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>
<u><em /></u>
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>
<em />
I hope it helps!
Answer:
tttttttttttthankkssss!!!!!
Nothing at all happens because pure water cannot conduct electricity
Reaction: 2K₍s₎ + 2H₂O₍l₎ → 2KOH₍aq₎ + H₂₍g₎.
K - potassium.
H₂O - water.
KOH - potassium-hydroxide.
H₂ - hydrogen.
s - solid phase.
l - liquid.
aq - disolves in water.
g - gas.
Reaction is exothermal (release of energy) and potassium burns a purple flame. H<span>ydrogen released during the reaction reacts with </span>oxygen<span> and ignites.</span><span>
</span>
