The answer is A. Solids only
Answer:
C
Explanation:
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Answer:
d) 2.7 mol
Explanation:
limit reagent is H2:
∴ Mw H2 = 2.016 g/mol
∴ Mw N2 = 28.0134 g/mol
⇒ moles NH3 = (4.0 moles H2)×(2 mol NH3/3mol H2)
⇒ moles NH3 = 2.666 mol
⇒ moles NH3 ≅ 2.7 mol
Answer:
2.94
Explanation:
There is some info missing. I think this is the original question.
<em>A solution is prepared at 25 °C that is initially 0.38 M in chloroacetic acid (HCH₂ClCO₂), a weak acid with Ka= 1.3 x 10⁻³, and 0.44 M in sodium chloroacetate (NaCH₂CICO₂). Calculate the pH of the solution. Round your answer to 2 decimal places.</em>
<em />
We have a buffer system formed by a weak acid (HCH₂ClCO₂) and its conjugate base (CH₂CICO₂⁻ coming from NaCH₂CICO₂). We can calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log [CH₂CICO₂⁻]/[HCH₂ClCO₂]
pH = -log 1.3 x 10⁻³ + log (0.44 M/0.38 M)
pH = 2.94
Convert grams to moles and use Avogadro's number 6.022x10tho the 23rd power. Hope this helps
