Answer:
Molarity = moles ÷ liters
to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)
Na = 22.989769
Br = 79.904
molar mass of NaBr = 102.893769
6.6g ÷ 102.893769 = 0.064143826 moles of NaBr
0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %
Answer:
Work done in Joules = -14865.432J
ΔU = -262.94KJ
Explanation:
The steps are as shown in the attachment
Answer:
the description of a mineral species usually includes its common physical properties such as habit, hardness, lustre, diaphaneity, colour, streak, tenacity, cleavage, fracture, parting, specific gravity, magnetism, fluorescence, radioactivity, as well as its taste or smell and its reaction to acid.
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
There are 4 significant figures! Start counting after the first non-zero digit :)
Hope this helps.
