Explanation:
the conductors are the three u have checked
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
The two oxygen atoms share two pairs of electrons, so two covalent bonds hold the oxygen molecule together
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
Answer:
1) Basic
2) Basic
3) Acidic
Explanation:
pH of a solution is a measure of the H3O+ ions in the solution and hence reflects its acidity.
![pH = -log[H3O+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH3O%2B%5D)
Solutions are classified as acidic, basic or neutral based on the pH range
-pH < 7, acidic
- pH = 7, neutral
- pH > 7, basic
1) [H3O+] = 2.5*10^-9M
![pH = -log[H3O+]=-log[2.5*10^{-9}]=8.60](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH3O%2B%5D%3D-log%5B2.5%2A10%5E%7B-9%7D%5D%3D8.60)
Since pH > 7, solution is basic
2)[OH-] = 1.6*10^-2M
![pOH = -log[OH-]=-log[1.6*10^{-2}]=1.80](https://tex.z-dn.net/?f=pOH%20%3D%20-log%5BOH-%5D%3D-log%5B1.6%2A10%5E%7B-2%7D%5D%3D1.80)
pH = 14 - pOH = 14 - 1.80 =12.2
Since pH > 7, solution is basic
3) [H3O+] = 7.9*10^-3M
![pH = -log[H3O+]=-log[7.9*10^{-3}]=2.10](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH3O%2B%5D%3D-log%5B7.9%2A10%5E%7B-3%7D%5D%3D2.10)
Since pH < 7, solution is acidic
