I'm pretty sure its D. or A. so choose which one u think
Answer:
[HAc] = 0.05M
[Ac⁻] = 0.20M
Explanation:
The Henderson-Hasselbalch formula for the acetic acid buffer is:
pH = pka + log₁₀ [Ac⁻] / [HAc]
Replacing:
5.36 = 4.76 + log₁₀ [Ac⁻] / [HAc]
3.981 = [Ac⁻] / [HAc] <em>(1)</em>
Also, as total concentration of buffer is 0.25M it is possible to write:
0.25M = [Ac⁻] + [HAc] <em>(2)</em>
Replacing (2) in (1)
3.981 = 0.25M - [HAc] / [HAc]
3.981 [HAc] = 0.25M - [HAc]
4.981 [HAc] = 0.25M
<em>[HAc] = 0.05M</em>
Replacing this value in (2):
0.25M = [Ac⁻] + 0.05M
<em>[Ac⁻] = 0.20M</em>
I hope it helps!
Answer:
84.4g of AgCl
Explanation:
Based on the reaction:
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
<em>2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca(NO₃)₂</em>
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100g of each reactant are:
AgNO₃: 100g × (1mol / 169.87g) = 0.589 moles
CaCl₂: 100g × (1mol / 110.98g) = 0.901 moles
For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901×2 = <em>1.802 moles of AgNO₃. </em>As there are just 0.589moles, <em>AgNO₃ is limitng reactant</em>
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0.589 moles of AgNO₃ produce:
0.589 moles × ( 2 moles AgCl / 2 moles AgNO₃) =
<em>0.589 moles of AgC</em>l. In mass:
0.589 moles of AgCl × (143.32g / mol) =<em> 84.4g of AgCl</em>
It is D: a spinning cloud of dust...