Answer:
0.8078 Kg
Explanation:
Pressure of water = 0.15 MPa = 1.5 bar
At critical point of water ,temperature = 647 K=374°C
From the ideal gas equation
P×V= m×R×T
Let us assume volume = 1 m^3
1.5 x 105 x 1 = m x 287 x 647
m= 0.8078 kg
the fraction of mass of liquid at 25°C.
We are given that 1 teaspoon is equivalent to 5 mL,
therefore 0.75 teaspoon is:
0.75 teaspoon * (5 mL / 1 teaspoon) = 3.75 mL
So the mass is density times volume:
mass = (12.5 mg/5 ml) * 3.75 mL
<span>mass = 9.375 mg</span>
Answer:
35.8 g
Explanation:
Step 1: Given data
Mass of water: 63.5 g
Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C
Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.
63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl
