I think this is how you do it:
Answer: 2.54g
Explanation:
Molar Mass of H2O2 = (2x1) + (2x16) = 34g/mol
1mole (34g) of H2O2 contains 6.02x10^23 molecules
Therefore Xg of H2O2 will contain 4.5x10^22 molecules i.e
Xg of H2O2 = (34x4.5x10^22)/6.02x10^23 = 2.54g
<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
<u>Explanation:</u>
The equation used to calculate the Vant' Hoff factor in dissociation follows:

where,
= degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:

The equation used to calculate the degee of dissociation follows:

Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
Hydrocarbons are molecules that contain only carbon and hydrogen. Due to carbon's unique bonding patterns, hydrocarbons can have single, double, or triple bonds between the carbon atoms. The names of hydrocarbons with single bonds end in "-ane," those with double bonds end in "-ene," and those with triple bonds end in "-yne". The bonding of hydrocarbons allows them to form rings or chains.