Red is the answer in the blank
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
Explanation:
The molecular mass of a monomer unit is:
C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u
For 1565 units,
Answer:
a) The concentration of drug in the bottle is 9.8 mg/ml
b) 0.15 ml drug solution + 1.85 ml saline.
c) 4.9 × 10⁻⁵ mol/l
Explanation:
Hi there!
a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml
b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:
0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.
The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)
If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.
To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline
c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.
Let´s convert it to molarity:
0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l
