<span>3.75 gallons need to be drained and replaced with bleach.
Change the problem to "What amounts of a 100% solution and a 4% solution is needed to make 90 gallons of a 8% solution?" Given that, we'll use the following values.
x = amount of 4% solution.
90-x = amount of a 100% solution.
The equation to solve then becomes.
0.04 x + (90-x) = 0.08 * 90
0.04 x + 90 - x = 7.2
Add x to both sides
0.04x + 90 = 7.2 + x
Subtract 0.04x from both sides
90 = 7.2 + 0.96x
Subtract 7.2 from both sides
82.8 = 0.96x
Divide both sides by 0.96
86.25 = x
So you now know that you need 86.25 gallons of the original 4% solution and (90-86.25) = 3.75 gallons of the bleach to make the desired 90 gallons.
So simply drain 3.75 gallons from the tank and replace with bleach.</span>
Answer:
Ksp = 1.64 ×10⁻⁵
Explanation:
Given data:
Mass of Ag₂SO₄ = 5.0 g
Volume of water = 1.0 L
Solubility product = Ksp = ?
Solution:
First of all we will calculate the molarity.
Molarity = Number of moles of solute / Volume of solution in L
Number of moles = mass/ molar mass
Number of moles = 5.0 g/311.74 g/mol
Number of moles = 0.016 mol
Molarity = 0.016 mol / 1 L
Molarity = 0.016 M
Dissociation equation:
Ag₂SO₄ ⇔ 2Ag⁺ + SO₄²⁻
Ksp = [Ag⁺]² [SO₄²⁻]
Ksp = [0.032]²[0.016]
Ksp = 1.64 ×10⁻⁵
Answer:
1.096g
Explanation:
You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:
Hydrogen: 1.008g/mol
Fluorine: 18.99g/mol
Sodium: 22.98g/mol
Next, find the composition percentage of NaF
22.98 + 18.99 = 41.97
Fluorine is 18.99/41.97 =45.25%
Sodium is 100-45.25 = 54.75%
Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g
Find comp. percentage of HF
18.99+1.008 = 19.998; H/total F/total
Hydrogen 5.041%
Fluorine 94.959%
Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g
Moles of calcium metal used = 100/40.1=2.5
Moles of HBr need to react = 5 moles
As the molar ratio is 1 is to 2 among them
so
Moles=molarity x volume
5=2.25 x volume
volume=2.22 litres of HBr required for this reaction
ANSWER IS 2.22 LITRES