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2 years ago

Swimming pool structures and structural _____ [680.26(B)(1) and (B)(2)] shall not be used as a grounding electrode.

1 answer:

7 0

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 reinforcing steel. 250.52(B)(3), [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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Why are the alkali metals and the halogens very reactive

jeyben [28]

Answer:

Answer in explanation

Explanation:

The reactivity or passiveness of an element depends solely on how close it is to attain a noble gas configuration. This means the closer an element is to attain a noble gas configuration, the greater its reactivity in both direction, positively or negatively.

Alkali metals belong to group 1 of the periodic table while halogens belong to group 17 of the periodic table. This means they are just one electron away from achieving the stability of a noble gas configuration. While alkali metals need to lose one electron to form a univalent positive ion, halogens news to gain one electron to form a univalent negative ion.

They tend to go about this vigorously and as such undergo several chemical reactions because of that single electron they neeed.

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2 years ago

How are the carbon, nitrogen, and oxygen cycles different?

k0ka [10]

The CNO cycle (for carbon–nitrogen–oxygen) is one of the two known sets of fusion reactions by which stars convert hydrogen to helium, the other being the proton–proton chain reaction. Unlike the latter, the CNO cycle is a catalytic cycle.

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3 years ago

You have 500 mL of 5 M HCl already made. You need to dilute the solution to 1 M HCl. How much water will you need to add?

hram777 [196]

Answer:

You need to add 400mL of water

Explanation:

500mL = 5 M HCI That means that if you divide both sides by 5

100mL = 1 M HCI If you need ot get rid of 4 M HCI then you add 400 mL of water because that is what it is equal to

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3 years ago

What is the molarity of a sodium hydroxide solution containing 2 moles of NaOH with a volume of .5L?

Maksim231197 [3]

Answer:

0.4M NaOH

Explanation:

Molarity, M, is an unit of concentration widely used defined as the ratio between moles of solute (In this case, NaOH) and volume of solution in liters.

As the solution contains 2 moles of NaOH-Moles of solute- in 5L of solution, the molarity is:

2 moles NaOH / 5L =

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3 years ago

Given the ion C2O4-2, what species would you expect to form with each of the following ions?

Ksivusya [100]

Answer:

A. K₂C₂O₄ Potassium oxalate

B. CuC₂O₄ Copper oxalate

C. Bi₂(C₂O₄)₃ Bismuth (III) oxalate

D. Pb(C₂O₄)₂ Lead (IV) oxalate

E. (NH₄)₂C₂O₄ Ammonium oxalate

F. HC₂O₄⁻ Acid oxalate

Explanation:

C₂O₄⁻² → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄ Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O

B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓

This is a precipitate.

C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓

This is a precipitate.

D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓

This is a precipitate.

E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻

HC₂O₄⁻ + H₂O ⇄ C₂O₄⁻² + H₃O⁺ Ka

HC₂O₄⁻ + H₂O ⇄ H₂C₂O₄ + OH⁻ Kb

HC₂O₄⁻ is an amphoteric compound

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3 years ago