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tiny-mole [99]
2 years ago
12

NH4Cl + AgNO3 → AgCl + NH4NO3

Chemistry
1 answer:
iVinArrow [24]2 years ago
7 0
<h3><u>A</u><u> </u><u>N</u><u> </u><u>S</u><u> </u><u>W</u><u> </u><u>E</u><u> </u><u>R</u><u> </u><u>:</u><u> </u><u>–</u><u> </u></h3>

  • It's a balanced equation!
  • This is a precipitation reaction.
  • AgCl is the formed precipitate.
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sammy [17]

Answer:

Just make it sound right please it don't even add up what was you even thinking like come on now...lol

Explanation:

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7 0
3 years ago
A chemist must prepare 900.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. She will do this in three steps: Fill a
Ierofanga [76]

Answer:

28.58 g of NaOH

Explanation:

The question is incomplete. The missing part is:

<em>"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"</em>

To do this, we need to know how much of the base we have to weight to prepare this solution.

First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:

NaOH -------> Na⁺ + OH⁻

As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.

This can be done with the following expression:

14 = pH + pOH

and pOH = -log[OH⁻]

So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:

pOH = 14 - 13.9 = 0.10

[OH⁻] = 10⁽⁻⁰°¹⁰⁾

[OH⁻] = 0.794 M

Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:

n = M * V

n = 0.794 * 0.9

n = 0.7146 moles

Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:

m = 39.997 * 0.7146

<h2>m = 28.58 g</h2>
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