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pentagon [3]
3 years ago
12

How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be usefu

l.ΔHvap = 40.7 kJ/molCliq = 4.18 J/g°CCgas = 2.01 J/g°CCsol = 2.09 J/g°CTmelting = 0°CTboiling = 100°C
Chemistry
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.

q 1= m c (T2-T1)

q1 = 36.0 g (4.18 J/gC) (100 - 65 C)

q1 = 5267 J =5.267kJ

next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2= 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3= 36.0 g (2.01 J/gC)(115-100C)

q3 = 1085 J =1.085kJ

qnet=q1 +q2 +q3

energy required=qnet=87.75kJ

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What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

 The  empirical  formula    of hydrocarbon is  CH2

The  molecular formula  of the  hydrocarbon is  C6H12


    <u><em>Explanation</em></u>

Hydrocarbon  is  made up  of carbon and hydrogen


<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




5 0
3 years ago
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