Answer: The law of conservation of mass is reffering to the fact that energy cannot be created or destroyed. So when you speak about atoms not being created or destroyed it is the same thing. Unless you're talking about an atomic bomb where the atoms are split.
Explanation:
the energy that that is needed to break a bond is called the bond energy or dissciation energy
Answer:
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Explanation:
Step 1: Data given
Kp = 4.7 x 10^3 at 400K
Pressure of CH3OH = 0.250 atm
Pressure of HCl = 0.600 atm
Volume = 10.00 L
Step 2: The balanced equation
CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)
Step 3: The initial pressure
p(CH3OH) = 0.250atm
p(HCl) = 0.600 atm
p(CH3Cl)= 0 atm
p(H2O) = 0 atm
Step 3: Calculate the pressure at the equilibrium
p(CH3OH) = 0.250 - X atm
p(HCl) = 0.600 - X atm
p(CH3Cl)= X atm
p(H2O) = X atm
Step 4: Calculate Kp
Kp = (pHO * pCH3Cl) / (pCH3* pHCl)
4.7 * 10³ = X² /(0.250-X)(0.600-X)
X = 0.249962
p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm
p(HCl) = 0.600 - 0.249962 = 0.350038 atm
p(CH3Cl)= 0.249962 atm
p(H2O) = 0.249962 atm
Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)
Kp = 4.7 *10³
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Answer:
option A = S(s) + O₂(g) → SO₂ (s)
Explanation:
Chemical equation:
S(s) + O₂(g) → SO₂ (s)
when sulfur burned in the presence of oxygen it produce sulfur dioxide. The sulfur dioxide can further react with oxygen to produce sulfur trioxide and then react with water to form sulfuric acid.
Uses of sulfur dioxde:
It is used as a solvent and reagent in laboratory.
Sulfur dioxide is used to produce sulfuric acid.
It is used as a disinfectant
It is also used as a reducing agent.
It is used to preserve the dry food.
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