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3 years ago

Explain why decomposition of sugar on heating is an irreversible change.

1 answer:

5 0

Sugar, sucrose (C12H22O11: a disaccharide, composed of the two monosaccharides: glucose and fructose), is odorless, that is, it lacks odor. When heated a phase change occurs resulting in melting of a thick syrup.

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Help please What percentage of the world does not have a reliable source for clean water?

disa [49]

Answer:9%

Explanation:

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3 years ago

At 9 a.m. on May 21, a body is discovered near a path in a park. The body is 10 feet from a tree beside the path. The body is pa

katrin2010 [14]

I'm pretty sure its D, Time of discovery of the body. Let me know (:

4 0

3 years ago

Read 2 more answers

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

pav-90 [236]

Answer: The initial concentration of hydrogen peroxide at the given temperature is 0.399 M

Explanation:

Decomposition of hydrogen peroxide is following first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $2.54\times 10^{-4}s^{-1}$

t = time taken for decay process = 855 s

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.321 M

Putting values in above equation, we get:

$2.54\times 10^{-4}s^{-1}=\frac{2.303}{855s}\log \frac{[A_o]}{0.321}$

$[A_o]=0.399M$

Hence, the initial concentration of hydrogen peroxide at the given temperature is 0.399 M

7 0

3 years ago

Delocalization which occurs through overlap between stigma and p orbitals

elixir [45]

Answer:

Single covalent bonds that form between nuclei are created from the "head-to-head" overlap of orbitals and are called stigma (s) bonds. ... Another type of bond, a pi (p) bond is formed when two p orbitals overlap. Pi bonds are found in double and triple bond structures

Explanation:

Delocalization happens when electric charge is spread over more than one atom. For example, bonding electrons may be distributed among several atoms that are bonded together.

7 0

3 years ago

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

Step 2: 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 = 2.412 * 10^-4 mol/L = solubility in water

Step 3: 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 = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: 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