Answer is <span>8.422
L.

The solving method as the attached pdf.</span>

**Answer:**

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**Explanation:**

**Answer:**

To produce 1 mole of N2O required 1 mole of NH4NO3 . Hence, 60 grams of NH4NO3 are required to produce 33.0 g N2O

**Explanation:**

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Volume required for neutralization V will be:

V * 0.2125 M HCl = 25 mL * 0.17 M

V = 20 ml

**First part:**

When 10 mL is added we can apply Henderson equation to get the result, so:

The pH will be of basic buffer

pOH = pKb + log(salt/base)

or pOH = 4.19 + log (0.2125*10 / 25*0.17 - 10*0.2125 )

pOH = 4.19 and pH = 14 - 4.19 = 9.81

**Second part:**

When 20 ml is added, there is only salt formed

The pH will be salt of strong acid and weak base

So pH = 7 - 0.5 pKb - 0.5 log C

where C is the concentration of the salt formed so:

pH = 7 - (0.5*4.19) - (0.5 log (25*0.17) / (25+20))

= 5.42

**Third part:**

When 30 ml of the acid has been added,

The pH will be of the remaining strong acid

pH = - log (0.2125*10 / 25 + 30 )

= 1.326