Chemistry, 15.03.2020 10:01 princesskoi1910

Hi:) evaporation is the process by which liquids changes to gas at temperatures lower than boiling point. But why can’t liquids evaporate at their boiling point? Is that because liquids will boil instead of evaporate? Anyone able to help I’m rlly confused.

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evaporation is the process by which liquids changes to gas at temperatures lower than boiling point. But why can’t liquids evaporate at their boiling point? Is that because liquids will boil instead of evaporate?

Evaporation only take place at a certain temperature or it is high at a particular temperature which makes it to be rapid then, when liquid is heated, at a temperature the vapor leaves and evaporates in gaseous form known as vapor to the atmosphere. Evaporation takes place below the temperature of boiling point and above it but the rapidity depends on the temperature, the higher the temperature the rapid the evaporation takes place

Explanation:

Answer from: Quest

the grams of barium that are in 2.3 x 10^23 atoms of barium is 52.4 g (answer c)

calculation

mass = moles of ba x molar mass of ba

molar mass of ba from periodic table= 137.3 g/mol

use the avogadro's law constant to calculate the moles of barium

that is 1 mole = 6.02 x 10^23 atoms

? moles = 2.3 x 10^23 atoms

by cross multiplication

=(1 mole x 2.3x 10^23 atoms) / 6.02 x10^23 atoms)= 0.382 moles

mass is therefore = 137.3 g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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