Fluorine has 9 protons and 9 electrons
Iodine has 53 protons and 53 electrons
When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:

The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
#SPJ1
The number of moles of moles of Magnesium,chlorine and oxygen atoms in 7.80 moles of Mg(ClO4)2 is calculated as below
find the total number of each atom in Mg(ClO4)2
that is mg = 1 atom
Cl = 1x2 = 2 atoms
O = 4 x2 = 8 atoms
then multiply 7.80 moles with total number of each atom , to get the number moles of each atom
that is
Mg = 7.80 x1= 7.80 moles
cl = 7.80 x2=15.6 moles
O = 7.80 x8= 62.4 moles
The number of moles of the magnesium (mg) is 0.00067 mol.
The number of moles of hydrogen gas is 0.0008 mol.
The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.
<h3>
Number of moles of the magnesium (mg)</h3>
The number of moles of the magnesium (mg) is calculated as follows;
number of moles = reacting mass / molar mass
molar mass of magnesium (mg) = 24 g/mol
number of moles = 0.016 g / 24 g/mol = 0.00067 mol.
<h3>Number of moles of hydrogen gas</h3>
PV = nRT
n = PV/RT
Apply Boyle's law to determine the change in volume.
P1V1 = P2V2
V2 = (P1V1)/P2
V2 = (101.39 x 146)/(116.54)
V2 = 127.02 mL
Now determine the number of moles using the following value of ideal constant.
R = 8.314 LkPa/mol.K
n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)
n = 0.0008
<h3>Volume of 1 mole of hydrogen gas at STP</h3>
V = nRT/P
V = (1 x 8.314 x 273) / (101.325)
V = 22.4 L
Learn more about number of moles here: brainly.com/question/13314627
#SPJ1