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Elis [28]
3 years ago
9

Calcular la presión final de un gas que inicialmente está a 21°C y 0,98 atm. Sabiendo que su temperatura aumenta a 37°C.

Chemistry
1 answer:
KatRina [158]3 years ago
7 0

Answer:

1.03 atm

Explanation:

Primero <u>convertimos 21 °C y 37 °C a K</u>:

  • 21 °C + 273.16 = 294.16 K
  • 37 °C + 273.16 = 310.16 K

Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la<em> ley de Gay-Lussac</em>:

  • T₁P₂ = T₂P₁

En este caso:

  • T₁ = 294.16 K
  • P₂ = ?
  • T₂ = 310.16 K
  • P₁ = 0.98 atm

Colocando los datos:

  • 294.16 K * P₂ = 310.16 K * 0.98 atm

Y <u>despejando P₂</u>:

  • P₂ = 1.03 atm
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A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k . what is the ratio of f at the h
lapo4ka [179]
Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..

Then you can find the ration between K's at two different temperatures as:

K1 = A * e ^ (-Ea / RT1)

K2 = A* e ^(-Ea / RT2)

=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }

=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }

=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }

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3 years ago
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An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g
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Answer:

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B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{1000 ml}{1 L}

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

4 0
3 years ago
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