Answer:
23.0733 L
Explanation:
The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

Mass = 62.5 g
Molar mass of
= 34 g/mol
The formula for the calculation of moles is shown below:
Thus, moles are:

Consider the given reaction as:

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.
Also,
1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.
So,
1.8382 moles of hydrogen peroxide decomposes to give ![\frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr The conversion of P(torr) to P(atm) is shown below: [tex]P(torr)=\frac {1}{760}\times P(atm)](https://tex.z-dn.net/?f=%5Cfrac%20%7B1%7D%7B2%7D%5Ctimes%201.8382%20mole%20of%20oxygen%20gas.%20%3C%2Fp%3E%3Cp%3EMoles%20of%20oxygen%20gas%20produced%20%3D%200.9191%20mol%3C%2Fp%3E%3Cp%3EGiven%3A%20%3C%2Fp%3E%3Cp%3EPressure%20%3D%20746%20torr%0A%3C%2Fp%3E%3Cp%3EThe%20conversion%20of%20P%28torr%29%20to%20P%28atm%29%20is%20shown%20below%3A%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%28torr%29%3D%5Cfrac%20%7B1%7D%7B760%7D%5Ctimes%20P%28atm%29)
So,
Pressure = 746 / 760 atm = 0.9816 atm
Temperature = 27 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (27 + 273.15) K = 300.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K
<u>⇒V = 23.0733 L</u>
See if the carbon atoms are SP2 or Sp they the coplanirty is more and if its Sp3 hybridization it cant be in coplanar as Sp3 is having Td shape where as sp2 and sp are not :)
Answers:
Density = 0.8 g/mol.
Given data:
v = 25 ml
m = 20 g
δ = ?
Solution:
Formula for calculating density is given as,
Density = Mass / Volume
putting values
Density = 20.0 g / 25 ml
Density = 0.8 g/mol.
a balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.
Answer:
D. The rate decreases as reactants are used up.
Explanation:
Initially, the rate increases until the reaction is at equilibrium. At equilibrium, the rate is constant.
As the reaction progresses, the rate decreases to zero when reactants are used up ( for irriversible reactions only )