Explain preparation of alkanes C--C from industrial source for industrial preparation of alkane
1 answer:
You might be interested in
Data:
m (<span>Sample Mass) = ?
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (
)
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass
= 401.18 + 70.906 = 472.086 ≈ 472.09<span> amu or 472.09 g/mol
</span>
Formula:
Solving:
Answer:
By approximation would be letter D) <span>337.2 g</span>
m (<span>Sample Mass) = ?
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass
</span>
Formula:
Solving:
Answer:
By approximation would be letter D) <span>337.2 g</span>