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
Since the density of water is 1 g /mL, hence there is 100
g of H2O. So total mass is:
m = 100 g + 5 g = 105 g
=> The heat of reaction can be calculated using the
formula:
δhrxn = m C ΔT
where m is mass, C is heap capacity and ΔT is change in
temperature = negative since there is a decrease
δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)
δhrxn = -1,009.47 J
=> However this is still in units of J, so calculate
the number of moles of NaCl.
moles NaCl = 5 g / (58.44 g / mol)
moles NaCl = 0.0856 mol
=> So the heat of reaction per mole is:
δhrxn = -1,009.47 J / 0.0856 mol
δhrxn = -11,798.69 J/mol = -11.8 kJ/mol
A polyatomic ion.
Examples: Sulfate, Sulfite, Nitrate, Nitrite...
Answer:
The active ingredients in baking soda (NaHCO3) are
and 
when Baking soda reacts with Acetic acid
Molecular equation
NaHCO3(aq) + CH3COOH(aq) → Na(CH3COO)(aq) + CO2(g) +H2O(l)
Ionic equation
→ 
as is present on both sides so it will cancel out and the net ionic equation will be
→ 
