The moles of ammonia in 1.20 x10^25 molecules of ammonia is calculated as follows
by use of Avogadro constant
1moles = 6.02 x10^23
what about 1.20 x10^25 moles
by use of cross multiplication
= 1 mole x ( 1.20 x10^25) /( 6.02 x10^23) = 19.93 moles
Living organisms contain relatively large amounts of oxygen, carbon, hydrogen, nitrogen, and sulfur (thesefive elements are known as the bulk elements), along with sodium, magnesium, potassium, calcium, chlorine, and phosphorus (these six elements are known as macrominerals).
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
A hydrate is a chemical that has water molecules loosely bonded to it. The water molecules are not ... You will be using the hydrate CuSO4 . ?H2O. Sample Calculation-. An empty crucible has a mass of 12.770 grams.
Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)
