Answer:
7.58
Explanation:
Given that
A buffer solution obtained by dissolving 13.0 g of KH₂PO₄ and 26.0 g of Na₂HPO₄ in water and then diluting to 1.00 L.
pKa = 7.21.
To get the PH we are going to use Henderson - Hasselblach equation:
PH = Pka + ㏒ [A/AH]
when the molar mass of Na2HPO4 = 142 g/mol
and A is the conjugate base HPO4-- ions so,
∴[A] = 32g / 142 g/mol
= 0.225 M
and when the molar mass of KH2PO4 = 136 g/mol
and AH is the weak acid H2PO4- ions so,
∴[AH] = 13 g / 136 g/mol
= 0.096 M
and when we have the Pka value of H3PO4 = 7.21
so, by substitution:
∴ PH = 7.21 + ㏒ (0.225 / 0.096)
= 7.58
If a 3.50g sample of the hydrate of copper (II) sulfate is heated to yield 2.10g of anhydrous copper (II) sulfate, the mass percent of water in the hydrate is 40%.
<h3>How to calculate mass percent?</h3>
The mass percent of hydrate in a sample can be calculated by dividing the mass of water in the sample by the mass of the hydrated compound.
According to this question, 3.50g sample of an hydrate of copper (II) sulfate is heated to yield 2.10g of anhydrous copper (II) sulfate. The mass percent of water is calculated as follows:
Mass percent of water = (3.5-2.1)/3.5 × 100
Mass percent of water = 40%
Therefore, if a 3.50g sample of the hydrate of copper (II) sulfate is heated to yield 2.10g of anhydrous copper (II) sulfate, the mass percent of water in the hydrate is 40%.
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Answer:
K= 44.018%; S= 19.957%; O= 36.025%
Explanation:
your welcome
Answer:
false i am pretty sure because dead flowers.
I’m pretty sure it would be:
NH4+Cn