We are asked to find the value of ΔG°rxn from the equilibrium concentrations of the reactants and products. We can use the following formula:
ΔG°rxn = -RTlnK
The value of R = 8.314 J/Kmol, T = 298.15 K and we are given the equilibrium constant Keq = 2.82.
The question provides equilibrium concentrations and then asks to find ΔG°rxn when more of a product is added to the reaction mixture. However, you are asked to find ΔG after the reaction has settled down and reached equilibrium once more. Therefore, we can simply use Keq = 2.82 still and solve for ΔG.
ΔG°rxn = -(8.314 J/Kmol)(298.15 K)(ln(2.82))
ΔG°rxn = -2570 J/mol
ΔG°rxn = -2.57 kJ/mol
Under equilibrium conditions at standard temperature and pressures, the value of ΔG°rxn = -2.57 kJ/mol.
Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol
9.
(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3
10.
Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:
NaHCO3 + H{+} = Na{+} + H2O + CO2
(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory
11.
n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2
12.
(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%
Explanation:
Answer:
Yes
Explanation:
They continue to split and grow and split again until the organism that is carrying them dies.
Sorry I don't really know how to explain:(
Answer:
2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.
Explanation:
<em>∵ pH = - log[H₃O⁺]</em>
∴ 4.6 = - log[H₃O⁺].
∴ log[H₃O⁺] = - 4.6.
∴ [H₃O⁺] = 2.51 x 10⁻⁵.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 2.51 x 10⁻⁵ M.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.51 x 10⁻⁵ M) = 3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.
<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>
