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Lana71 [14]
3 years ago
11

A 10.0 g sample of propane, C3H8, was combusted in a constant-volume bomb calorimeter. The total heat capacity of the bomb calor

imeter and water was 8.0 kJ/°C. The molar heat of combustion of propane is -2 222 KJ/mol. If the starting temperature of the water was 20 °C, what will be the final temperature of the bomb calorimeter?
Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

83ºC

Explanation:

A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.

The reaction of combustion of propane is:

C₃H₈ +  5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol

<em>1 mole of propane release 2222kJ</em>

10.0g of propane (Molar mass: 44.1g/mol).

10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>

If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):

0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.

Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:

504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.

As initial temperature was 20ºC, final temperature will be:

<h2>83ºC</h2>
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El potencial celular estándar, E_{cell} is +0.46 V

Explanation:

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Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

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Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V

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3 years ago
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garri49 [273]

Answer:

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater (m_{sw}), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

m_{sw} = \rho_{sw}\cdot V_{sw}

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\rho_{sw} - Density of seawater, measured in grams per liters.

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If V_{sw} = 1.5\times 10^{21}\,L and \rho_{sw} = 1030\,\frac{g}{L}, then:

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The total mass of sodium chloride is determined by the following ratio:

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m_{NaCl} = r\cdot m_{sw}

Given that m_{sw} = 1.545\times 10^{24}\,g and r = 0.035, the total mass of sodium chloride in all the seawater in the world is:

m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)

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3 years ago
the reaction of Al with HCl to produce hydrogen gas. What is the pressure of H2 if the hydrogen gas collected occupies 14 L at 3
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