ will be:

$= Molarity\times Volume$

$= 0.050\times 0.030$

$=0.0015 \ mol$

(a)

The reaction is:

⇒ $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b)

1 mole of NaCl react with,

= 1 mol of $AgNO_3$

0.0015 mol $AgNO_3$ needs,

= $0.00150 \ mol \ NaCl$

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= $0.00147\times \frac{1}{1}\times \frac{143.32}{1}$

= $0.211 \ g \ AgCl$

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3 years ago

Draw the organic product formed when the structure shown below undergoes a reaction with nanh2. there is no aqueous workup in th

blagie [28]

Answer:

Prop-1-yn-1-ide

Explanation:

In this case, we will have an acidic hydrogen upon the left carbon in the alkyne. In the presence of a base, we can remove this hydrogen. In this reaction the base will be $NH_2^-$ . The amidure ion will remove the hydrogen of the alkyne to produce a carboanion. (See figure)

5 0

3 years ago