Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
- A weak acid and its conjugate base.
- A weak base and its conjugate acid.
Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
I believe the answer is letter C
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Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.
<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>
C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)
<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)
<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)
<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)
<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)
<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
