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How many moles of ammonium ions are in 125 mL of 1.40 M NH4NO3 solution? ________ moles (give answer with correct sig figs in un

Sholpan [36]

The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

Volume = 125 mL = 125 / 1000 = 0.125 L

Molarity = 1.40 M

Mole = Molarity x Volume

Mole of NH₄NO₃ = 1.40 × 0.125

Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:

NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)

From the balanced equation above,

1 mole of NH₄NO₃ contains 1 mole of NH₄⁺

Therefore,

0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺

Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

Learn more: brainly.com/question/25469095

3 0

2 years ago

Help me please!!! HELP ME

Neporo4naja [7]

Answer:

Its a

Explanation:

3 0

2 years ago

If the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in

anygoal [31]

Answer:

d. supersaturated.

Explanation:

A solution naturally contains a solute and a solvent. The solute is the solid substance that dissolves in the solvent, which is usually a liquid substance. A solution has a maximum amount of solute that can dissolve in its constituent solvent.

However, when the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in the solution at that temperature, the solution is said to be SUPERSATURATED. This means that the solution contains more than the maximum amount of solute.

5 0

3 years ago

In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq) Sn2 (aq)→H2

Sav [38]

Answer:

SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O

Explanation:

At first calculate the oxidation state of that element which undergoes oxidation as well as reduction.

for SO₄²⁻ the oxidation state of sulphur is +6 and H₂SO₃ the oxidation state of sulphur is +4

So balance equation is

(Reduction) SO₄²⁻ + 4H⁺+ 2e⁻ → H₂SO₃ + H₂O.........................................(1)

(oxidation) Sn²⁺ → Sn⁴⁺ + 2e⁻ .............................................................(2)

Adding equation 1 & 2

we get

SO₄²⁻(aq) +Sn²⁺(aq) +4H⁺ → H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O

6 0

3 years ago

Calculate the pH of a buffer solution prepared by mixing 60.0 mL of 1.00 M lactic acid and 25.0 mL of 1.00 M sodium lactate.

marshall27 [118]

This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:

pH = pKa + log[(salt/acid]

Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).

Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L

Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M

Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48

4 0

3 years ago