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3 years ago

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Chemistry

1 answer:

Zina [86]3 years ago

5 0

Answer:

Explanation:

n CaCO3 = mass / m.wt

= 500 /( 40 + 12 + 16x 3)

= 5 mole

n CaO = 5 moles ( from the balanced equation we have 1:1 moles )

mass of CaO = nCaO X m.wt

5 x( 40 +16 )

= 280 grams

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Answer:

2 molecules

Explanation:

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Ii) During burning, elements in the coal are converted to compounds called ????

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3 years ago

1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.

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Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

$H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ$ ..[1]

$CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ$ ..[2]

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3 years ago

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4 years ago

Read 2 more answers

4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o

Vilka [71]

Answer: The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = $\frac{1}{1}\times 0.0711=0.0711mol$ of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

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3 years ago