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:
Option C. 4.03 g
Explanation:
Firstly we analyse data.
12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.
Density is the data that indicates grams of solution in volume of solution.
We need to determine, the volume of solution for the concentration
Density = mass / volume
1.05 g/mL = 100 g / volume
Volume = 100 g / 1.05 g/mL → 95.24 mL
Therefore our 12 g of solute are contained in 95.24 mL
Let's finish this by a rule of three.
95.24 mL contain 12 g of sucrose
Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g
Answer:
Beryllium (Be) : 9.01 g/mol
Silicon (Si) : 28.09 g/mol
Calcium (Ca) : 40.08 g/mol
Rhodium (Rh) : 102.91 g/mol
Explanation:
Answer:
Explanation:
A single replacement or single displacement reaction is a reaction in which one substance replaces another.
A + BC → AC + B
The replacement of an ion in solution by a metal higher in the activity series is a special example of this reaction type.
The relative positions of the elements in the activity series provides the driving force for single displacement reactions.
A double replacement reaction is one in which there is an actual exchange of partners between reacting species. This reaction is more common between ionic substances;
AB + CD → AC + BD
Such reactions are usually driven by;
- formation of precipitation
- formation of water and a gaseous product
So,
With addition, we the last digit we keep will be the one which is known for both individual values.
We know 2.13 to the hundredths, but we only know 1 to the ones. Therefore, we will round off in the ones place.
2.13 + 1 = 3.13 (unrounded)
= 3 (rounded)
Hope this helps!
