Answer:
2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)
Explanation:
A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.
The only equation that meets those conditions is the last one.
A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.
C is wrong because Li⁺ and SO₄²⁻ are not elements.
D is wrong because it shows the formation of 8 mol of Li₂SO₄.
15 grams of NH3 can be dissolved
<h3>Further explanation</h3>
Given
50 grams of water at 50°C
Required
mass of NH3
Solution
Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility
- 1. Temperature:
- 2. Surface area:
- 3. Solvent type:
- 4. Stirring process:
We can use solubility chart (attached) to determine the solubility of NH3 at 50°C
From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g
So that the solubility in 50 grams of water is:
= 50/100 x 30
= 15 grams
Answer:
(1) addition of HBr to 2-methyl-2-pentene
Explanation:
In this case, we will have the formation of a <u>carbocation</u> for each molecule. For molecule 1 we will have a <u>tertiary carbocation</u> and for molecule 2 we will have a <u>secondary carbocation</u>.
Therefore the <u>most stable carbocation</u> is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)
<span> iron (Fe), ruthenium (Ru), osmium (Os) and hassium (Hs). They are all transition metals.</span>
Answer:
It effects all of the cellular respiration process
Explanation:
It inhibits the Glycolysis. It replaces the phosphate groups that is needed for making Pyruvate and ATP.