I'm assuming that by "miles" you mean moles.
If O2 is the excess reactant, that means Fe is the limiting reactant. That means that the amount of product being formed depends on the amount of Fe reactant present. To calculate the moles of Fe2O3 formed, start with the given 6.4 moles of Fe and use the mole to mole ratio given by the reaction as shown below:
6.4 mol Fe x
=
3.2 mol Fe2O3
The pressure of the gas = 40 atm
<h3>Further explanation</h3>
Given
200 ml container
P = 2 atm
final volume = 10 ml
Required
Final pressure
Solution
Boyle's Law
At a fixed temperature, the gas volume is inversely proportional to the pressure applied
Input the value :
P₂ = P₁V₁/V₂
P₂ = 2 x 200 / 10
P₂ = 40 atm
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the 
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17% concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
Answer:
The formal charge on nitrogen in 
is +1.
Explanation:
The structure of is as follows.
(In attachment)
From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.
![Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1](https://tex.z-dn.net/?f=Formal%5C%2Ccharge%5C%2Con%5C%2Cnitrogen%20%3D%205-%5B0%2B%20%5Cfrac%7B8%7D%7B2%7D%5D%3D%20%2B1)
Therefore, The formal charge on nitrogen in is +1.
