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4 years ago

Combustion analysis of a hydrocarbon produced 33.01 g co2 and 13.52 g h2o. calculate the empirical formula of the hydrocarbon.

1 answer:

6 0

Answer is: the empirical formula of the hydrocarbon is CH₂.
Chemical reaction: CₓHₐ + O₂ → xC + a/2H₂O.
m(CO₂) = 33.01 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 33.01 g ÷ 44.01 g/mol.
n(CO₂) = n(C) = 0.75 mol.
m(H₂O) = 13.52 g.
n(H₂O) = 13.52 g ÷ 18 g/mol.
n(H₂O) = 0.75 mol.
n(H) = 2 · n(H₂O) = 1.5 mol.
n(C) : n(H) = 0.75 mol : 1.5 mol /0.75 mol.
n(C) : n(H) = 1 : 2.

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3 years ago

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3 years ago

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A sample of h gas (2.0 l) at 3.5 atm was combined with 1.5 l of n gas at 2.6 atm pressure at a constant temperature of 25c into

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3 years ago

Calculate the molarity of 198 g of barium iodide (Bal2) in 2.0 l of solution

nikitadnepr [17]

Answer: The molarity of 198 g of barium iodide $(BaI_{2})$ in 2.0 L of solution is 0.253 M.

Explanation:

Given: Mass = 198 g

Volume = 2.0 L

Molarity is the number of moles of solute present in liter of a solution.

Moles is the mass of a substance divided by its molar mass. So, moles of barium iodide (molar mass = 391.136 g/mol) is calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{198 g}{391.136 g/mol}\\= 0.506 mol$

Now, molarity is calculated as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.506 mol}{2.0 L}\\= 0.253 M$

Thus, we can conclude that the molarity of 198 g of barium iodide $(BaI_{2})$ in 2.0 L of solution is 0.253 M.

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3 years ago

Complete the table:

Stells [14]

Answer:

Mass solution = 28.65 g

Mass percent = 9.14 %

Mass solute = 1.95 g

Mass solution = 48.75 g

Mass solvent = 24.52 g

Mass percent = 6.41 %

Mass solution = 476.8 g

Mass solvent =450.1 g

Explanation:

Hello,

In this case, we define the mass percent as:

$\%m/m=\frac{m_{solute}}{m_{solute}+m_{solution}} *100\%$

Wherein we use the mass of the solute and the mass of the solvent in order to compute it, thus, for each case we have:

Mass solution: 2.65 g + 26.0 g = 28.65 g

Mass percent: 2.62 g / (2.65 g + 26.0 g) x 100% = 9.14 %

Mass solute: 4.0 % x 46.8 g / (1 - 4.0 %) = 1.95 g

Mass solution: 1.95 g + 46.8 g = 48.75 g

Mass solvent: 26.2 g - 1.68 g = 24.52 g

Mass percent: 1.68 g / (1.68 g + 24.52 g) x 100% = 6.41 %

Mass solution: 26.7 g / 5.6 % = 476.8 g

Mass solvent: 476.8 g - 26.7 g =450.1 g

Best regards.

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3 years ago