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.
Answer:
Acid + Oxide or Hydroxide
Many oxide, hydroxide and carbonate compounds
are insoluble in water, but do react with acid.
Acid + Oxide → Salt + Water
Acid + Hydroxide → Salt + Water
MgO (s) + HNO3 (aq) → Mg(NO3)2 (aq) + H2O (l)
CuOH (s) + HCl (aq) → CuCl (aq) + H2O (l)
Explanation:
lead compounds are the oxides: lead monoxide, PbO, in which lead is in the +2 state; lead dioxide, PbO2, in which lead is in the +4 state; and trilead tetroxide, Pb3O4. Lead monoxide exists in two modifications, litharge and massicot. Litharge, or alpha lead monoxide, is a red or reddish…
Answer:
Molarity = 2.3 M
Explanation:
Molarity can be calculated using the following rule:
Molarity = number of moles of solute / volume of solution
1- getting the number of moles:
We are given that:
mass of solute = 105.96 grams
From the periodic table:
atomic mass of carbon = 12 grams
atomic mass of hydrogen = 1 gram
atomic mass of oxygen = 16 grams
Therefore:
molar mass of C2H6O = 2(12) + 6(1) + 16 = 46 grams
Now, we can get the number of moles as follows:
number of moles = mass / molar mass = 105.96 / 46 = 2.3 moles
2- The volume of solution is given = 1 liter
3- getting the molarity:
molarity = number of moles of solute / volume of solution
molarity = 2.3 / 1
molarity = 2.3 M
Hope this helps :)
Answer:
Answer Below! : )
Explanation:
Groundwater is stored in the tiny open spaces between rock and sand, soil, and gravel. How well loosely arranged rock (such as sand and gravel) holds water depends on the size of the rock particles.
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