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Quantitative data is numerical.
Qualitative data is non-numerical.
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MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)
Using the standard enthalpies of formation given in the source below:
(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
Answer:
1.3 × 10²³ Atoms of Mercury
Solution:
Step 1: Calculate Mass of Mercury using following formula,
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume
Putting values,
Mass = 13.55 g.cm⁻³ × 3.2 cm³ ∴ 1 cm³ = 1 cc
Mass = 43.36 g
Step 2: Calculating number of Moles using following formula;
Moles = Mass ÷ M.mass
Putting values,
Moles = 43.36 g ÷ 200.59 g.mol⁻¹
Moles = 0.216 mol
Step 3: Calculating Number of Atoms using following formula;
Number of atoms = Moles × 6.022 ×10²³
Putting value of moles,
Number of Atoms = 0.216 mol × 6.022 × 10²³
Number of Atoms = 1.3 × 10²³ Atoms of Hg
