Answer:
Explanation:
El peso equivalente del metal se calcula por regla de tres simple:
Answer:
El peso equivalente del metal es 1/20 g
Explanation:
Una muestra contiene 160 g de bromo y 2 g de hidrógeno. Otra muestra de bromuro metálico contiene 4 g de bromuro y 1 g de metal. Encuentra el peso equivalente del metal
Aquí tenemos la primera muestra que tiene la relación de bromo a hidrógeno de 160 g de bromo a 2 g de hidrógeno.
Eso es 160 g de bromo está presente con 2 g de hidrógeno
Dividiendo la combinación anterior por 160 para encontrar la cantidad de hidrógeno que se combina con 1 g de bromo, tenemos
(160 g) / 160 de bromo está presente con (2 g) / 160 de hidrógeno
1 g de bromo está presente con 1/80 g de hidrógeno
Por lo tanto, 4 g de bromo están presentes con 4/80 g de hidrógeno o tenemos;
4 g de bromo están presentes con 1/20 g de hidrógeno
Por lo tanto, el peso equivalente del metal = 1/20 g.
(2) They differ in their properties, only.
(3) They differ in their molecular structure and properties.
(4) They do not differ in their molecular structure or properties.
Answer: Option (3) is the correct answer.
Explanation:
When an element exists in two or more different physical forms then they are known as allotropes.
For example, diamond and graphite are both allotropes of carbon.
In graphite, carbon atoms are joined together in sheets of six sided lattice. Whereas in diamond, carbon atoms are joined together in four cornered lattice.
Therefore, as these allotropes are made up of same element so they have similar chemical properties but different physical properties like melting point, boiling point etc.
Also, both of them have different molecular structure.
Thus, we can conclude that the statement the differ in their molecular structure and properties correctly describes diamond and graphite.
B) KBr
C) CH3CH2OH
D) HCl
E) C6H6
How is this determined?
Using the periodic table, determine which material is most likely to be a good insulator.
the correct answer is 4. sulfur
Answer:
Empirical formula
Explanation:
Castle learning
(2) density and specific heat capacity
(3) malleability and thermal conductivity
(4) solubility and molecular polarity
Answer:
The correct option is: (4) solubility and molecular polarity
Explanation:
Chromatography is an analytical technique which is used for the separation of various components in a given mixture.
In this process, the mixture to be separated is first dissolved in the mobile phase, which carries it through the stationary phase, thus leading to separation of the various components.
This differential separation is the result of the differential migration which is based upon the solubility and polarity of the individual components of of the given mixture.