Ask question

Ask question

All categories

3 years ago

15

Alkanes with a branch coming off the straight chain are identified as substituted alkanes. what would be the name of a straight

chain made of 8 carbons with – ch3 groups on the second and fifth carbons?

1 answer:

Elden [556K]3 years ago

8 0

That would be 2,5-dimethyl octane

You might be interested in

Using the equations and enthalpy values provided, which mathematical expression can be used to determine the unknown enthalpy ch

vlabodo [156]

In order to solve this, we need to make use of Hess' Law.

We are already given the equations and their corresponding deltaH. Using Hess' Law, we can generate this equation:
104 kJ = x - (-1182 kJ) - (-1144 kJ)

Among the choices, the answer is
<span>B.104 = x - [(-1182) + (-1144)]
</span>

3 0

3 years ago

Read 2 more answers

An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal was formed and 5.20 L of oxygen was released at

Yakvenalex [24]

Answer:

<u>68.3g</u>

Explanation:

<u>1. Word equation:</u>

<em>mercury(II) oxide → mercury + oxygen </em>

<u>2. Balanced molecular equation:</u>

2HgO → 2Hg + O₂(g)

<u>3. Mole ratio</u>

Write the ratio of the coefficients of the substances that are object of the problem:

$2molHgO/1molO_2$

<u>4. Calculate the number of moles of O₂(g)</u>

Use the equation for ideal gases:

$pV=nRT\\\\\\n=\dfrac{pV}{RT}\\\\\\n=\dfrac{0.970atm\times5.20L}{0.08206atm.L/K.mol\times 390.0K}\\\\\\n=0.1576mol$

<u>5. Calculate the number of moles of HgO</u>

$\dfrac{2molHgO}{1molO_2}\times 0.1576molO_2=0.315molHgO$

<u>6. Convert to mass</u>

mass = # moles × molar mass

molar mass of HgO: 216.591g/mol

mass = 0.315mol × 216.591g/mol = 68.3g

7 0

3 years ago

Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of hcl with 1,3-pentadiene

GrogVix [38]

1,3-pentadiene has two double bonds which are conjugated, which undergo electrophilic addition reaction on reacting with $HCl$ .

The structure of 1,3-pentadiene is shown in the image.

When strong acid such as $HCl$ reacts with 1,3-pentadiene, the electrophilic addition reaction can occur either on double bond at 1,2-position or at 3,4-position. The reaction that occurs is shown in the image.

3 0

3 years ago

What does condensation mean in the water cycle?

tester [92]

Condensation<span> is the process by which </span>water<span> vapor in the air is changed into liquid </span>water<span>. </span>Condensation<span> is crucial to the </span>water cycle<span> because it is responsible for the formation of clouds. ... </span>Condensation<span> is the opposite of evaporation.</span>

4 0

3 years ago

Read 2 more answers

Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

<u>Step 2:</u> Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = <u>2.412 * 10^-4 mol/L = solubility in water</u>

<u>Step 3</u>: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = <u>3.32*10^-9 = solubility in 0.130 M NaOH </u>

<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0

3 years ago