Answer:
Mixtures can be either homogeneous or heterogeneous. A mixture in which its constituents are distributed uniformly is called homogeneous mixture, such as salt in water. A mixture in which its constituents are not distributed uniformly is called heterogeneous mixture, such as sand in water.
Explanation:
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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Explanation:
(1). It is known that in a reaction equation, reactants are placed or written on left hand side and products are written on the right hand side.
For example, 
Hence, in a reaction equation you start with the reactants and end up with the products.
(2). The number of atoms in a reaction will remain the same because according to the law of conservation of mass, mass of reactants will be equal to the mass of products.
Therefore, number of atoms on the reactant side will be equal to the number of atoms on product side.
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
