En este caso, y basado en las normas IUPAC para la escritura de las fórmulas moleculares, es necesario primero escribir el catión a la izquerda, seguido del anión a la derecha, tal y como se muestra en los siguientes ejemplos, recordando que el catión es el ion cargado positivamente y el anión, negativamente:

$K^+Cl^-\\\\Ag_2^+(SO_4)^{2-}$

Los cuales son cloruro de potasio y sulfato de plata respectivamente. También es necesario tener en cuenta que los metales tienden a ser cationes por su capacidad de perder electrones, mientras que los no metales a ganarlos y por ende resultar como aniones.

¡Un gusto ayudarte!

6 0

3 years ago

pls this is my second time asking these questions, i'm sorry, but i really need help!!! 1. what happens to atoms when there is a

dusya [7]

Answer:

the atoms of the original substances gain, lose and even share their very own electrons. write the symbols of the elements that form the compound. Write down the valency, and lastly go over valencies. This is all I got so far. I hope this helps

Explanation:

8 0

3 years ago

In a reaction vessel, 17.6 g of solid chromium(III) oxide, Cr2O3, was allowed to react with excess carbon tetrachloride in the f

notsponge [240]

Answer:

72.53% is the yield of CrCl3

Explanation:

Given

Reaction:

Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)

CCl4 is in excess and 17.6g Cr2O3 present

The reaction yields 26.6g of CrCl3

To Find:

% yields of the reaction

Also given

Molar mass of CrCl3 = 158.35g/mol

Molar mass of Cr2O3 = 152.00 g/mol

By the stoichiometry of the reaction

1 mole of Cr2O3 gives 2 moles of CrCl3

1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3

= 1 52 g of Cr2O3 gives 316.70 g of CrCl3

17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3

= 36.67 g CrCl3

but actual yield is only 26.6g

so % yield is (26.6 ÷÷ 36.67) × 100

= 72.53% is the yield of CrCl3

8 0

3 years ago