When acids react with water, H ions are released which then combine with water molecules to form H₃O⁺
Answer:
Explanation:
1) Chemical formula of sodium carbonate: <em>Na₂CO₃</em>
2) Ratio of carbon atoms:
- The number of atoms of C in the unit formula Na₂CO₃ is the subscript for the atom, which is 1 (since it is not written).
Hence, the ratio is 1 C atom / 1 Na₂CO₃ unit formula.
This is, there is 1 atom of carbon per each unit formula of sodium carbonate.
3) Calculate the number of moles in 1.773 × 10⁷ carbon atoms
- Divide by Avogadro's number: 6.022 × 10²³ atoms / mol
- number C moles = 1.773 × 10⁷ atoms / (6.022 × 10²³ atoms/mol)
- number C moles = 2.941 × 10⁻¹⁷ mol
Since, the ratio is 1: 1, the number of moles of sodium carbonate is the same number of moles of carbon atoms.
(sample g/1) X (1 mole/40.078(MW of Ca)) = moles of sample (moles of sample)(6.022 x 10^23( no of atoms)/ 1 mole) = # of atoms in a 120 g sample of calcium Avogadro's number=6.022x 10^23 atoms in 1 mole
M(P)=3.72 g
M(P)=31 g/mol
m(Cl)=21.28 g
M(Cl)=35.5 g/mol
n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol
n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol
P : Cl = 0.12 : 0.60 = 1 : 5
PCl₅ - is the empirical formula
Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
