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
The correct option is D.
The reactants that combine together to form glucose are carbon dioxide, water and energy from the sun. Six molecules of carbon dioxide combine with six molecules of water in the presence of sunlight to form glucose. The chemical equation for the reaction is given below"
6CO2 + 6H2O + Sunlight = C6H12O6.
Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series
Explanation:
Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.
Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.
In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .
Thus for mass number : 235 = 207+4X
4X= 28
X = 7
Thus for atomic number : 92 = 82+2X-Y
2X- Y = 10
2(7) - Y= 10
14-10 = Y
Y= 4
Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series
Answer:
(A) N4H6 (B) H2O (C) LiH (D) C12H26
Explanation:
The given compounds have been arranged from left to right in order of increasing percentage by mass of hydrogen.
The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.
