The volume of H₂ evolved at NTP=0.336 L
<h3>Further explanation</h3>
Reaction
Decomposition of NH₃
2NH₃ ⇒ N₂ + 3H₂
conservation mass : mass reactants=mass product
0.28 NH₃= 0.25 N₂ + 0.03 H₂
2 g H₂ = 22.4 L
so for 0.03 g :
Answer:
37.1 calories.
Approximately, 37.1 = 40 calories.
Explanation:
So, without mincing words let's dive straight into the solution to the question above.
We are given the following parameters which are going to help in solving this particular Question.
The mass of broccoli = 86g of broccoli, mass of carbohydrates present = 6g of carbohydrates, the mass of protein present = 2.6g of protein and the mass of fat present = 0.3g of fat.
Therefore, the nutritional energy content (in Calories) = (6 × 4) + (2.6 × 4) + (0.3 × 9) = 10.4 + 24 + 2.7 = 37.1
Hence, the nutritional energy content (in Calories) = 37.1 calories.
Approximately, 37.1 = 40 calories.
Answer:
CaO + SO 3 → CaSO. 4
This is an acid-base reaction (neutralization): CaO is a base, SO 3 is an acid.
Producers are the foundation of every food web in every ecosystem—they occupy what is called the first tropic level of the food web. The second trophic level consists of primary consumers—the herbivores, or animals that eat plants. At the top level are secondary consumers—the carnivores and omnivores who eat the primary consumers. Ultimately, decomposers break down dead organisms, returning vital nutrients to the soil, and restarting the cycle. Another name for producers is autotrophs, which means “self-nourishers.” There are two kinds of autotrophs. The most common are photoautotrophs—producers that carry out photosynthesis. Trees, grasses, and shrubs are the most important terrestrial photoautotrophs. In most aquatic ecosystems, including lakes and oceans, algae are the most important photoautotrophs.
Answer:
When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Explanation:
Ksp of BaF₂ is:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.
As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
1.7x10⁻⁶ = [0.0144M] [F⁻]²
1.18x10⁻⁴ = [F⁻]²
0.0109M = [F⁻]
That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
