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
Electrons, everything is pretty much based around the likeliness of electrons to be swapped or shared between atoms
It showed that atoms can be divided into smaller parts.
It showed that all atoms contain electrons.
Explanation:
The experiment carried out by J.J Thomson on the gas discharge tube by passing electric current through a tube filled with many different gases provided a good insight into the structure of an atom.
This experiment led to the development of the plum pudding model of the atom.
- Cathode rays and it properties were discovered in this set up.
- It furnished the scientific community with evidences that atoms can be divided into smaller parts.
- Since atoms now contain some subatomic particles, they can be broken down in like manner into further bits.
- The cathode rays which were later termed electrons became a fundamental particles known for every atom.
learn more:
Rutherford's model of the atom brainly.com/question/1859083
#learnwithBrainly
d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
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