Answer:
- <em>The net ionic equation is: </em><u>Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s)</u>
Explanation:
<u>1) Start by writing the total ionic equation:</u>
The total ionic equation shows each aqueous substance in its ionized form, while the solid or liquid substances are shown with their chemical formula.
These are the ionic species:
- AgF (aq) → Ag⁺ (aq) + F⁻ (aq)
- NH₄Cl (aq) → NH₄⁺ (aq) + Cl ⁻ (aq)
- NH₄F(aq) → NH₄⁺ (aq) + F⁻ (aq)
Then, replace each chemical formula in the chemical equation by those ionic forms:
- Ag⁺ (aq) + F⁻ (aq) + NH₄⁺ (aq) + Cl ⁻ (aq) → AgCl (s) + NH₄⁺ (aq) + F⁻ (aq)
That is the total ionic equation.
<u>2) Spectator ions:</u>
The ions that appear in both the reactant side and the product side are considered spectator ions (they do not change), and so they are canceled.
In our total ionic equation they are F⁻ (aq) and NH₄⁺ (aq).
After canceling them, you get the net ionic equation:
<u>3) Net ionic equation:</u>
- Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s) ← answer
Precision relates to how close the answers are to each other, so I’d think it would be D because of the limited range between data points.
Proton:
Positively charged
Inside nucleus
Mass - 1
Electrons:
Negatively charged
Outside the nucleus
Mass - 1/2000
Answer:
a) Keq = 4.5x10^-6
b) [oxaloacetate] = 9x10^-9 M
c) 23 oxaloacetate molecules
Explanation:
a) In the standard state we have to:
ΔGo = -R*T*ln(Keq) (eq.1)
ΔGo = 30.5 kJ/moles = 30500 J/moles
R = 8.314 J*K^-1*moles^-1
Clearing Keq:
Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6
b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])
4.5x10^-6 = ([oxaloacetate]/(0.20*10)
Clearing [oxaloacetate]:
[oxaloacetate] = 9x10^-9 M
c) the radius of the mitochondria is equal to:
r = 10^-5 dm
The volume of the mitochondria is:
V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L
1 L of mitochondria contains 9x10^-9 M of oxaloacetate
Thus, 4.18x10^-42 L of mitochondria contains:
molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules
