Context:
175 kilograms of methane (CH4) is to be converted into hydrogen cyanide (HCN)
The equation that balances this reaction is listed here:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To find the quantities of ammonia and oxygen needed, we will use 175 kg of CH4 as our reference.
Molar masses are as follows:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
For ammonia: mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
This results in 185.94 kg of NH3 required
For oxygen: mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
So the mass of O2 needed equals 525 kg
To derive the mass of oxygen: mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
This gives a mass of O equal to 131.25 kg O
Refer to the explanation and the attached image for further details. When AlCl3 reacts with (CH3)3CCH2Cl, a primary carbocation is produced. This primary carbocation then undergoes a 1,2-alkyl shift leading to the formation of a tertiary carbocation, which subsequently bonds with the benzene ring as depicted in the attached image.
Response: The volume of the cube that is underwater is 29.8 mL
Clarification:
Initially, we need to find the mass of the ice.
Equation applied:
Provided information:
Ice density = 
Volume of ice = 45.0 mL
The cube displaces 40.5 g of liquid to float.
Next, we will ascertain how much of the cube is submerged in the liquid.
Therefore, the submerged portion of the cube is 29.8 mL
