Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
Hello!
The balanced equation for the
neutralization of KOH is the following:
HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)
To calculate the
volume of HCl required, we can apply the following equation:
So, the required volume of HCl is
541,54 mLHave a nice day!
An electron. Electrons are found on the outer shells.
Hey there!
In order to solve for the percentage of water in the compound, you will first need to find its total molar mass. You can do this by adding up the molar masses of each individual element in the compound. Then, you will divide the mass that you find of the water molecules by the total mass to get the percentage.
→ Na₂CO₃ ×<span> 10 H</span>₂<span>O
</span>→ Na₂ = 22.9898 × 2 = 45.9796
→ C = 12.0107
→ O₃ = 15.999 × 3 = 47.997
→ 10 H₂O = 18.015 × 10 = 180.15
Now, just add all of those numbers up for the total molar mass.
→ 45.9796 + 12.0107 + 47.997 + 180.15 = <span>286.1373
</span>
The last step is to divide the molar mass of the 10 water molecules by the total mass.
→ 180.15 ÷ 286.1373 = <span>0.62959 </span>≈ 0.63
Your answer will be about 63%.
Hope this helped you out! :-)
Answer:
I think it's the second answer --If you increase the acidity..
I hope answer I can answer your question!
