Answer:
- About 18 g of NH₄Cl will precipitate.
Explanation:
The <em>table G</em> is the graph of the solubility curves for several solutes which is attached.
The second picture identifies the solubilities for the NH₄Cl at 50ºC and 10ºC.
The solubility of NH₄Cl at 50ºC is about 52 g/ 100 g of water.
The solubility of NH₄Cl at 10ºC is about 34 g / 100 g of water.
Then, at 50ºC 100 g of water saturated with NH₄Cl contains about 52 g of NH₄Cl and 100 g of water saturated with NH₄Cl contains 34 g of NH₄Cl.
The difference, 52g - 34 g of NH₄Cl shall precipitate:
52 g - 34 g = 18g ← answer
Answer:
ethanol
Explanation:
CO2 is carbon dioxide
C2H5OH is ethanol and we know that because that is what is left after taking out the CO2
The answer is b, because if it gets colder then means more heat is exiting than it is entering.
Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
Hey there!:
Molar mass H3PO4 = <span>97.9952 g/mol
Atomic Masses :
H = </span><span>1.00794 a.m.u
</span>P = <span>30.973762 a.m.u
</span>O = 15.9994 a.m.u<span>
H % = [ ( 1.00794 * 3 ) / </span> 97.9952 ] * 100
H% = <span>3.0857 %
P % = [ ( </span>30.973762 * 1 ) / 97.9952 ] * 100
P% = <span>31.6074 %
O % = [ ( </span>15.9994 * 4 ) / 97.9952 ] * 100
O% = <span>65.3069 %
Hope this helps!</span>
