Answer:
1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.
Explanation:
The reaction that takes place is:
Mn⁺² + S⁻² ⇄ MnS(s)
ksp = [Mn⁺²] [S⁻²]
If the pksp of MnS is 13.500, then the ksp is:

From the problem we know that [S⁻²] = 0.0900 M
We use the ksp to calculate [Mn⁺²]:
3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M
[Mn⁺²] = 3.514 * 10⁻¹³ M.
Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:
3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.
<u>Answer:</u> The
of the reaction at given temperature is -12.964 kJ/mol.
<u>Explanation:</u>
For the given chemical reaction:

The expression of
for the given reaction:

We are given:

Putting values in above equation, we get:

To calculate the Gibbs free energy of the reaction, we use the equation:

where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 
Putting values in above equation, we get:

Hence, the
of the reaction at given temperature is -12.964 kJ/mol.
You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles
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Answer: B. photosynthesis
Explanation: Glucose and oxygen react together in plant cells to produce energy.