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ivann1987 [24]
3 years ago
6

Ne, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete comb

ustion of butane is 2c4h10(g)+13o2(g)→8co2(g)+10h2o(l) at 1.00 atm and 23 ∘c, what is the volume of carbon dioxide formed by the combustion of 2.20 g of butane?
Chemistry
1 answer:
Simora [160]3 years ago
4 0
<span>3.68 liters First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286 Looking at the balanced equation for the reaction which is 2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l) It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have: 0.037851286 mol * 4 = 0.151405143 mol The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) ) T = absolute temperature (23C + 273.15K = 296.15K) So let's solve the formula for V and the calculate using known values: PV = nRT V = nRT/P V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm) V = (3.679338871 L*atm)/(1 atm) V = 3.679338871 L So the volume of CO2 produced will occupy 3.68 liters.</span>
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Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe
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The amount of heat to absorb is 6,261 J

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Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

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Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 \frac{J}{g*C} and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 \frac{J}{g*C} *20 C and solving: Q=420 J

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The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}= 3.333 kJ= 3,333 J (being kJ=1,000 J)

  • Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 \frac{J}{g*C} and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 \frac{J}{g*C} *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

<u><em> The amount of heat to absorb is 6,261 J</em></u>

<u><em></em></u>

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