The catalytic decomposition of hydrogen peroxide yields oxygen gas and water, according to the reaction given below. in an experiment, the decomposition of hydrogen peroxide yielded 75.3 ml of gas collected over water at 25°c and 0.976 atm. (water = 0.032 atm at 25°c) 2 h2o2(aq) → 2 h2o(l) + o2(g) if the original h2o2 solution had a volume on 500 ml, what was the molar concentration of hydrogen peroxide prior to decomposition?

The molar concentration of hydrogen peroxide is 0.01164 M

Explanation:

Step 1: Data given

Volume of gas yielded = 75.3 mL = 0.0753 L

Temperature = 25.0 °C

Atmosphere = 0.976 atm

(Water = 0.032 atm at 25°C)

The original volume of H2O2 is 500 mL

Molar mass of O2 =32 g/mol

Step 2: The balanced equation

2H2O2(aq) → 2 H2O(l) + O2(g)

Step 3: Calculate pressure of O2

P = P(02) + P(water)

P(O2) = P - P(water)

P(O2) = 0.976 atm - 0.032 atm

P(O2) = 0.944 atm

Step 4: Calculate moles O2

p*V=n*R*T

⇒ with p = the pressure of O2 gas = 0.944 atm

⇒ with V= the volume of O2 = 0.0753 L

⇒ with n = the number of moles of O2

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 25°C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.944 * 0.0753)/(0.08206*298K)

n = 0.00291 moles O2

Step 5: Calculate moles of H2O2

For 1 mole of O2 produced, we need 2 moles of H2O2

For 0.00291 moles O2 we need 2*0.00291 = 0.00582 moles H2O2

Step 6: Calculate molar concentration of H2O2

Molar concentration = moles / volume

Molar concentration = 0.00582 moles / 0.500L

Molar concentration = 0.01164 M

The molar concentration of hydrogen peroxide is 0.01164 M

can you list the answers so i can answer

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied