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Sophie [7]
3 years ago
14

Please help answer these

Biology
1 answer:
san4es73 [151]3 years ago
7 0

Answers:

  1. C3H8 + 5 O2 = 3 CO2 + 4 H2O
  2. Al2(SO3)3 + 6 NaOH = 3 Na2SO3 + 2 Al(OH)3
  3. 4 Al2O3 + 9 Fe = 3 Fe3O4 + 8 Al
  4. 2 KClO3 = 2 KCl + 3 O2
  5. NH4NO3 = N2O + 2 H2O
  6. 2 NaHCO3 = Na2O + 2 CO2 + H2O
  7. P4O10 + 6 H2O = 4 H3PO4
  8. 2 Al + 3 H2SO4 = Al2(SO4)3 + 3 H2
  9. Be2C + 4 H2O = 2 Be(OH)2 + CH4
  10. S + 6 HNO3 = H2SO4 + 6 NO2 + 2 H2O
  11. 2 NH3 + 3 CuO = 3 Cu + N2 + 3 H2O
  12. 3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2 NO + 4 H2O
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5 0
1 year ago
A certain buffer is made by mixing a weak acid HA and its conjugate base A–. When HA is present at a concentration of 0.5 mM and
MAVERICK [17]

Answer:The ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02 is 0.69

Explanation:

The dissociation constant for formic acid = K_a=1.8\times 10^{-4}

Concentration of HA= 0.5 mM

Concentration of A^-= 0.1 mM

pH = 6.16

First we have to calculate the value of K_a.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[A^-}{[HA]}

Now put all the given values in this expression, we get:

6.16=pK_a+\log (\frac{[0.1]}{0.5})

pK_a=6.86

To calculate the ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02.

Using Henderson Hesselbach equation :

7.02=6.86+\log \frac{[A^-]}{[HA]}

7.02=6.86+\log \frac{[A^-]}{[HA]}

\frac{[A^-]}{[HA]}=1.44

\frac{[HA]}{[A^-]}=0.69

Thus the ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02 is 0.69

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