Why not search it up and figure it out and then write it how you would?
The subscript is the amount of atoms in each molecule and the coefficient is the amount of molecules. there are 4 Hydrogen, 2 Sulfur, and 8 Oxygen in this particular substance.
Additional pressure required : 175 mm
<h3>Further explanation</h3>
Boyle's Law
At a constant temperature, the gas volume is inversely proportional to the pressure applied

V₁=250 ml
P₁=700 mmHg


Additional pressure :

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:
(1) < (5) < (3) < (4) < (2)
Explanation :
<u>The formula used for isothermally irreversible expansion is :</u>

where,
w = work done
= external pressure
= initial volume of gas
= final volume of gas
<u>The expression used for work done in reversible isothermal expansion will be,</u>

where,
w = work done = ?
n = number of moles of gas = 1 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 
= initial volume of gas
= final volume of gas
First we have to determine the work done for the following process.
(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.



(2) A free isothermal expansion from 1 L to 100 L.



(3) A reversible isothermal expansion from 0.5 L to 4 L.



(4) A reversible isothermal expansion from 0.5 L to 40 L.



(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.



Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:
(1) < (5) < (3) < (4) < (2)
Answer:
3.0×10⁻¹³ M
Explanation:
The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.
<h3>Equation</h3>
The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.
Ksp = [Al₃⁺³]×[OH⁻]³
<h3>Solving for [Al₃⁺³]</h3>
The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.
This means the solubility product equation is ...
Ksp = x(10⁻⁷ +3x)³
For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...
3.0×10⁻³⁴ = x(10⁻⁷)³
Then x = [Al₃⁺³] will be ...
![[\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}](https://tex.z-dn.net/?f=%5B%5Ctext%7BAl%7D_3%5E%7B%5C%2C%2B3%7D%5D%3D%5Cdfrac%7B3.0%5Ctimes10%5E%7B-34%7D%7D%7B10%5E%7B-21%7D%7D%5Capprox%20%5Cboxed%7B3.0%5Ctimes10%5E%7B-13%7D%5Cqquad%5Ctext%7Bmoles%20per%20liter%7D%7D)
We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.
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<em>Additional comment</em>
The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.