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marissa [1.9K]
1 year ago
11

Combustion analysis of a hydrocarbon produced 33.01 g of CO2 and 6.75 g of H2O.Calculate the empirical formula of the hydrocarbo

n
Chemistry
1 answer:
Basile [38]1 year ago
7 0

Given, 33.019 mol of co₂ of and 6.75g of  h₂o

mol of co₂ = mass/molar mass = 33.01g/44gmol = 0.75 mol

Since 1mol have co₂ 1 mol c

Hence 0.75 mol have = 0.75 mol

Now,

mol of H₂O = 6.75g/18gmol = 0.375 mol

Since 1mol H20 have = 2 mol H

Hence 0.375 mol H20 have = 2x0.375 mol H = 0.75 mol H.

Now

mol of H = [0.75 mol x4] = 3mol

mol of c = [0.75 molx4] = 3 mol

Hence

Empirical Formula = CH.

This is done by dividing the atomic mass by the molecular mass and then multiplying it by the mass of the compound is made. Therefore, the molecular formula for this hydrocarbon is CH3 because it has a ratio of 1 carbon atom to 3 hydrogen atoms. The general formula for saturated hydrocarbons with one ring is CnH2n. Most compounds require information other than the chemical formula to elucidate the structure. However, the CH ratio in the chemical formula can provide information about the chemical structure.

Learn more about Hydrocarbon here:-brainly.com/question/19453390

#SPJ1

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When the volume of a gas is
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Answer:

\boxed {\boxed {\sf 232.9 \textdegree C}}

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

250 \ mL * T_2 = 137 \textdegree C * 425 \ mL

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

The units of milliliters (mL) cancel.

T_2=\frac{ 137 \textdegree C * 425 }{250 }

T_2= \frac{58225}{250} \textdegree C

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The temperature changes to <u>232.9 degrees Celsius.</u>

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This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

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