Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Conduction is heat tranfer through physical contact. Hope this helps. :)
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
To calculate this, we need the Molarity formula. This formula tell us that Molarity, which is a concentration unit, is equal to the number of moles divided by the volume. In this question we already have the Molarity and the Volume, so let's build our equation:
C = n/V (You can see Molarity with the letter "C" because it means concentration)
3 = n/1
n = 1 * 3
n = 3 moles of NaOH
The digestive system takes the nutrients from the food you eat as well as the water. The leftover matter is excreted through the body as feces, or poop. Hope I helped :)
