Answer:
Explanation:
- The chemical reaction for cellular respiration is as follows -
C₆H₁₂O₆ ( Glucose ) + 6 O₂ ( Oxygen ) --------------> 6CO₂ ( Carbon dioxide ) + 6H₂O ( Water ) + 38 ATP molecules ( Energy ) .
This is an example of oxidation of carbon .
- The chemical reaction for photosynthesis is as follows -
6CO₂ ( Carbon dioxide ) + 6H₂O ( Water ) --------------> C₆H₁₂O₆( Glucose ) + 6 O₂ ( Oxygen ).
This is an example of reduction of carbon .
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
We have that every gas satisfies the fundamental gas equation, PV=nRT where P is the Pressure, V is the volume of the gas, n are the moles of the gas, R is a universal constant and T is the Temperature in Kelvin. We have that PV/T=nR and during our process, the moles of the gas do not change (no argon enters or escapes our sample). See attached.
Molality is the moles of solute per kg of solvent.
Moles of NH₄Cl = 2.4 / (14 + 4 x 1 + 35.5)
= 0.0448 mole
Molality = 0.0448 / (19.4 / 1000)
= 2.31 m
