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3 years ago

What information does the formula for an ionic compound provide?

1 answer:

7 0

It provides us with the knowledge of what the cation and anion of the compound are, as well as how many atoms of each are present.

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Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl

Mumz [18]

Answer:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

Explanation:

The products of this reaction are given by:

$K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)$

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

$Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)$

The following steps should be taken:

balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)$

balance oxygen atoms by adding 7 water molecules on the right:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the hydrogen atoms by adding 14 protons on the left:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

Similarly, tin(II) cation becomes tin(IV) cation:

$Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-$

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

$3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-$

Add them together:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)$

Adding the ions spectators:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

7 0

3 years ago

Cuántos gramos de hidróxido de sodio (NaOH) se debe de agregar a un litro de agua para obtener una disolución al 25% en masa

Tamiku [17]

Answer:

= 250 gramos

Explanation:

No ha preguntado sobre peso / peso o peso / volumen.

Para w / v:

Simplemente tome el 25% de 1000 y luego agréguelo en 1L de agua

La masa de NaOH necesaria es de 250 g.

Para w / w:

250 gramos de NaOH y 750 g de agua, hará que la solución sea 1L.

5 0

3 years ago

Lets say you were doing an experiment on two different chemicals in two different test tubes. One chemical is giving off oxygen

boyakko [2]

Answer:

Tests for gases

Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.

Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.

Oxygen. A glowing wooden splint relights in a test tube of oxygen.

hope it will help

5 0

2 years ago

The equation for the chemical reaction shown below is not balanced. What number should replace the question mark to balance this

malfutka [58]

B. the number 3.
there are 2Al's on both sides and 6 Cl's on the right side so to balance it, you multiply Cl2 by 3 to get 6 Cl's.

6 0

3 years ago

What is an echo? (1 point)

zaharov [31]

Answer:

The reflection of sound waves

Explanation:

An echo is a reflection of sound that bounces of of one thing to another.

4 0

3 years ago

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