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Naily [24]
3 years ago
10

A solution is prepared at that is initially in nitrous acid , a weak acid with , and in sodium nitrite . Calculate the pH of the

solution. Round your answer to decimal places.
Chemistry
1 answer:
bija089 [108]3 years ago
7 0

Answer:

4.07

Explanation:

There is some info missing. I think this is the original question.

<em>A solution is prepared at 25 °C that is initially 0.057 M in nitrous acid (HNO₂), a weak acid with Ka = 4.5 × 10⁻⁴, and 0.30 M in sodium nitrite (NaNO₂). Calculate the pH of the solution. Round your answer to 2 decimal places.</em>

<em />

Nitrous acid is a weak acid and nitrite (coming from sodium nitrite) is its  conjugate base. Together, the form a buffer system. We can calculate its pH using the Henderson-Hasselbach equation.

pH = pKa + log [base]/[acid]

pH = -log 4.5 × 10⁻⁴ + log 0.30/0.057

pH = 4.07

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ANSWER PLEASE!!
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3 years ago
How many molecules of NH3 are produced from 4.72x10 negative 4 power g of H2?
Delvig [45]
<span>9.40x10^19 molecules.
   The balanced equation for ammonia is:
 N2 + 3H2 ==> 2NH3
   So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
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7 0
3 years ago
Electromagnetic radiation with _______ wavelengths is called ultraviolet light
andre [41]

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10nm

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4 0
4 years ago
Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the
Oksanka [162]

<u>Answer:</u> The \Delta G for the reaction is -806.86 kJ

<u>Explanation:</u>

We are given:

\Delta H^o_{rxn}=-803kJ=-803000J      (Conversion factor:  1 kJ = 1000)

\Delta S^o_{rxn}=-4.05J/K

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}

Putting values in above equation, we get:

\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J

For the given chemical equation:

CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)

The expression for K_c is given as:

K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}

We are given:

[H_2O]=6.41M

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[O_2]=9.27M

Putting values in above equation, we get:

K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}

K_c=0.126

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_c

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = 8.314J/K mol

T = Temperature = 293 K

K_c = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))

\Delta G=-806859.46J=-806.86kJ

Hence, the \Delta G for the reaction is -806.86 kJ

8 0
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