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RUDIKE [14]
3 years ago
5

What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

n.Your answer should have two significant figures.Use −890.8kJmol for the molar heat of combustion of methane.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol

Mass of methane burnt = 530 g

Moles of methane burnt = \frac{530 g}{16 g/mol}=33.125 mol

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ

-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

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I'd say 624.2^3 inches.

Explanation:

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How many moles of agcl are contained in 244 ml of 0.135 magcl solution? the density of the solution is 1.22 g/ml?
spin [16.1K]
Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:

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Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
3 0
3 years ago
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If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of
Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.1\ g}{186.04\ g/mol}

Moles\ of\ ferrocene= 0.0005375\ mol

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

\rho=\frac{Mass}{Volume}

or,  

Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.044\ g}{78.49\ g/mol}

Moles\ of\ acetyl\ chloride= 0.0005606\ mol

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms  0.0005375 mole of monoacetylferrocene

Moles of product formed =  0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0
3 years ago
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