The term that best described a 10 gram of KClO3 per 100 grams of water solution at 30 degree Celcius is Saturated. The solubility chart is needed for this work. If the solubility chart is drawn for KClO3, it will be observed that the proportion of KClO3 that is needed to dissolve in 100ml of water to make the solution saturated is 10 grams at 30 degree Celcius.
Answer:
6.9 ml of concentrate
Explanation:
100 ml of .1 M will require .01 moles
from a 1.45 M solution, .01 mole would be
.01 mole / ( 1.45 mole / liter) = 6.9 ml of the concentrate then dilute to 100 ml
Answer:
4.5 g/L.
Explanation:
- To solve this problem, we must mention Henry's law.
- Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
- It can be expressed as: P = KS,
P is the partial pressure of the gas above the solution.
K is the Henry's law constant,
S is the solubility of the gas.
- At two different pressures, we have two different solubilities of the gas.
<em>∴ P₁S₂ = P₂S₁.</em>
P₁ = 525.0 kPa & S₁ = 10.5 g/L.
P₂ = 225.0 kPa & S₂ = ??? g/L.
∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.
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Water Is A Liquid And Gravity Pulls Objects Up
