Answer: 0.0007 moles of 
are released when the temperature rises.
Explanation:
To determine the moles, we utilize the ideal gas law, as follows:
where,
P = gas pressure = 1.01 bar
V = gas volume = 1L
R = gas constant = 
- Calculated moles at T = 20° C
The gas temperature = 20° C = (273 + 20)K = 293K
Substituting values into the equation gives:
- Calculated moles at T = 25° C
The gas temperature = 25° C = (273 + 25)K = 298K
Substituting values into the equation gives:
- Released moles =
Therefore, 0.0007 moles of are released when the temperature increases from 20° C to 25° C.
The mass calculated for the copper piece is 290 grams. The formula for mass is given by mass = density × volume, where the density of copper is 8.96 grams per mL. The volume of the copper piece, determined by the increase in volume, equals 137 mL - 105 mL = 32 mL. Multiplying the volume by the density gives us the mass of copper: mass = 8.96 g/mL × 32 mL = 286.72 grams. Since the volume is presented with two significant figures, rounding the mass to two significant figures results in 290 grams.
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
Answer:
14.5L
Explanation:
The following measurements were taken:
V1 = 14.1L
T1 = 13.9°C = 286.9K
T2 = 22°C = 295K
V2 =?
By applying Charles' Law: V1/T1 = V2/T2, we calculate the new volume:
14.1/286.9 = V2/295
Cross-multiplying gives:
286.9 × V2 = 14.1 × 295
Then dividing both sides by 286.9:
V2 = (14.1 × 295)/286.9
V2 = 14.5L
Thus, the updated volume is 14.5L
Answer:
To lower the temperature of the solution from 25.0°C to 5.0°C, it is necessary to use 35.2g of NH₄NO₃ for every 100.0g of water.
Explanation:
In order to cool down the solution, we need:
4.184 J/g°C × (5.0°C - 25.0°C) × (100.0g + X) = -Y
8368 J + 83.68 J/gX = Y (1)
Here, x represents the grams of NH₄NO₃ required, and Y represents the energy needed to remove heat.
Furthermore, the energy Y becomes:
Y = 25700 J/mol × 
X
Y = 321 J/g X (2)
Substituting (2) into (1)
8368 J + 83.68 J/g X = 321 J/g X
8363 J = 237.32 J/gX
X = 35.2g
This means 35.2g of NH₄NO₃ must be used for every 100.0g of water to achieve a temperature decrease from 25.0°C to 5.0°C.
I trust this information will be useful!
