Topsoil is considered to be most fertile when it has a

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

Ipatiy [6.2K]

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

$Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)$

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the net ionic equation is written above.

8 0

3 years ago

2. Which one of the following chemical equations is balanced? A. 2NaCl + H2SO4 → HCL + NaSO4 B. NH3 + H2O → 2NH4OH C. 2Na + S →

Lady_Fox [76]

D. KOH + H₂SO₄ → KHSO₄ + H₂O

6 0

3 years ago

Read 2 more answers

7.66 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water,

White raven [17]

Explanation:

(a) potassium oxide with water

$K_2O(s)+H_2O(l)\rightarrow 2KOH(aq)$

According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.

(b) diphosphorus trioxide with water

$P_2O_3(s)+3H_2O(l)\rightarrow 2H_3PO_3(aq)$

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.

(c) chromium(III) oxide with dilute hydrochloric acid,

$Cr_2O_3(s)+6HCl(aq)\rightarrow 2CrCl_3(aq)+3H_2O(l)$

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.

(d) selenium dioxide with aqueous potassium hydroxide

$SeO_2 (s)+2KOH (aq)\rightarrow K_2SeO_3(aq)+H_2O(l)$

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.

6 0

3 years ago

The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0

3 years ago

How many moles are in 29.5 grams of Ax?

JulsSmile [24]

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

Ar = 39.95 g/mole

The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

$= \frac{29.5}{39.95} \\\\= 0.74 \ mole$

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.

"Your question seems to be missing the correct symbol for the element"

Argon = Ar

Learn more here:brainly.com/question/4628363

6 0

2 years ago