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3 years ago

What structural feature is indicated by the suffix -ene in the name of a hydrocarbon

Chemistry

1 answer:

vaieri [72.5K]3 years ago

7 0

Answer:

A double bond between carbon atoms

Explanation:

The suffix -ene is used for alkenes, which tells you that the molecule has a carbon-carbon double bond.

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In a 1.0-liter container there are, at equilibrium, 0.10 mole h2, 0.20 mole n2, and 0.40 mole nh3. what is the value of kc for t

bogdanovich [222]

The reaction involved here would be written as:

2N2 + 3H2 = 2NH3

The equilibrium constant of a reaction is the ratio of the concentrations of the products and the reactants when in equilibrium. The expression for the equilibrium constant of this reaction would be as follows:

Kc = [NH3]^2 / [N2]^2[H2]^3
Kc = 0.40^2 / (0.20)^2 (0.10)^3
Kc = 4000

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3 years ago

Which of the following is an example of a buoyant force acting on a piece of

hodyreva [135]

Answer:

D. The piece of bread travels with straight- line horizontal acceleration.

Explanation

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2 years ago

Which statement is true of diffusion?

raketka [301]

Answer:

Molecules move from areas of high concentration to areas of low concentration.

4 0

2 years ago

Someone help answer this

alexandr402 [8]

it is water energy :) this is easy

6 0

2 years ago

Read 2 more answers

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

3 years ago