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3 years ago

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Chemistry

1 answer:

hichkok12 [17]3 years ago

5 0

Answer: B

Explanation: I answered randomly

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Calculate the mass of methane that must be burned to provide enough heat to convert 242.0 g of water at 26.0°C into steam at 101

Anarel [89]

Answer: The mass of methane burned is 12.4 grams.

Explanation:

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2(g))})+(2\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(CH_4(g))})+(2\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(CH_4(g))}=-74.81kJ/mol\\\Delta H^o_f_{O_2}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-393.51))+(2\times (-241.82))]-[(1\times (-74.81))+(2\times (0))]\\\\\Delta H^o_{rxn}=-802.34kJ$

The heat calculated above is the heat released for 1 mole of methane.

The process involved in this problem are:

$(1):H_2O(l)(26^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(101^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_p,l\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water = 242.0 g

$C_{p,l}$ = specific heat of water = 4.18 J/g°C

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $26^oC$

Putting all the values in above equation, we get:

$q_1=242.0g\times 4.18J/g^oC\times (100-(26))^oC=74855.44J$

For process 2:

$q_2=m\times L_v$

where,

$q_2$ = amount of heat absorbed = ?

m = mass of water or steam = 242 g

$L_v$ = latent heat of vaporization = 2257 J/g

Putting all the values in above equation, we get:

$q_2=242g\times 2257J/g=546194J$

For process 3:

$q_3=mC_p,g\times (T_2-T_1)$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of steam = 242.0 g

$C_{p,g}$ = specific heat of steam = 2.08 J/g°C

$T_2$ = final temperature = $101^oC$

$T_1$ = initial temperature = $100^oC$

Putting all the values in above equation, we get:

$q_3=242.0g\times 2.08J/g^oC\times (101-(100))^oC=503.36J$

Total heat required = $q_1+q_2+q_3=(74855.44+546194+503.36)=621552.8J=621.552kJ$

To calculate the number of moles of methane, we apply unitary method:

When 802.34 kJ of heat is needed, the amount of methane combusted is 1 mole

So, when 621.552 kJ of heat is needed, the amount of methane combusted will be = $\frac{1}{802.34}\times 621.552=0.775mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of methane = 16 g/mol

Moles of methane = 0.775 moles

Putting values in above equation, we get:

$0.775mol=\frac{\text{Mass of methane}}{16g/mol}\\\\\text{Mass of methane}=(0.775mol\times 16g/mol)=12.4g$

Hence, the mass of methane burned is 12.4 grams.

8 0

3 years ago

What do the two arrows in the diagram most likely represent?

elixir [45]

I think radiation and energy transfer!

8 0

3 years ago

Which energy transformation takes place in a gasoline lawn mower?

fenix001 [56]

Hey there!!

A lawn mower works by converting chemical energy (energy stored in the gas) to mechanical energy.

Hope this helps you and have a great day!!!! I wish you the best of luck (:

6 0

3 years ago

Read 2 more answers

What is the mole fraction of each component in a solution made by mixing 300 g of ethanol(C2H5OH) and 500 g of water?

Snowcat [4.5K]

Answer:

The mole fraction of ethanol and water in the solution are 0.19 and 0.81 respectively.

Explanation:

The Molar Fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution, which are calculated by adding the moles of solute (s) and solvent.

It is calculated by dividing the number of moles of one of the components by the total number of moles of the solution.

The sum of all the molar fractions of the substances present in a solution is equal to 1.

So, the expression to calculate the mole fraction is:

$Molar fraction (Xi)=\frac{moles of sudstance}{total moles of the solution}$

Being the molar mass of the compounds:

Ethanol 46 $\frac{g}{mol}$

Water 18 $\frac{g}{mol}$

the number of moles that represent the mass quantities is calculated as:

Moles of ethanol: 300 grams* $\frac{1 mol}{46 grams}$ = 6.52 moles
Moles of water: 500 grams* $\frac{1 mol}{18 grams}$ = 27.78 moles

So the total moles in solution are 6.52 moles + 27.78 moles = 34.3 moles

The mole fraction of ethanol in the solution is calculated as:

$mole fraction of ethanol=x_{ethanol} =\frac{6.52 moles}{34.3 moles}$

$x_{ethanol} =0.19$

The mole fraction of water in the solution is calculated as:

$mole fraction of water=x_{water} =\frac{27.78 moles}{34.3 moles}$

$x_{water} =0.81$

Then:

$x_{water} +x_{ethanol} = 0.81 + 0.19$

$x_{water} +x_{ethanol} =1$

The mole fraction of ethanol and water in the solution are 0.19 and 0.81 respectively.

8 0

3 years ago

A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the

Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047 mol/dm³

The concentration of the sulfuric acid is 0.047 mol/dm³

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7 0

3 years ago