How many moles is 118 grams of Argon
1 answer:
Answer:
Explanation:
To convert from grams to moles we must use the molar mass, which is found on the Periodic Table.
- Argon: 39.9 g/mol
Use the molar mass as a ratio.
Multiply by the given number of grams: 118
Flip the fraction so the grams of argon will cancel each other out.
The original measurement of grams had 3 significant figures, so our answer must have 3 sig figs.
For the number we calculated, that is the hundredth place. The 7 in the thousandth place tells us to round the 5 to a 6.
There are about <u>2.96 moles of Argon</u> in 118 grams.
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Answer:
1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.
Explanation:
The reaction that takes place is:
Mn⁺² + S⁻² ⇄ MnS(s)
ksp = [Mn⁺²] [S⁻²]
If the pksp of MnS is 13.500, then the ksp is:
From the problem we know that [S⁻²] = 0.0900 M
We use the ksp to calculate [Mn⁺²]:
3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M
[Mn⁺²] = 3.514 * 10⁻¹³ M.
Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:
3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.
Answer: ✅
The main difference between boiling point and melting point is that the melting point is defined as the temperature at which solid and liquid phases are in equilibrium, whereas the boiling point is the temperature at which the vapour pressure of a liquid is equal to the external pressure.
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