Answer: The concentration of 
ions in vinegar is 0.001 M.
Explanation:
Given: pH = 3.0
pH is the negative logarithm of concentration of hydrogen ion.
The expression for pH is as follows.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the value into above expression as follows.
![pH = - log [H^{+}]\\3.0 = - log [H^{+}]\\conc. of H^{+} = antilog (- 3.0)\\= 0.001 M](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5C3.0%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%20antilog%20%28-%203.0%29%5C%5C%3D%200.001%20M)
Thus, we can conclude that the concentration of ions in vinegar is 0.001 M.
Answer:
The reaction is not spontaneous in the forward direction, but in the reverse direction.
Explanation:
<u>Step 1: </u>Data given
H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol
Temperature = 25°C = 25+273 = 298 Kelvin
The initial pressures are:
pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
<u>Step 2</u>: Calculate ΔG
ΔG = ΔG° + RTln Q
with ΔG° = 2.60 kJ/mol
with R = 8.3145 J/K*mol
with T = 298 Kelvin
Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]
with pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
Q = (3.10²)/(1.5*1.75)
Q = 3.661
ΔG = ΔG° + RTln Q
ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)
ΔG =5815.43 J/mol = 5.815 kJ/mol
To be spontaneous, ΔG should be <0.
ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.
The reaction for magnesium iodide when put into water is as below
MgI2(s) → Mg^2+(aq) + 2I^-(aq)
when magnesium iodide but into water it dissociate/ ionize completely into Mg^2+ and 2l^- ions. Magnesium iodide dissociate/ionize completely because magnesium iodide is a strong electrolyte which dissociate/ ionize completely into their ions when it is put into water .
The answer to the question is- Fe2O3
Answer:
B
Explanation:
it shows that the solution is neutral.
I hope this helps
