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tamaranim1 [39]
3 years ago
8

Write the balanced chemical equation for the combustion of ethane, , and answer these questions. (Use the lowest possible coeffi

cients. Omit states of matter.) How many molecules of oxygen would combine with 16 molecules of ethane in this reaction
Chemistry
1 answer:
horrorfan [7]3 years ago
3 0

Answer:

1. 2C2H6 + 7O2 —> 4CO2 + 6H2O

2. 56moles of O2

Explanation:

Ethane undergo Combustion to produce CO2 and H20 according the equation below:

C2H6 + O2 —> CO2 + H2O

Let us balance the equation. There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 3 in front of H2O as shown below:

C2H6 + O2 —> CO2 + 3H2O

There are 2 atoms of C on left and 1atom on the right. It can be balanced by putting 2 in front of CO2 as shown below:

C2H6 + O2 —> 2CO2 + 3H2O

There are a total of 7 atoms of O on the right and 2 atoms on the left. It can be balanced by putting 7/2 in front of O2 as shown below:

C2H6 + 7/2O2 —> 2CO2 + 3H2O

Now we multiply through by 2 to remove the fraction as shown below

2C2H6 + 7O2 —> 4CO2 + 6H2O

Now the equation is balanced

2. 2C2H6 + 7O2 —> 4CO2 + 6H2O

From the equation above,

2 moles of ethane(C2H6) combined with 7moles of O2.

Therefore, 16moles of ethane(C2H6) will combine with = (16x7)/2 = 56moles of O2

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An organic solvent has a density of 1.31 g/mL. What volume is occupied by 57.5 g of the liquid?
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Answer:

<h3>The answer is 43.89 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 57.5 g

density = 1.31 g/mL

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volume =  \frac{57.5}{1.31}  \\  = 43.89312977...

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What is the unknown metal if the temperature of a beaker of 100ml of water was raised 17c to 19 c when 21 grams of the metal at
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Answer:

The metal has a heat capacity of 0.385 J/g°C

This metal is copper.

Explanation:

<u>Step 1</u>: Data given

Mass of the metal = 21 grams

Volume of water = 100 mL

 ⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams

Initial temperature of metal = 122.5 °C

Initial temperature of water = 17°C

Final temperature of water and the metal = 19 °C

Heat capacity of water = 4.184 J/g°C

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<u>Step 2: </u>Calculate the specific heat capacity

Heat lost by the metal = heat won by water

Qmetal = -Qwater

Q = m*c*ΔT

m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)

21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)

-2173.5 *c(metal) = -836.8

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The metal has a heat capacity of 0.385 J/g°C

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