An inhibitor, which slows down the reaction enough to measure the release of gas.
The grams of Fe₂O₃ that are formed is 47.68 g
<u><em>calculation</em></u>
Step 1: write the equation for reaction
4 Fe +3O₂ → 2 Fe₂O₃
Step 2: find the moles of Fe
moles = mass÷ molar mass
= 33.4 g÷55.8 g/mol =0.5986 moles
Step 3 : use the mole ratio to determine the moles of Fe₂O₃
That is from equation above Fe:Fe₂O₃ is 4:2 therefore the moles of Fe₂O₃ is = 0.5986 moles x 2/4 =0.2993 moles
Step 4 : find the mass of Fe₂O₃
mass = mass x molar mass
The molar mass of Fe₂O₃ = (55.8 x 2 +(15.9 x3) = 159.3 g/mol
mass is therefore = 0.2993 moles x 159.3 g/mol =47.68 g
First question I believe is B
The second one is also B
I really hope it helped
Answer:
80ml
Explanation:
you have your initial concentration to be 0.25 mole on your final volume to be 250 ml and your final concentration to be 0.8 0.08 molar you don't have your initial volume sotify your initial volume you use the expression see 1 * 21 equals see two times between you make when when the subject then 1 equals to 2 x 2/3 one you know substitute your values into it to get being one that's your original volume to be at the latest or 80 ml
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
1.75 M x V1 = 0.100 M x 2.0 L
V1 = -.11 L
