One mole of h2s gas escapes from a container by effusion in 77 seconds how long would it take one mole of nh3 gas to escape from

Sagot:

0.1 M NaCl

Paliwanag:

Ang tanong na ito ay nagpapaalala sa atin ng mga patakaran sa solubility. Alalahanin natin na ang lahat ng chlorides ay natutunaw maliban sa mga ng lead, mercury II at silver na hindi natutunaw sa tubig.

Ang sumusunod na reaksyon ay mangyayari na humahantong sa pagbuo ng isang precipitate;

Pb(NO3)2(aq) + 2NaCl(aq) -------> 2NaNO3(aq) + PbCl2(s)

Ang puting precipitate na nabuo ay  PbCl2.

Answer:

The adjustable legs along with the sand table.

Note: The question is incomplete. The full question is presented below.

Using Models to Address Questions Regarding Systems

Armando’s class was examining images of rivers shaped by flowing water. Most rivers appeared wide and shallow, except for one, which was narrow and deep. The students theorized that this river's narrowness and depth are due to:

• the steepness of the hill from which the water descends, or
• the diminutive size of the sand grains the water flows through.

To explore the answer to the question of why this river is so narrow and deep, Armando created the model outlined below.

Explanation:

The model constructed by Armando will facilitate addressing the question due to specific features:

1. Adjustable leg - as one theory proposed by the class suggests that the steep hill affecting the water's path could be the reason for the river's dimensions, the adjustable legs are designed to be raised or lowered to alter the slope, allowing testing of this theory.

2. Sand table - this acts as the streambed. By modifying the size of the sand grains, students can examine the second hypothesis that smaller sand grains contribute to the river's narrowness and depth.

The outcomes of their experimentation will lead them to a conclusion.

Answer:

a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.

b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.

d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.

(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.

g. HS⁻ is amphoteric.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

This is similar to the case of bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hydrazinium acts as an acid, making hydrazine its conjugate base.

Noble gas notation serves as a condensed form of indicating electron configurations. This notation employs the symbol for the preceding noble gas in the electron configuration of an element. For antimony, the noble gas prior is Kr, which means Xe is not used in its electron configuration. Similarly, for radium, the prior noble gas is Rn, whereas, for uranium, it is also Rn. However, for cesium, the preceding noble gas is Xe, thus it is utilized in the noble gas notation for Sb, specifically written as: Cs: [Xe] 6s.

Answer: cesium

Solution:

Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.

Thus,

Molality = moles of solute / kg of solvent.

Therefore, kg of solvent = moles of solute / molality.

moles of solute = mass / molar mass

= 25.31 g / 101.1 g/mole

= 0.2503 mole.

kg of solvent = 0.2503 mole / 0.1982 m

= 1.263 kg

= 1263 g.

This is the final answer.