Amount of a substance (called the solute) that dissolves in a unit volume of a liquid substance (called the solvent) to form a saturated solution under specified conditions of temperature and pressure.Solubility is expressed usually as moles of solute per 100 grams of solvent.
Answer:
0.01836 M
Explanation:
Again the reaction equation is;
Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)
E°cell= 0.77 V
Ecell= 0.78 V
[Mn2+] = 0.040 M
[Fe2+] = the unknown
n=2
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]
0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]
0.01/ -0.0296= log [Fe2+] /[0.040]
-0.3378= log [Fe2+] /[0.040]
Antilog(-0.3378) = [Fe2+] /[0.040]
0.459= [Fe2+] /[0.040]
[Fe2+] = 0.459 × 0.040
[Fe2+] = 0.01836 M
Answer and Explanation:
The options aren't listed in your question, but here are some units that are regularly and normally used (in the classroom and in the outside world):
(The SI unit of distance and displacement is the meter. The SI unit of time is the second.)
<u>Meters per Second (m/s)</u>
kilometers per hour (km/hr)
kilometers per second (km/sec)
To find the average speed, you do distance divided by time.
To find the average velocity, you do the final position minus the initial position, divided by the final time minus the initial time.
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
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<em><u>I hope this helps!</u></em>
Answer:
pressure and temperature are directly proportional.
Explanation:
At constant temperature and pressure the volume of a gas is directly proportional to the number of moles of gas. Also given a constant temperature and volume the pressure of a gas is directly proportional to the number of moles of gas.
This means that if V is constant then
P = n (RT/V) then n= PV/RT. As P is increased, T is also increased thereby decreasing the value of n since pressure and temperature are directly proportional.
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
