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slavikrds [6]
3 years ago
14

On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i

s the new volume of the balloon? Selected Answer:
Chemistry
1 answer:
ddd [48]3 years ago
8 0

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

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Answer:

Answer:

Number of Significant Figures: 5

The Significant Figures are 3 0 6 7 0

Explanation:

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Balance this equation. Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)
const2013 [10]

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Explanation:

Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

This is how it starts out.

Left:

  • 2 NO3s
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Right

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So the place to start with this equation is to bring the Cls up to 2

Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

But the Nas are now out of kilter.

Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

Now the right has a problem. There's only 1 Na

Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

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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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