Two things
Ferrous Ammonium Sulfate is a pale green or blue-green powder or sand-like solid. It is used in photography, analytical chemistry, and Iron-plating baths.
An FAS-DPD titration is as simple as a test for total alkalinity or calcium hardness. A buffered DPD indicator powder is added to a water sample and reacts with chlorine to produce the pink color characteristic of the standard DPD test.
Hope this it gang
Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.
next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.
use the mols of the limiting reagent to compare with the mols of the product.
take the mols of the product/mr of the product.
this will give u the mass.
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:
The expression for 
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:
The expression of 
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)
Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
Answer:
1.4 × 10² mL
Explanation:
There is some info missing. I looked at the question online.
<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>
Step 1: Given data
- Initial volume (V₁): 215 mL
- Initial pressure (P₁): 625 mmHg
- Final pressure (P₂): 1.3 atm
Step 2: Convert 625 mmHg to atm
We will use the conversion factor 1 atm = 760 mmHg.
625 mmHg × 1 atm/760 mmHg = 0.822 atm
Step 3: Calculate the final volume of the air
Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL
