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3 years ago

an aqueous solution contains 1.2mM of total ions. if the solution is NaCl (aq), what is the concentration of chloride ions?

Chemistry

1 answer:

Genrish500 [490]3 years ago

5 0

Explanation:

As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.

As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.

$NaCl \rightarrow Na^{+} + Cl^{-}$

So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.

Therefore, concentration of chlorine ions will be calculated as follows.

Concentration of $Cl^{-}$ ions = $\frac{1.2mM}{2}$

= 0.6 mM

Thus, we can conclude that the concentration of chloride ions is 0.6 mM.

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KFell Fe"(CN), + e + Nat → KNaFe'Fe(CN)6

Alinara [238K]

Answer:

Most common oxidation states: +2, +3

M.P. 1535º

B.P. 2750º

Density 7.87 g/cm3

Characteristics: Iron is a gray, moderately active metal.

Characteristic reactions of Fe²⁺ and Fe³⁺

The [Fe(H2O)6]3+ ion is colorless (or pale pink), but many solutions containing this ion are yellow or amber-colored because of hydrolysis. Iron in both oxidation states forms many complex ions.

Aqueous Ammonia

Aqueous ammonia reacts with Fe(II) ions to produce white gelatinous Fe(OH)2, which oxidizes to form red-brown Fe(OH)3:

Fe2+(aq)+2NH3(aq)+3H2O(l)↽−−⇀Fe(OH)2(s)+2NH+4(aq)(1)

Fe3appt.gif

Aqueous ammonia reacts with Fe(III) ions to produce red-brown Fe(OH)3:

Fe3+(aq)+3NH3(aq)+3H2O(l)↽−−⇀Fe(OH)3(s)+3NH+4(aq)(2)

Fe3bppt.gif

Both precipitates are insoluble in excess aqueous ammonia. Iron(II) hydroxide quickly oxidizes to Fe(OH)3 in the presence of air or other oxidizing agents.

Sodium Hydroxide

Sodium hydroxide also produces Fe(OH)2 and Fe(OH)3 from the corresponding oxidation states of iron in aqueous solution.

Fe2+(aq)+2OH−(aq)↽−−⇀Fe(OH)2(s)(3)

Fe4appt.gif

Fe3+(aq)+3OH−(aq)↽−−⇀Fe(OH)3(s)(4)

Fe4bppt.gif

Neither hydroxide precipitate dissolves in excess sodium hydroxide.

Potassium Ferrocyanide

Potassium ferrocyanide will react with Fe3+ solution to produce a dark blue precipitate called Prussian blue:

K+(aq)+Fe3+(aq)+[Fe(CN)6]4−(aq)↽−−⇀KFe[Fe(CN)6](s)(5)

Fe5a1ppt.gif

With Fe2+ solution, a white precipitate will be formed that will be converted to blue due to the oxidation by oxygen in air:

2Fe2+(aq)+[Fe(CN)6]4−(aq)↽−−⇀Fe2[Fe(CN)6](s)(6)

Fe5a2ppt.gif

Many metal ions form ferrocyanide precipitates, so potassium ferrocyanide is not a good reagent for separating metal ions. It is used more commonly as a confirmatory test.

Potassium Ferricyanide

Potassium ferricyanide will give a brown coloration but no precipitate with Fe3+. With Fe2+, a dark blue precipitate is formed. Although this precipitate is known as Turnbull's blue, it is identical with Prussian blue (from Equation 5).

K+(aq)+Fe+2(aq)+[Fe(CN)6]3−(aq)↽−−⇀KFe[Fe(CN)6](s)(7)

Fe5b.gif

Potassium Thiocyanate

KSCN will give a deep red coloration to solutions containing Fe3+:

Fe+3(aq)+NCS−(aq)↽−−⇀[FeNCS]+2(aq)(8)

Fe5cppt.gif

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3 years ago

A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp

MArishka [77]

Answer:

V = 11.2L are produced

Explanation:

... <em>Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..</em>

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

<em>Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.</em>

<em />

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

<em>Moles Al-Molar mass: 26.98g/mol-:</em>

27g * (1mol / 26.98g) = 1mol of Al

<em>Moles HCl:</em>

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and <em>HCl is limiting reactant.</em>

<em />

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

<em>Where V is volume</em>

<em>n are moles: 0.5mol</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 273.15K</em>

<em>P is pressure: 1atm.</em>

<em />

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>

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3 years ago