MgSO4 + Na3PO4 = Na2SO4 + Mg3(PO4)2
Answer: The products of Na3PO4 + MgSO4 are Na2SO4 + Mg3(PO4)2
Explanation:
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
Answer:
The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams
Explanation:
The chemical equation for the reaction is
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas
The molar mass of CO₂ = 44.01 g/mol
There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles
However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆
and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of
mass of 1 mole C₆H₁₂O₆ = 180.2 g
mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams
Mass of glucose produced = 90.1 grams
Phenolphthalein
In acid it is colourless and in water also colourless
