Answer:
C₆H₁₂O₆ (glucose)
A molecular substance.
Explanation:
C₆H₁₂O₆ does not dissociate. It is not a ionic type.
There is no metal in the compound, it is not a metallic type.
Glucose is an organic molecule, a type of carbohidrate, the most common from the monosaccharides type.
Answer : The value of 
for this reaction is, 
Explanation :
The given chemical reaction is:
Now we have to calculate value of 
.
![\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BHCH_3CO_2%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28HCH_3CO_2%28g%29%29%7D%5D-%5Bn_%7BCH_3OH%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CH_3OH%28g%29%29%7D%2Bn_%7BCO%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CO%28g%29%29%7D%5D)
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
![\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B1mole%5Ctimes%20%28-389.8kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-163.2kJ%2Fmol%29%2B1mole%5Ctimes%20%28-137.2kJ%2Fmol%29%5D)
The relation between the equilibrium constant and standard Gibbs, free energy is:
where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature = 
= equilibrium constant = ?
Now put all the given values in this expression, we get:
Thus, the value of for this reaction is,
There are O-H bonds in H2O. They have the intramolecular force of polar covalent bond.
Answer:
Option A. 9.4 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 8 L
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 343 K
Final volume (V₂) =?
V₁ / T₁ = V₂ / T₂
8 / 293 = V₂ / 343
Cross multiply
293 × V₂ = 8 × 343
293 × V₂ = 2744
Divide both side by 293
V₂ = 2744 / 293
V₂ = 9.4 L
Therefore, the final volume of the gas is 9.4 L
Answer: (a) There are 0.428 moles present in 12 g of 
molecule.
(b) There are 2 moles present in 
particles of oxygen.
Explanation:
(a). The mass of nitrogen molecule is given as 12 g.
As the molar mass of is 28 g/mol so its number of moles are calculated as follows.
So, there are 0.428 moles present in 12 g of molecule.
(b). According to the mole concept, 1 mole of every substance contains 
atoms.
Therefore, moles present in particles are calculated as follows.
So, there are 2 moles present in particles of oxygen.
