Explanation:
1)Mass of CO when 210.3 g of Fe produced.
Number of moles of 
in 210.3 g=
According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : 
of CO that is 5.64 moles.
Mass of CO in 5.64 moles =
2)Mass of CO when 209.7 g of Fe produced.
Number of moles of in 209.7 g=
According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : 
of CO that is 5.625 moles.
Mass of CO in 5.625 moles =
Chemical formula of the glucose: C₆H₁₂O₆
We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u
atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol
2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g---------------------- x
x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles
we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)
3)We calculate the number of molecules:
Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles-------- x
x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.
4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²
number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .
number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²
Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
First, let's start off by finding the mass of this whole hydrate.
(Note: the unit of measurement for mass will be amu)
Let's find the molecular mass of each element.
Now, let's find the mass of each compound.
We have 6 molecules of H2O, so multiply 18.015 by 6 then add that with the weight of CoCl2.
Now divide 108.09 (mass of all the H2O in the hydrate) by 237.923 (total mass of hydrate).
Turn that into a percentage and you get 45.431%.
Hope this helps! :)
Answer : The volume of hydrogen gas at STP is 4550 L.
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 100.0 atm
= final pressure of gas at STP = 1 atm
= initial volume of gas = 50.0 L
= final volume of gas at STP = ?
= initial temperature of gas = 
= final temperature of gas at STP = 
Now put all the given values in the above equation, we get:
Therefore, the volume of hydrogen gas at STP is 4550 L.
Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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