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1 year 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]1 year 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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Read 2 more answers

A 2 mole sample of F2(g) reacts with excess NaOH(aq) according to the equation above. If the reaction is repeated with excess Na

77julia77 [94]

Answer:

The amount of NaF produced is doubled.

(d) is correct option.

Explanation:

Given that,

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The balance equation is

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1 year ago

At a given set of conditions 241.8 kJ of heat is released when one mole of H2O forms from its elements. Under the same condition

yan [13]

Answer:

44Kj

Explanation:

These are the equations for the reaction described in the question,

Vaporization which can be defined as transition of substance from liquid phase to vapor

H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H

-241.8kj -------eqn(1)

H2(g)+ 1/2 O2(g) ------>H2O(l).

Δ H =285.8kj ---------eqn(2)

But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as

H2O(l) ------>>H2O(g)---------------eqn(3)

But the equation from eqn(2) the eqn does go with vaporization so we can re- write as

H2O ------> H2(g)+ 1/2 O2(g)

Δ H= 285.8kj ---------------eqn(4)

To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)

Δ H=285.8kj +(-241.8kj)= 44kj

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2 years ago