Balance the equation first:
2 Fe+6 HNO3→2 Fe(NO3)3+3H2
Then calculate mass of Iron :
4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe
Now use Stoichiometry:
371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3
Convert moles of nitric acid to grams
19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
It's an alkali metal so one, they sit on the very first of the column in the periodic table :DDDD
I'm pretty sure (C) Nitrogen is a compound.
