Answer;
= 64561.95 g/mole
Explanation;
mass of Fe in 100g = .346g
= .346 / 55.8452 moles
= 0.0061957 moles
These represent 4 moles of Fe in the molecule so moles of hemaglobin
= 0.0061957/4
= 0.0015489 moles
these are in 100 g so mass of 1 mole = 100 / 0.0015489
= 64561.95 g / mole
molar mass of hemoglobin = 64561.95 g/mole
Add a coefficient 2 before H2O and a 2 before SO2.
Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
Answer:
Q = 1379.4 J
Explanation:
Given data:
Mass of water = 22 g
Initial temperature = 18°C
Final temperature = 33°C
Heat absorbed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 J/g.
°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 33°C - 18 °C
ΔT = 15°C
Q = 522 g ×4.18 J/g.°C× 15°C
Q = 1379.4 J
