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3 years ago

Which of the following represents an organic compound? A. CN B. CH4 C. NaCl D. H2O?

Chemistry

2 answers:

Kipish [7]3 years ago

8 0

Answer: Option (B) is the correct answer.

Explanation:

A compound that contains both carbon and hydrogen atoms is known as an organic compound.

For example, $CH_{4}$ is an organic compound.

Whereas a compound that does not contain a carbon and hydrogen atom is known as an inorganic compound.

For example, $H_{2}O$ , CN, NaCl is an inorganic compound.

Thus, we can conclude that out of the given options $CH_{4}$ represents an organic compound.

olga55 [171]3 years ago

6 0

CH4 Methane which is the prototypical organic molecule and the others are chemical not sure about water

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3 years ago

What mass of TiCl4 must react with an excess of water to produce 50.0g of TiO2 if the reaction has a 78.9% yield

Kipish [7]

Answer:

\large \boxed{\text{150 g TiCl}_{4}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 189.68 79.87

TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl

m/g: 50.0

To solve this stoichiometry problem, you must

Convert the actual yield to the theoretical yield
Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂
Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄
Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄

1. Theoretical yield of TiO₂

$\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}$

2. Moles of TiO₂

$\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}$

3, Moles of TiCl₄

The molar ratio is 1 mol TiO₂:1 mol TiCl₄.

$\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}$

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$\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}$

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