How are the particles in a solid arranged differently from the particles in a liquid
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Explanation:
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In this case, since we can see that the heat released by the combustion of the mayonnaise is absorbed by the bomb calorimeter and water, we can set up the following equation:
Whereas q for mayonnaise stands for the heat due to its usage; thus, we plug in to obtain:
Now, if 10.0 g are eaten, the energy provided by it turns out:
And we need that value in dietary calories, 1 kcal:
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Taking into account the reaction stoichiometry, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Al₂(SO₄)₃ + 6 KOH → 2 Al(OH)₃ + 3 K₂SO₄
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂(SO₄)₃: 1 mole
- KOH: 6 moles
- Al(OH)₃: 2 moles
- K₂SO₄: 3 moles
The molar mass of the compounds is:
- Al₂(SO₄)₃: 342 g/mole
- KOH: 56.1 g/mole
- Al(OH)₃: 78 g/mole
- K₂SO₄: 174.2 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
- KOH: 6 moles ×56.1 g/mole= 336.6 grams
- Al(OH)₃: 2 moles ×78 g/mole= 156 grams
- K₂SO₄: 3 moles ×174.2 g/mole= 522.6 grams
Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume:
Molarity= number of moles÷ volume
<h3>Mass of Al(OH)₃ formed</h3>In firts place, you know that 48.6 mL (0.0486 L) of 0.15 M KOH react with excess Al₂(SO₄)₃.
Replacing in the definition, you can calculate the amount of moles of KOH that react:
0.15 M= number of moles÷ 0.0486 L
Solving:
0.15 M× 0.0486 L= number of moles
<u><em>0.00729 moles= number of moles</em></u>
Then, 0.00729 moles of KOH react with excess Al₂(SO₄)₃. So, following rule of three can be applied: if by reaction stoichiometry 6 moles of KOH form 156 grams of Al(OH)₃, 0.00729 moles of KOH form how much mass of Na₂SO₄?
<u><em>mass of Al(OH)₃= 0.18954 grams</em></u>
Then, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.
Learn more about
the reaction stoichiometry:
molarity:
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