Describing a solution as concentrated tells that the solution has a relative large concentration, but it is a qualitative description, not a quantitative one, so this does not tell really how concentrated the solution is. This is, the term concentrated is a kind of vague; it just lets you know that the solution is not very diluted, but, as said initially, that there is a relative large amount (concentration) of solute.

One conclusion, of course, is that the solute is soluble: else the solution were not concentrated.

On the other hand, the terms saturated and supersaturated to define a solution are specific.

A saturated solution has all the solute that certain amount of solvent can contain, at a given temperature. A supersaturated solution has more solute dissolved than the saturated solution at the same temperature; superstaturation is a very unstable condition.

From above, there is no way that you can conclude whether a solution is supersaturated or not from the statement that a solution is concentrated, so the answer is none of the above.

5 0

3 years ago

Please help me with the questions?

weqwewe [10]

Answer:

1. Forecast

2. Cold front

3. Shelf Cloud

4. Barometer

5. Pressure

6 Prevailing Westerlies

7. Thermometer

8. Weather

6 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$