Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) S₈
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
Molar mass of S₈ = 8 * 32 g/mol. = 256 g/mol.
(b) C₂H₁₂
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
Carbon , C = 12 g/mol
Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol
(c) Sc₂(SO₄)₃
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
oxygen , O = 16 g/mol.
scandium , Sc = 45 g/mol.
Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol
(d) CH₃COCH₃ (acetone)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol
(e) C₆H₁₂O₆ (glucose)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.
Answer:
HCl + Ca(OH)2 = CaCl2 + H2O - Chemical Equation Balancer.
Let us assume that there is a 100g sample of Opal. The masses of each element will be:
29.2g Si
33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
Now we divide by the smallest number and obtain:
Si: 1
O: 2
H2O: 2
Thus, the empirical formula of Opal is:
SiO2 . 2H2O
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
