<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.
<u>Explanation:</u>
We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.
To calculate the volume of KBr, we use the formula:
Mass of the solution = 100 grams
Density of KBr solution = 1.03g/mL
Volume of the solution = ? mL
Putting values in above equation, we get:
Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.
In 97.08 mL of solution, mass of KBr present is 8.3 grams.
So, 41.2 mL of solution will contain = 
of KBr.
Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.
There are a couple of ways todetermine if a reaction is exothermic or endothermic. Endothermic meaning that heat is added to the reaction to make the reactants interact and exothermic meaning heat is released during the reaction between the two reactants.
In endothermic reactions you can find a triangle above the arrow.
Answer:
See explanation
Explanation:
Extraction has to do with the separation of the components of a mixture by dissolving the mixture in a set up involving two phases. One phase is the aqueous phase (beneath) while the other is the organic phase (on top). The solvents used for the two phases must not be miscible. Water commonly is used for the aqueous phase.
Ethanol is an important solvent in chemistry but the solvent is miscible with water in all proportions. As a result of this, ethanol is a poor solvent for carrying out extraction.
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
Because the resulting hybridized orbitals are more stable
