A grounding electrode is any object that directly links to the earth. They are most times used to divert electricity from the elements.
- Swimming pool structures and structural <u>reinforcing steel. 250.52(B)(3)</u><u>,</u> [680.26(B)(1), and (B)(2)] shall not be used as a grounding electrode.
In code 250.52(B)(3) it is clearly specified that the bonding grid and reinforcing steel that is related to a pool should not be used as grounding electrodes.
This is essential because when a metal that lies beneath a swimming pool is used as a grounding electrode, current from nearby electrical systems can be introduced into the pool.
This could cause the electrocution of anybody in the swimming pool at that time.
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Answer:- 2.39 mL are required.
Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:

Where,
and
are molarities of concentrated and diluted solutions and
and
are their respective volumes.
= 1.10M
= 5.00mM = 0.005M (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)
= ?
= 525 mL
Let's plug in the given values in the formula:



So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.
Answer: -112200J
Explanation:
The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of water vapour = 30.0g
C = 187 J/ G°C
Φ = (Final temperature - Initial temperature)
= 100°C - 120°C = -20°C
Then apply the formula, Q = MCΦ
Q = 30.0g x 187 J/ G°C x -20°C
Q = -112200J (The negative sign does indicates that heat was released to the surroundings)
Thus, -112200 joules of heat is released when cooling the superheated vapour.
Answer:
pro
Explanation:
c3h8 is propane
3 carbons makes it PROpane
the ANE come from all single bonds