Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.
Proton plus neutron is the correct answer. Protons and neutrons have a mass of 1 and electrons have a mass of 0. So in order to find the mass of an atom you need to add the number of protons and the number of neutrons.
Answer:
The percent yield reaction is 64.3%
Explanation:
This is the ballanced reaction
Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)
Let's determine the moles of our reactants:
Mass / Molar mass = Mol
26.8 g / 310.18 g/m = 0.0864 moles of phosphate.
54.3 g / 98.06 g/m = 0.554 moles of sulfuric
1 mol of phosphate reacts with 3 mol of sulfuric so
0.0864 mol of PO₄⁻³ will react with (0.0864 .3)/1 = 0.259 moles
I have 0.554 of sulfuric, so this is the reactant in excess.
The limiting reagent is the Phosphate.
1 mol of PO₄⁻³ produces 2 mol of phosphoric
0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles
Mol . molar mass = Mass
0.173 m . 97.98g/m = 16.95 g (This is the theoretical yield)
Percent yield = (Produced / Theoretical) .100
(10.9 g / 16.95 g) . 100 = 64.3 %
A flood, if it hits the environment of the natural rubbers, would destroy how the rubber is being produced. to have a large amount of limitation, the flood would destroy a large percentage of rubber trees. This natural rubber is needed to make synthetic polymers. Without the rubber (because of damages to it's ecosystem through the flood), there would be a limited supply, and a substancial drop on synthetic polymers.
hope this helps
Answer : The value of equilibrium constant for this reaction at 262.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 262.0 K is
