Answer:
Being a weak acid and a strong base, where it is diluted in a neutral medium such as water, the basic medium predominates, almost alkaline pH.
Explanation:
The acidity of the solution, being weak, means that its pH is not so low, therefore it will be easier to reach the values of 7 or 7 where alkalinity or basity is indicated.
First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
Sodium , bromine zinc magnesium sulphur nitrogen potassium oxygen lead
Answer is: mass of unused sulfur is 5.87 grams.
Balanced chemical reaction: C + 2S → CS₂.
m(C) = 12.0 g; mass of carbon.
m(S) = 70.0 g; mass of sulfur.
n(C) = m(C) ÷ M(C).
n(C) = 12 g ÷ 12 g/mol.
n(C) = 1 mol; amount of substance.
n(S) = m(S) ÷ M(S).
n(S) = 70 g ÷ 32.065 g/mol.
n(S) = 2.183 mol.
From chemical reaction: n(C) : n₁(S) = 1 : 2.
n₁(S) = 1 mol · 2 = 2 mol.
Δn(S) = n(S) - n₁(S).
Δn(S) = 2.183 mol - 2 mol.
Δn(S) = 0.183 mol; amount of unused sulfur.
Δm(S) = 0.183 mol · 32.065 g/mol.
Δm(S) = 5.87 g.
