Using the solubility graph place the substances in order from most soluble (#1) to least soluble at 40°C. KNO3, NaClO3,KBr, NaCl
1 answer:
Answer:
- NaClO₃ > KBr > KNO₃ > NaCl.
Explanation:
The attached file contains the graph with the solubility curves for the four substances, KNO₃, NaClO₃, KBr, NaCl.
To determine the solubility of each salt at a certain temperature, you read the temperature on the horizontal axis, labeled Temperature (ºC), and move upward up to intersecting the curve of the corresponding salt. Then, move horizontally up to insersceting the vertical axis, labeled Solubility (g/100g of H₂O), to read the solubility.
The higher the reading on the vertical axis, the higher the solubility.
The red vertical line that I added is at a temperature of 40ºC.
The number in blue indicate the order in which the solubility curves are intersected at that temperature:
- 4: NaCl: this is the lowest solubility
- 3: KNO₃: this is the second lowest solubility
- 2: KBr: this is the third lowest solubility
- 1: NaClO₃: this is the highest solubility.
Thus, the rank, from most soluble to least soluble is:
- NaClO₃ > KBr > KNO₃ > NaCl.
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Explanation:
The given data for case (1) is as follows.
h = 20 cm = 0.2 m
Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.
Q =
where, A = area
L = length
k = thermal conductivity = 0.8 W/m
= change in temperature.
Therefore, putting the given values into the above formula as follows.
Q =
=
= 168 W
For case (2), h = 180 cm = 1.8 m
Therefore, heat lost will be calculated as follows.
Q =
=
= 18.67 W
Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.
Energy lost to condense = 803.4 kJ
<h3>Further explanation</h3>Condensation of steam through 2 stages:
1. phase change(steam to water)
2. cool down(100 to 0 C)
1. phase change(condensation)
Lv==latent heat of vaporization for water=2260 J/g
2. cool down
c=specific heat for water=4.18 J/g C
Total heat =