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!
Answer:
Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ
2. Mass of O₂
A) the average global temp. Would decrease
Answer:
See explanation
Explanation:
The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-
The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.
