Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
Basis: 100 mL solution
From the given density, we calculate for the mass of the solution.
density = mass / volume
mass = density x volume
mass = (1.83 g/mL) x (100 mL) = 183 grams
Then, we calculate for the mass H2SO4 given the percentage.
mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams
Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles
Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M
Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m
Oxygen 4. The fatty acid chain of the lipids is often referred to as a hydrocarbon chain.
Convert each amount of grams into moles:
I: 23.24g x 1 mol / 126.90g = 0.1831 mol I
C: 2.198 x 1 mol / 12.01g = 0.1830 mol C
N: 2.562 x 1 mol / 14.01g = 0.1829 mol N
Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1
Therefore the empirical formula is ICN
Answer:
Option B
Explanation:
We will check the solubility graph for potassium nitrate, KNO
3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.
Hence, option B is correct
