According to the periodic table:
the molar mass of barium is 137.2 grams
the molar mass of oxygen is 16 grams
the molar mass of hydrogen is 1 gram
The molar mass of Ba(OH)2 can be calculated as 137.2 + 2(16) + 2(1) = 171.2 grams.
The molar mass of 4H2O is computed as 4 [2(1) + 16] = 72 grams.
Consequently, the molar mass of Ba(OH)2·4H2O is 171.2 + 72 = 243.2 grams.
Therefore, a sample weighing 243.2 grams of <span>barium hydroxide tetrahydrate includes 72 grams of water, meaning that within 92.8 grams, the mass of water would be:
mass of water in 92.8 grams = (92.8 x 72) / 243.2 = 27.474 grams.
Thus, when heating a 92.8 gram sample of Ba(OH)2·4H2O (barium hydroxide tetrahydrate), 27.474 grams of water will be emitted.</span>
Answer:
The temperature increase of the calorimeter, which is missing in the problem, is necessary for the calculation.
Explanation:
Since the temperature rise (X) is unspecified, we'll express the calculation in terms of X, and demonstrate with an example value.
1) Calorimeter details:
- Temperature increase: X °C
- Heat capacity ratio: 4.87 J / 5.5 °C (given)
- Energy absorbed by calorimeter at X °C rise:
(4.87 J / 5.5 °C) × X
2) Reaction data:
- Heat released: 362 kJ per mole of reactant
- Number of moles consumed: n
- Total energy from reaction:
362 kJ/mol × 1000 J/kJ × n = 362,000 n J
3) Using energy conservation, assuming no heat loss to surroundings, the energy from the reaction equals the energy absorbed by the calorimeter:
- 362,000 n = (4.87 J / 5.5 °C) × X
- n = [(4.87 / 5.5) × X] / 362,000
n = 0.000002446 × X
This means for each degree Celsius rise in calorimeter temperature, 0.000002446 moles of reactant were consumed.
Example:
If the calorimeter temperature increases by 100 °C, then:
- n = 0.000002446 × 100 = 0.0002446 mol
Answer:
0.133
Explanation:
The reaction that occurs between KIO3 and KI in an acidic medium is described as
IO3⁻ +5I⁻ +6h⁺ → 3I₂ + 3H₂O
I₂ subsequently reacts with sodium thiosulfate
NaS₂O₃ → 2Na⁺ + S₂O₃²⁻
The overall reaction can be summarized as
IO⁻₃ + 6H⁺ + 6S₂O₃³⁻ → I⁻ + 3S₄O₆²⁻ + 3H₂O
The mole of KIO₃
is computed using molarity multiplied by volume
which equals 0.00002mol
One mole of KIO₃ reacts with 6 moles of S₂O₃²⁻
which gives 2x6x10⁻⁵
= 0.00012 mol
The volume is 0.90 ml
1 ml equals 0.001L
0.90ML is 0.0009L
To find concentration,
molarity/volume
= 0.00012/0.0009
= 0.133m
