The wave length of light. This determinds the color/type of light.
Na 2Co3
1*23=23. 2*12=24. 6*16=96
23+24+96=143
(23*100)/143. (24*100)/143. (96*100)/143
=16%. =16.7%. =67.1%
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.
Explanation:
conduction involves the transfer of molecules due to the movement of particles
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
