Use Boyle's Law of Pressure: P1 x V1 = P2 x V2. Givens: P1=0.9 atm V1= 4 P2= 0.9 atm Find: V2 Equation: 0.9 atm x 4 x 4 L = 0.20 atm x V2Solve: 36 atmL= 0.20 atm x V2 18 : = V2 Short answer: The new volume is 104 ml.
Explanation:
Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.
As a result, formula of magnesium oxide is MgO and not
.
The student write the equation as
, it is not correct.
Therefore, given equation will be balanced as follows.

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.
Answer:
you sure it is the right sign your using
Answer:
2nd option is NOT a postulate of kinetic theory of gases
Explanation:
actually particles move in straight random motion and create bombardment with the molecules and container
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.