Answer:
9.9652g of water
Explanation:
The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:
n = PV / RT
Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)
Replacing:
n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K
n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:
1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>
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As the initial mass of water was 10g, the mass of water that remains in liquid phase is:
10g - 0.0348g = <em>9.9652g of water</em>
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I hope it helps!
Sodium, Atomic mass: 22.989769 g
You can see in a periodic table
Scale and measuring tape.
To determine the mass and volume, you can find the density. The block will float if its density is less than that of water.
Answer:
0.85 Molar Na2O
Explanation:
Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).
(10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.
Molar is a measure of concentration. It is defined as moles/liter. A 1 M solution contains 1 mole of solute per liter of solvent. [200 ml water = 0.2 Liters water.]
In this case, we have 0.161 moles Na2O in 0.200 L of solvent.
(0.161 moles Na2O)/(0.200 L) = 0.85 Molar Na2O
Answer:
48.5 mL , just did it on edge
