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In-s [12.5K]
3 years ago
8

What is heat of vaporization?

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
7 0

Answer:

Explanation:

Heat of vaporization is the amount of energy (enthalpy) that must be added to a liquid substance, to transform a quantity of that substance into a gas. Heat of Vaporization is also known as heat of evaporation

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C. 4 decorated cupcakes.

Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries
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2. What effect does adding a neutron have on the atom’s mass?
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It does<span>, however, change the </span>mass<span> of the nucleus. </span>Adding<span> or removing </span>neutrons<span>from the nucleus are how isotopes are created. Protons carry a positive electrical charge and they alone determine the charge of the nucleus.</span>
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3 years ago
Calculate the standard entropy change for the following reactions at 25°C.
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Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

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Ca(OH)2 is an Arrhenius base and increases the concentration of OH when added to water
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What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?
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3.3256 Liters

See the image I have shared to you above

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