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Montano1993 [528]
2 years ago
13

A 3.00 g sample of unknown hydrocarbon is prepared for combustion analysis. after the hydrocarbon undergoes complete combustion,

8.80 g of co2 and 5.40 g of h2o are produced. what is the empirical formula of the unknown?
Chemistry
1 answer:
zmey [24]2 years ago
7 0

Answer:

CH₃.

Explanation:

  • The complete combustion of hydrocarbon in an excess of oxygen produces CO₂ and H₂O according to the general reaction equation:

CₐHₓ + O₂(excess) → b CO₂ + c H₂O,

Where a is the no. of C atoms in the unknown hydrocarbon.

x is the no. of H atoms in the unknown hydrocarbon.

b is the no. of moles of CO₂ produced from the combustion of unknown hydrocarbon.

c is no. of moles of H₂O produced from the combustion of unknown hydrocarbon.

  • We can calculate b and c using the relation:

n = mass / molar mass.

  • n of CO₂ = (8.80 g) / (44.0 g/mol) = 0.2 mol.

Every molecule of CO₂ contains 1 C atom, so no. of moles of C atoms in CO₂ and thus in the original hydrocarbon is 0.2 mol.

  • n of H₂O = (5.40 g) / (18.0 g/mol) = 0.3 mol.

Every molecule of H₂O contains 2 H atoms, so no. of moles of H atoms in H₂O and thus in the original hydrocarbon is (2 x 0.3) 0.6 mol.

  • The ratio of C to H in the unknown hydrocarbon is (0.2): (2 x 0.3) = 0.2: 0.6.
  • Dividing by the lowest no. of moles (0.2) that of C.

∴ The ratio of C to H in the unknown hydrocarbon is 1: 3.

<em>So, the empirical formula of the unknown hydrocarbon is CH₃.</em>

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Which ion is the only negative ion produced by an Arrhenius base in water?
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Pressure can be defined as the force acting on a perpendicular surface per unit area.

Force exerted by a man of mass 100 kg wearing snow shoes = m.a

Where m = mass of the man = 100 kg

a = acceleration due to gravity= 9.8 m/s^{2}

Force exerted by the man of mass 100 kg = 100 kg(9.8 m/s^{2}) = 980 N

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Force exerted by 100 kg man is greater than that exerted as 60 kg woman. The area on which this force is acting determines the pressure. Pressure is inversely proportional to the area on which the force acts. Therefore, the pressure exerted by 100 kg man wearing snow shoes is less than the pressure exerted by a 60 kg woman woman wearing high heels as the force acts over a larger area when the man wears snow shoes when compared to the force exerted over a smaller area in case of the woman wearing high heels.

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Use this equation for the following problems: 2NaN3 --&gt; 2Na+3N2
olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

<em>2NaN₃ → 2Na + 3N₂,</em>

  • It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?​ </em>

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

  • Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

  • We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

  • We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

<em><u>using cross multiplication:</u></em>

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

  • Finally, we can get the grams of NaN₃ needed:

<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>

<em />

<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>

  • We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

  • We can get the no. of moles of N₂ produced with 0.36 mol of Na:

<em><u>using cross multiplication:</u></em>

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

  • We can get the mass of 0.54 mol of N₂:

mass = no. of moles  x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

  • Now, we can get the mL of 15.12 g of N₂:

<em><u>using cross multiplication:</u></em>

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>

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