Answer:
Number of moles nitric acid in the cylinder is 400.539g/mol.
Explanation:
From the given,
Weight of empty gas cylinder ![W_{1}= 90.0 lb= 40823.3 gramsWeight of full cylinder[tex]W_{2} = 116.5 lb= 52843.511 gramsThe critical temperature = 287 KThe critical pressure 54.3 LMolar mass of nitric acid = [tex]M_{NO}](https://tex.z-dn.net/?f=W_%7B1%7D%3D%2090.0%20lb%3D%2040823.3%20grams%3C%2Fp%3E%3Cp%3EWeight%20of%20full%20cylinder%5Btex%5DW_%7B2%7D%20%3D%20116.5%20lb%3D%2052843.511%20grams%3C%2Fp%3E%3Cp%3EThe%20critical%20temperature%20%3D%20287%20K%3C%2Fp%3E%3Cp%3EThe%20critical%20pressure%2054.3%20L%3C%2Fp%3E%3Cp%3EMolar%20mass%20of%20nitric%20acid%20%3D%20%5Btex%5DM_%7BNO%7D)
= 30.01 g/mol
Number of moles nitric acid = 
=?
The mass of nitric acid in the cylinder = 
Number of moles of nitric acid =
Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.
Answer:
Ensure that the glassware is designed for heating
Check that there are no cracks in the glassware
Inspect the hot plate for frayed cords
Explanation:
All except measuring the height and width of the glassware could cause hazards within the lab.
Answer:
0.036 M
Explanation:
To do this, let's mark the dye as D and bleach as B.
We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.
The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.
To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:
moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles
moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles
Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:
moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles
These are the moles presents of B after the reaction has been made. The concentration of the same will be:
[B] = 3.597x10⁻⁴ / 0.010
[B] = 0.0357
With 2 SF it would be:
[B] = 3.6x10⁻² M
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Answer:
11482 ppt of Li
Explanation:
The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:
<em>Moles Mg²⁺:</em>
0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA
<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>
0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium
That means mass of lithium is (Molar mass Li=6.941g/mol):
4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:
0.00287g * (1000000μg / g) = 2870μg of Li
As ppt is μg of solute / Liter of solution, ppt of the solution is:
2870μg of Li / 0.250L =
<h3>11482 ppt of Li</h3>
