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3 years ago

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The π bond in ethylene, H2C=CH2, results from the overlap of ________. A) sp3 hybrid orbitals B) s atomic orbitals C) sp hybrid

orbitals D) sp2 hybrid orbitals E) p atomic orbitals

1 answer:

IgorC [24]3 years ago

6 0

Answer:

<h3>The answer is option E.</h3>

Hope this helps you

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Samira has a son who is 24 months old. Her sister Paulina also has a son who is one and a half years old. Whose son is older? Ex

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Samira’s son is older because there are 12 months in a year and we can gather that 12+12=24 so we know that Paulina’s son is younger because he is only 18 months.

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3 years ago

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____is not a resource that can be used to help meet our energy needs

Bezzdna [24]

I believe the answer to this is A.

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4 years ago

Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp

notka56 [123]

Answer:

0.00 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ 30.07 32.00

2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g: 1.50 11.

2. Calculate the moles of each reactant

$\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}$

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

$\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}$

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

$\text{Moles of CO}_{2} = \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

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3 years ago

1 Identify Draw an example of a force acting on an object.

Alisiya [41]

Answer:

Answer below

Explanation:

Just draw a photo of someone pushing an object across a table. Your push is the force acting on the object you're pushing.

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2 years ago

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 97. g of octane is mi

zhannawk [14.2K]

Answer: 61 grams

Explanation:

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

$\text{Number of moles of octane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{97g}{114g/mol}=0.85moles$

$\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{150g}{32g/mol}=4.69moles$

The chemical equation for the combustion of octane in oxygen follows the equation:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

By stoichiometry of the reaction;

25 moles of oxygen react with 2 moles of octane

4.69 moles of oxygen react with= $\frac{2}{25}\times 4.69=0.37$ moles of octane

Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.

25 moles of oxygen produce 18 moles of water

4.69 moles of oxygen produce= $\frac{18}{25}\times 4.69=3.38$ moles of water.

Mass of water produced= $moles\times {\text{Molar mass}}=3.38\times 18g/mol=61g$

The maximum mass of water that could be produced by the chemical reaction is 61 grams.

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3 years ago