The volume of H₃PO₄ : 13.33 ml
<h3>Further explanation</h3>
Given
0.003 M Phosphoric acid-H₃PO₄
40 ml of 0.00150 M Calcium hydroxide-Ca(OH)₂
Required
Volume of H₃PO₄
Solution
Acid-base titration formula
Ma. Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence (amount of H⁺/OH⁻)
H₃PO₄⇒3H⁺ + PO₄³⁻ ⇒ 3 H⁺ = valence = 3
Ca(OH)₂⇒Ca²⁺ + 2OH⁻⇒ 2 OH⁻ = valence = 2
Input the value :
a = H₃PO₄, b = Ca(OH)₂
0.003 x Va x 3 = 0.0015 x 40 x 2
Va = 13.33 ml
Answer: 8.7 grams
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.
According to stoichiometry:
4 moles of aluminium give = 2 moles of 
Thus 0.17 moles of aluminium give=
Mass of 
Thus the mass of is 8.7 grams
Answer:
A
Explanation:
First, let's find the molar mass of CO₂. This is 12 + 2(16) = 44 g/mole.
Now we can write 100g * (1 mole / 44g) = 2.27 mol, or A. Hope this helps!
An: Calculate the molarity of a solution made by adding 120 g of NaOH (40.00 g/mol) to enough water to make 500.0 mL of solution. a) 4.0 M b) 6.0 M c) 1.0 ...
Explanation:
