<span>1962.252(a)
whenever the material is dropped from 20feet to any poiny lying outside the exterior wall of building, an enclosed chute of wood, or equivalent material. For the purpose of paragraph, an enclose chute is a slide, closed in all sides, through which material is placed from high plac eto lower one.
1962.252(b)
when the disposal is dropped through holes in the floor without the use of chute, the area onto which the material is dropped shall be completely enclosed with barricades not less than 42 inch high and not less than 6 feet back from the projected edge of of the opening valve. signs warning of hazards of falling material shall be posted at each level.
1962.252(c)
All scrap lumper, waste material, and rubbish shall be removed from the immediate work area as the work progresses.
1962.252(d)
Disposal of waste material or debris by burning shall comply with local fire regulation
1962.252(e)
All sovent waste, oily rags, and flammable liquid should be kept in fire resistant covered container until removed from work site.</span>
I believe you are referring zero as the exponent. <span>Any number (except 0) with exponent 0 is defined to mean 1.
</span>
For one thing, there is a rule:
<span> a^m/ a^m = a^m-m = a^0
</span>But (when a is not equal to <span>0),
</span>
a^m/ a^m = 1
Therefore, we must define a^0 as 1.
Answer: The equivalent mass of the acid is 83.16 grams
Explanation:
To calculate the number of moles for given molarity, we use the equation:
Molarity of 
solution = 0.1165 M
Volume of solution = 24.68 mL = 0.02468 L
Putting values in equation 1, we get:
( as acidity of NaOH is 1)
For end point: gram equivalents of acid = gram equivalents of base = 
Mass of acid=
Thus equivalent mass of the acid is 83.16 grams
2Fe + Cl₂ -> 2FeCl
4Fe + 3O₂ -> 2Fe₂O₃
Answer:
-122 J/K
Explanation:
Let's consider the following balanced reaction.
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
We can calculate the standard reaction entropy (ΔS°) using the following expression.
ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r
where,
- η: stoichiometric coefficients of products and reactants
- Sf°r: entropies of formation of products and reactants
ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol
ΔS° = -121.77 J/K ≈ -122 J/K
