Answer: He(g)
Explanation: I had the same question and I got the answer right
Answer:
HgBr₂.
Explanation:
- To determine the empirical formula of the mercury bromide that is produced, we should calculate the no. of moles of Hg (1.750 g) and Br (1.394 g) used to prepare the mercury bromide.
no. of moles of Hg = mass/atomic mass = (1.75 g)/(200.59 g/mol) = 0.00872 mol.
no. of moles of Br = mass/atomic mass = (1.394 g)/(79.904 g/mol) = 0.01744 mol.
- We divide by the lowest no. of moles (0.00872) to get the mole ratio:
The mole ratio of (Br: Hg) is: (2: 1).
<em>So, the empirical formula is: HgBr₂.</em>
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Answer:
Mass = 239.62 g
Explanation:
Molality is given as,
m = Moles of Solute / Kg of Solvent --- (1)
Data Given;
Molality = 1.5 m
Mass of Solvent = 750 g = 0.75 Kg
Solving eq. 1 for Moles,
Moles = Molality × Mass of Solvent
Putting values,
Moles = 1.5 mol.Kg⁻¹ × 0.75 Kg
Moles = 1.125 mol
Calculating mass from moles,
Moles = Mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 1.125 mol × 213 g.mol⁻¹
Mass = 239.62 g