The absolute pressure inside the ball is 1.25atm.
<u>Explanation:</u>
Given-
Volume, V = 5..27L
Temperature, T = 27°C
Gauge Pressure, Pg = 0.25 atm
Atmospheric pressure, Patm = 1 atm
Absolute pressure, Pabs = ?
We know,
Therefore, absolute pressure inside the ball is 1.25atm.
6.96 * 10^5 km for sun radius
0.00019 mm for bacterial cells
It is the boilimg point now aa
Answer: -
4C₂H₅ + 13 O₂ → 8 CO₂ + 10 H₂O
Explanation: -
The given equation is
C₂H₅ + O₂ → CO₂ + H₂O
Balancing for H,
2C₂H₅ + O₂ → CO₂ + 5H₂O
Balancing for C
2C₂H₅ + O₂ → 4CO₂ + 5H₂O
Now right hand side we are getting 13 O which is an odd number.
In order to balance O, we first try to even out the number of O on right side by multiplying whole equation by 2.
4C₂H₅ + 2O₂ → 8CO₂ + 10H₂O
Now balancing for O
4C₂H₅ + 13 O₂ → 8 CO₂ + 10 H₂O
Answer:
0.168mL of the 1.0M KI solution must be added
Explanation:
The PbI₂ is in equilibrium with water as follows:
PbI₂ ⇄ Pb²⁺ + 2I⁻
Where Ksp is:
Ksp = 1.4x10⁻⁸ = [Pb²⁺] / [I⁻]²
As molarity of Pb²⁺ is 0.11M:
1.4x10⁻⁸ = [0.11M] / [I⁻]²
1.27x10⁻⁷ = [I⁻]²
3.57x10⁻⁴M = [I⁻]
Thus, to precipitate all Pb²⁺ you need to add 3,57x10⁻⁴M of I⁻. As volume of the solution is 470.0mL = 0.47L, you need to add:
0.47L * (3,57x10⁻⁴moles / L) = 1.68x10⁻⁴ moles of I⁻ = Moles of KI.
That comes from the 1.0M KI. You need to add:
1.68x10⁻⁴ moles of KI * (L / 1.0 mol) = 1.68x10⁻⁴L =
<h3>0.168mL of the 1.0M KI solution must be added</h3>