Answer:
moles Ar in 20g = 0.500 mole Ar
Explanation:
moles = grams given / formula weight = 20g / 39.948g·mol⁻¹ = 0.500 mole Ar
<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>
Answer:
Explanation:
a )
energy produced per second = 500 J
Heat produced = 500 x .8 = 400 J per second.
If m be the mass of water evaporated per unit hour
m x latent heat = 400 x 60 x 60
= m x 2.42 x 10⁶ = 1.44 x 10⁶
m = .595 kg per hour
b )
volume of water = 595 mL
bottles = 595 / 750
.8 or 4/5 of bottle. per hour.
Answer:- Formula of the hydrate is 
and it's name is Iron(III)sulfate pentahydrate.
Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.
Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.
We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:
= 
= 
Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.
Answer:
1) 1.15 mol
2) M=0.45
3) 22.5 mL
4) 6.25 mL
Explanation:
1)
550 mL= 0.55 L
M= mol solute/ L solution
mol solute= M * L solution
mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute
2)
155 mL = 0.155 L
80 g -> 1 mol NH4NO3
5.61 g -> x
x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3
M=(0.07 mol NH4NO3)/0.155 L M=0.45
3) M1V1=M2V2
V1= M2V2/M1
V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL
4) M1V1=M2V2
V1= M2V2/M1
V1= (0.25 M * 0.45 L)/ 18.0 M
V1=6.25 x 10^-3 L = 6.25 mL
