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Hitman42 [59]
3 years ago
15

On hot days, you may have noticed that potato chip bags seem to "inflate", even though they have not been opened. If I have a 25

0 mL bag at a temperature of 19C, and I leave it in my car which has a temperature of 60C, what will the new volume of the bag be? (round your answer to the nearest tenth)
Chemistry
1 answer:
Slav-nsk [51]3 years ago
7 0

The new volume of the bag will be 789.5 mL.

<u>Explanation:</u>

As per the Charles law, at constant pressure the volume of the gas is directly related to its temperature in Kelvin (K). That is as the temperature increases, the gas expands and vice-versa.

$\frac{V1}{T1} = \frac{V2}{T2}\\

V1 = 250 ml

V2 = ?

T1 = 19° C

T2 = 60° C

Now we have to rewrite the equation to get the new volume as,

V2 = $\frac{V1T2}{T1}

    =$\frac{250\times60}{19}

<em>     </em>= 789.47 ≈ 789.5 ml.

So the new volume of the bag will be 789.5 mL.

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How do you think the crocodile might affected if it wandered near a strong magnet?
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a gas occupies 600.0 ml at 20.00 degrees C and 70.00kPa what will be the pressure at 40.00 degrees C and 150.0mL?
Zina [86]

Answer:

P₂ = 299.11 KPa

Explanation:

Given data:

Initial volume = 600 mL

Initial pressure = 70.00 KPa

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 40°C (40+273 = 313 K)

Final volume = 150.0 mL

Final pressure = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ =  70 KPa × 600 mL × 313 K / 293K ×150 mL  

P₂ = 13146000 KPa .mL. K /43950 K.mL

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4 0
3 years ago
When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms. (a) Enter the balan
Damm [24]

Answer:

a) Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) 67.6%

Explanation:

a) Nitrate is the ion NO₃⁻, and lead(II) forms the ion Pb⁺², so the compound lead(III) nitrate is Pb(NO₃)₂. (First, the cation, then the anion, with charges replaced without the signal).

Chromate is the ion Cr₂O₄⁻² and sodium forms the ion Na⁺, so the sodium chromate is Na₂Cr₂O₄. Both of them are in solutions, so they will be in an aqueous state.

In the reaction, the anions and cations will replace and will form: NaNO₃ and Pb(Cr₂O₄). The nitrates formed by metals from group 1, such as sodium, are soluble, so it will not forme a precipitated. So, the precipitated is PbCr₂O₄, and the balanced reaction is:

Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)

b) The molar masses are: Pb(NO₃)₃ = 331,2 g/mol; Na₂Cr₂O₄ = 162 g/mol; Pb(Cr₂O₄) = 323,2 g/mol.

First, let's find what is the limiting reactant, doing the stoichiometry calculus for the reactants. Let's suppose that Na₂Cr₂O₄ is the limiting so:

1 mol of Pb(NO₃)₂ ------------------------------ 1 mol of Na₂Cr₂O₄

Transforming to mass (mass = moles * molar mass):

331,2 g of Pb(NO₃)₂ ------------------------- 162 g/mol of Na₂Cr₂O₄

x ------------------------- 12.38

By a simple direct three rule:

162x = 4100.256

x = 25.3 g of Pb(NO₃)₂

This is higher than what is put in the reaction, so Pb(NO₃)₂ is the limiting reactant, and Na₂Cr₂O₄ is in excess. So, let's do the stoichiometric calculus for the limiting reactant and the solid formed:

1 mol of Pb(NO₃)₂ ----------------------- 1 mol of Pb(Cr₂O₄)

Transforming to mass:

331.2 g of Pb(NO₃)₂ ------------------- 323.2 g of Pb(Cr₂O₄)

11.39 g ------------------- y

By a simple direct three rule:

331.2 y = 3681.248

y = 11.115 g

The yield is the mass formed divided by the stoichiometric result multiplied by 100%:

yield = (7.52/11.115)*100% = 67.6%

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