I guess you could call them that. In chemistry, we call them Metalloids though.
pH of the buffer solution is 1.76.
Chemical dissociation of formic acid in the water:
HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)
The solution of formic acid and formate ions is a buffer.
[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions
[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate
[HCOOH] = 1.45 M - 0.015 M
[HCOOH] = 1.435 M; equilibrium concentration of formic acid
pKa = -logKa
pKa = -log 1.8×10⁻⁴ M
pKa = 3.74
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)
pH = 3.74 + log (0.015 M/1.435 M)
pH = 3.74 - 1.98
pH = 1.76
More about buffer: brainly.com/question/4177791
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Answer:
C
Explanation:
It afffects changes in pressure and temperature not melting and boiling points
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
