Answer:
The second one is the answer
Answer:
The equation is Fe₂O₃ + CO ⇒ Fe + CO₂.
The balanced reaction equation is Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂.
Explanation:
First, we have to write our equation. It's actually pretty straightforward - first we look for our reactants (looks like it's Fe₂O₃ and CO), then we look for our products (Fe and CO₂). Then, we have to balance it so that both sides have the same number of both element.
Currently, we have the equation Fe₂O₃ + CO ⇒ Fe + CO₂. There are 2 Fe atoms, 4 O atoms, and 1 C atom on the left side. There is 1 Fe atom, 2 O atoms, and 1 C atom on the right side.
First thing we can do is give our Fe on the right side a coefficient of 2. This will make it equivalent to the 2 Fe atoms on the left side:
Fe₂O₃ + CO ⇒ 2Fe + CO₂
Next, we need to make sure that we have the same number of C and O atoms on each side. This takes a little bit of thinking, but what we have to do is give CO a coefficient of 3 and CO₂ a coefficient of 3. This gives us 6 O atoms on the left side (when we include the O₃) and 6 O atoms on the right side (since there are 3 O₂ atoms and 3 times 2 is 6). Here's what that looks like:
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
And that's how I balanced the equation. It can be confusing, but with enough practice, it will get easier and easier. :)
Answer:
Check the explanation
Explanation:
When talking about our universe there are 5 d orbitals. The element of first transition series moves away from the universal principles of Hund's rule and Aufbav's principle. So in order to attain stability these elements tend to form half or full filled orbitals.
In our universe the ground state electronic configuration of sixth transition metal, Iron (Fe) : [Ar] 
and the electronic configuration of seventh transition metal, Cobalt (Co) : [Ar] 
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In universe L there are seven orbitals.
Ground state electronic configuration of sixth and seven transition element.
Sixth transition metal: [Ar] ![3d^{7} 4s^1 or [X] 3d^{7} 4s^1](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E1%20or%20%5BX%5D%203d%5E%7B7%7D%204s%5E1)
Seventh transition metal: [Ar] ![3d^{7} 4s^{2}or [X] 3d^{7} 4s^{2}](https://tex.z-dn.net/?f=3d%5E%7B7%7D%204s%5E%7B2%7Dor%20%5BX%5D%203d%5E%7B7%7D%204s%5E%7B2%7D)
1 mole ------------ 6.02 x10²³ molecules
? mole ----------- 1.505 x10²³ molecules
1.505x10²³ / 6.02x10²³ => 0.25 moles
hope this helps!
P₁ = 0.90 atm
V₁ = 50.0 mL
T₁ = 298 K
P₂ = 1 atm
T₂ = 273 K
V₂ = P₁ x V₁ x T₂ / T₁ x P₂
V₂ = 0.90 x 50.0 x 273 / 298 x 1
V₂ = 12285 / 298
V₂ = 41 mL
Answer (1)
hope this helps!
