C, because specific heat is measured in Joules/grams°C
(Can I have Brainlist)?
Answer:
2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction
8 (4 moles CO2 per mole butane)
Explanation:
could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O
Given:
175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)
The balanced chemical equation is shown below:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.
Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
mass of NH3 = 185.94 kg NH3 needed
mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg
mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O
Answer:
8510 mol
Explanation:you must divide both of the molar masses into each other
The rate constant for 1st order reaction is
K = (2.303 /t) log (A0 /A)
Where, k is rate constant
t is time in sec
A0 is initial concentration
(6.82 * 10-3) * 240 = log (0.02 /A)
1.63 = log (0.02 /A)
-1.69 – log A = 1.63
Log A = - 0.069
A = 0.82
Hence, 0.82 mol of A remain after 4 minutes.
