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2 years ago

Which substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes? a)CH4 b)H2S c)CO2 d)NaCl

Chemistry

1 answer:

schepotkina [342]2 years ago

8 0

Answer:

$H_2S$

Explanation:

Given the amount of heat absorbed and the amount of substance in moles, we may calculate the heat of vaporization. Heat of vaporization is defined as the amount of heat per 1 mole of substance required to evaporate that specific substance.

Based on the value of heat of vaporization, we will identify the substance. Firstly, let's calculate the heat of vaporization:

$\Delta H^o = \frac{58.16 kJ}{3.11 mol} = 18.7 kJ/mol$

Secondly, let's use any table for heat of vaporization values for substances. We identify that the heat of vaporization of $H_2S$ is 18.7 kJ/mol

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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

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Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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2 years ago

Write a net ionic equation for the reaction that occurs when nickel(ii carbonate and excess hydrobromic acid (aq are combined.

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First, we write the reaction equation:
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Overall equation:
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We can use the heat equation,
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By applying the formula,

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(T - 22) = 1200 J / (36 g x 4.186 J/g °C)

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Answer and Explanation:

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