Answer:
Please see the attached picture for the complete answer.
Explanation:
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!
Your answer would be a change in odor! Hope this helps! ;D
Molar mass :
Cl₂ = 71.0 g/mol Na = 23.0 g/mol
<span>2 Na + Cl</span>₂<span> = 2 NaCl
</span>
2 x 23 g Na -------> 71.0 g Cl₂
96.6 g Na ----------> ?
Mass Cl₂ = ( 96.6 x 71.0 ) / ( 2 x 23 )
Mass Cl₂ = 6858.6 / 46
= 149.1 g of Cl₂
hope this helps!
