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likoan [24]
3 years ago
7

The complete combustion of a sample of propane produced 2.641 grams of carbon dioxide and 1.442 grams of water as the only produ

cts. Find the empirical formula of propane.
thnx!
20pts on this one.
Chemistry
1 answer:
ohaa [14]3 years ago
8 0
A general equation for a combustion reaction would be expressed as follows:

CxHy + (x+y/2)O2 = xCO2 + y/2H2O

Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.

moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C

moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H

Then, we divide the smallest amount to the each mole of the atoms. We do as follows:

C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67

Then we multiply a number in order to obtain a whole number ratio between the atoms.

1        CH2.67
2       C2H5.34
3       C3H8   <-------- empirical formula
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If there are 80 liters of argon gas in a piston and the piston expands the gas until its volume is 2000 liters with a pressure o
Mice21 [21]

Answer: The correct answer is A. 11.5 atm. The temperature is held constant at 293 K, therefore, we can use Boyle's Law to determine the initial pressure. Boyle's Law states that there is an inverse relationship between pressure and volume of gases. Therefore, as volume increases, the pressure will decrease and vice versa.

Further Explanation:

Boyle's Law can be mathematically expressed as:

P_{initial}V_{initial}\ = P_{final}V_{final}

In this problem, we are given the values:

P(initial) = ?

V(initial) = 80 L

P (final) = 0.46 atm

V (final) = 2000 L

Plugging in these values into the equation:

P_{initial}(80 \ L) = \ (0.46 \ atm)(2000 \ L)\\P_{initial} \ = \frac{(0.46 \ atm) \ (2000 \ L)}{80 \ L}\\P_{initial}\ = 11.5 atm\\

The initial pressure was 11.5 atm. Since the volume increased or expanded, the space where the gas particles move is bigger, so the frequency of collisions with the wall of the container and with other particles are effectively decreased. This, therefore, decreases the pressure from 11.5 to 0.46 atm.

Learn More

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Keywords: gas, Boyle's Law, Ideal Gas Law

7 0
3 years ago
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) T
brilliants [131]

Answer:

ΔHrxn = [(1) -1675.5 ( kJ/mole) + (2) 0 ( kJ/mole)] - [(1) -824.3 ( kJ/mole) + (2) 0 ( kJ/mole)]

Explanation:

ΔHrxn = 2ΔHf (Al₂O₃)  - ΔHf (Fe₂O₃)

Remember that for pure elements in their standard state of temperature and pressure by definition their standard heats of formation are zero.

ΔHrxn = 2(-1675.7) - (-824.3) kJ/mol

ΔHrxn = 2527 kJ/mol

6 0
3 years ago
Which of these elements has four valence electrons
exis [7]
What are the elements?

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