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mafiozo [28]
3 years ago
9

A sample is left on the desk over several hours. On examination, the crystals appear moist, andliquid is forming around them. Th

e compound is classified as:
a. Stableb.
b. Deliquescent
c. Efflorescentd.
d. Hygroscopic
Chemistry
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

c. Efflorescent

Explanation:

Some hydrated salts, when exposed to the atmosphere, give off part of all of their water of crystallization spontaneously to form a lower dehydrated or anhydrous salt. Such salts are known as efflorescent and the phenomenon is called Efflorescence.

Efflorescence occurs when the hydrate's aqueous vapor pressure is greater than the water vapor's partial pressure in the air. For example, because the vapor pressures of washing soda (Na2CO3·10H2O) and Glauber's salt (Na2SO4·10H2O) normally exceed that of the water vapor in the atmosphere, these salts effloresce (i.e., lose all or part of their water of hydration), and their surfaces assume a powdery appearance.

You might be interested in
The soil of an agricultural field is too basic what can we do to correct it
katen-ka-za [31]

Answer:

you can correct it by adding an acid to make it neutral or less basic

3 0
3 years ago
1. When an acid reacts with a base, what compounds are formed? (1 point) a salt only water only metal oxides only a salt and wat
scoundrel [369]

Answer: 1) only a salt and water

An acid and base reacts together to undergo neutralization to form salt and water.

HCl+NaOH\rightarrow NaCl+H_2O

where HCl is an acid and NaOH is a base to form salt (NaCl) and water (H_2O).

Answer: 2) The formula of the hydrogen ion is often written as H^+.

All acids dissociate in water to give  H^+ ions.

Answer: 3) Arrhenius acids are substances that ionizes to yield protons in aqueous solution.

HCl\rightarrow H^++Cl^-

Arrhenius bases are substance that ionizes to yield hydroxide ions in aqueous solution.

NaOH\rightarrow Na^++OH^-

Answer: 4) A conjugate acid base pair  transfers hydrogen ion.

CH_3COO^-+H_2O\rightarrow CH_3COOH+H_3O^+

Here CH_3COO^- accepts a proton and thus act as a base and the corresponding CH_3COOH is its conjugate acid.

Answer: 5) Bronsted Lowry acid is a substance that donates protons.  Bronsted Lowry base is a substance that accepts protons.

NH_4^+ and NH_3

NH_3+H^+\rightarrow NH_4^+

NH_3 is a acid which accepts proton and thus acts as base to form conjugate acid NH_4^+.


4 0
2 years ago
Read 2 more answers
Answer the question
Ostrovityanka [42]
It is an adaptation
3 0
3 years ago
As you move across the period on the periodic table what pattern can be described about the electron configurations?
xeze [42]

Answer:

Elements in the same period have same number of electronic shell and electron is increased by one in every coming element with in same electronic shell.

Explanation:

Consider the second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

Electronic configuration of lithium:

Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell and number of electron increased by 1.

4 0
3 years ago
The standard reduction potentials of lithium metal and chlorine gas are as follows:Reaction Reduction potential(V)Li+(aq)+e−→Li(
meriva

Answer:

A) E° = 4.40 V

B) ΔG° = -8.49 × 10⁵ J

Explanation:

Let's consider the following redox reaction.

2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)

We can write the corresponding half-reactions.

Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq)      E°red = 1.36 V

Anode (oxidation):  2 Li(s) → 2 Li⁺(aq) + 2 e⁻         E°red = -3.04

<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>

The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.

E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V

<em>B) Calculate the free energy ΔG° of the reaction.</em>

We can calculate Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n.F.E°

where,

n are the moles of electrons transferred

F is Faraday's constant

ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J

8 0
3 years ago
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