Answer:
5 L.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 10 L
Initial pressure (P1) = 2.5 atm
Final pressure (P2) = 5 atm
Final volume (V2) =.?
Since the temperature is constant, we shall apply the Boyle's law equation to determine the new volume of the gas. This can be obtained as follow:
P1V1 = P2V2
2.5 × 10 = 5 × V2
25 = 5 × V2
Divide both side by 5
V2 =25/5
V2 = 5 L
Thus, the new volume of the gas is 5 L
<span>As we know that
1 cu cm H2O = 1 mL H2O = 1g H2O
now
Heat of fusion of water = 79.8 cal/g
and
Heat of vaporization of water = 540 cal/g
Atomic weight of water : H=1 O=16 H2O=18
now by calculating and putting values
65.5gH2O x 79.8cal/gH2O x 1gH2O/540cal = 9.68g H2O (steam)
9.68gH2O x 1molH2O/18gH2O x 22.4LH2O/1molH2O = 12.0 L H2O
hope it helps</span>
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The relation between density and mass and volume is
the dose required is 2.5 tsp
each tsp contain 5mL
So dose required in mL = 2.5 X 5 = 12.5 mL
the mass will be calculated using following formula
The mass of dose in grams will be 15.38 g
It increases as temperature rises.