Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
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Answer:
Less than 7 ph value are acidic. Greater than 7 ph value are basic.
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.
where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
Learn more: brainly.com/question/2292439
Answer : The molecular weight of a substance is 157.3 g/mol
Explanation :
As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.
Mass of solute = 7 g
Mass of solution = 100 g
Mass of solvent = 100 - 7 = 93 g
Formula used :
where,
= change in freezing point
= temperature of pure water = 
= temperature of solution = 
= freezing point constant of water = 
m = molality
Now put all the given values in this formula, we get
Therefore, the molecular weight of a substance is 157.3 g/mol
Answer:
Hexaflourine Pentaiodide?
Explanation:
f = flourine (6 = hexa)
i = iodine (5 = penta) + ide
