Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai

An r-selected species has a significantly faster reproductive rate compared to K-selected species.

The focus of r-selected species is on quick maturation and reproduction. They are likely to breed during short periods when water supply is available, thus enhancing their survival chances.

Conversely, K-selected species prioritize nurturing their young and tend to reproduce later. Due to the longer maturation time before breeding, by the time K-selected species are ready, the water supply may be depleted, leading to lower survival odds.

Hope this clarifies!
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~Collinjun0827, Junior Moderator

One electron is involved. Explanation: In redox reactions, determining the equivalents requires knowledge of the number of transferred electrons. In this specific case, one equivalent corresponds to a transfer of a single electron.

7.35 moles of oxygen. Initially, for each mole of H₂CO₃, there are 3 moles of oxygen, as derived from the acid's formula. For 2.45 moles of the compound stated in the problem, which is carbonic acid, we calculate: If 1 mole of H₂CO₃ corresponds to 3 moles of oxygen, then for 2.45 moles of H₂CO₃, we have X moles of oxygen. Thus, X = (3 × 2.45) / 1 = 7.35 moles of oxygen.

The molecular formula is C6H12O6.

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %