Answer:
Fermentation is where all alcohol is created, distillation is where the alcohol is separated and removed. In order for fermentation to occur, two things are needed: a raw material in liquid form that contains sugar, followed by the addition of yeast.
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2, valence shell: 4s2 3d10 4p2
18.998403
Explanation:
The internet told me
70.33 L is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm.
<h3>What is volume?</h3>
Volume is the percentage of a liquid, solid, or gas's three-dimensional space that it occupies.
Liters, cubic metres, gallons, millilitres, teaspoons, and ounces are some of the more popular units used to express volume, though there are many others.
We will use ideal gas law to find the volume
PV = nRT
Can also be written as
V = (nRT)/P
Where,
P = pressure
V = volume
n = amount of substance
R = ideal gas constant
T = temperature
Here, we have given
P = 3.5 atm
V = to find
n = 10 moles
R = 0.08206 L⋅atm/K⋅mol
T = 300k
Lets substitute the values
V = (10 × 0.08206 × 300)/3.5
V = 70.33 L
Learn more about volume
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<u>Answer:</u> The 
for HCN (g) in the reaction is 135.1 kJ/mol.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ](https://tex.z-dn.net/?f=-870.8%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%29%7D%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28-80.3%29%29%2B%283%5Ctimes%20%280%29%29%2B%282%5Ctimes%20%28-74.6%29%29%5D%5C%5C%5C%5C%5CDelta%20H_f_%7B%28HCN%29%7D%3D135.1kJ)
Hence, the for HCN (g) in the reaction is 135.1 kJ/mol.
