Answer:
A total of 2667 tires are required to satisfy the annual power needs of ten homes.
Explanation:
According to the Second Law of Thermodynamics, not all energy produced when tires are incinerated can be effectively used due to losses associated with finite temperature differences. The energy obtainable from a tire when burned, measured in kilowatt-hours (
), can be calculated using the efficiency definition:
Where:
- Efficiency, which is dimensionless.
- Energy released from burning, measured in kilowatt-hours.
Taking into account 
and 
, the yearly energy yield from a tire amounts to:
Thus, the number of tires necessary to meet the electricity demand of ten homes for one year is:
A total of 2667 tires are necessary to satisfy the annual power needs of ten homes.
The answer is a total of 74,844 calories.
Hi,
Due to calcium hydroxide being a strong base, its full dissociation will yield both calcium and hydroxyl ions:
Thus, the concentration of hydroxyl ions mirrors that of the calcium hydroxide, allowing for the calculation of pOH as demonstrated below:
Now, pH relates to pOH as:
Consequently, the final pH is achieved.
Best regards.
Answer:
Explanation:
Considering the reaction: 2X + 3Y = 3Z, combining 2.00 moles of X with 2.00 moles of Y results in the production of 1.75 moles of Z.
2 mol 2 mol 1.75 mol
2X + 3Y = 3Z
2 mol is required with 3 mol to yield 3 mol.
3 mol Z / 3 mol Y = 1 to 1
should yield 2 mol Z
1.75 / 2 = 87.5 % production yield
For the first-order decomposition, the equation is: ln(x0 / x) = kt. At t = 200, x = 0.0300 M, we have ln(x0 / 0.03) = 200k. At t = 400, when x = 0.0200 M, we utilize ln(x0 / 0.02) = 400k. By multiplying the first equation by 2, we get 2ln(x0 / 0.03) = 400k, which aligns with the second equation, leading us to conclude that 2ln(x0 / 0.03) = ln(x0 / 0.02). This suggests (x0 / 0.03)^2 = x0 / 0.02, allowing us to find x0 = 0.045 M as the initial concentration. Plugging this back into the first equation yields: ln(0.045 / 0.03) = 200k, from which it follows that k = 0.0020273 (rate constant). The half-life can be calculated with x = 0.5x0: ln(x0 / 0.5x0) = 0.0020273t, resulting in ln(2) = 0.0020273t, which simplifies to t = 341.90 minutes (half-life).
