So for D you have to find energy right
from c you get wavelength Lambda
so on D use this
E = Hc / lambda
c is given 1.5 x 10 ^20
h = 6.624 x 10^-34
and then you get answer for energy
140 g of nitrogen (N₂)
Explanation:
We have the following chemical equation:
N₂ + 3 H₂ -- > 2 NH₃
Now, to find the number of moles of ammonia we use the Avogadro's number:
if 1 mole of ammonia contains 6.022 × 10²³ molecules
then X moles of ammonia contains 6.022 × 10²⁴ molecules
X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³
X = 10 moles of ammonia
Taking in account the chemical reaction we devise the following reasoning:
If 1 mole of nitrogen produces 2 moles of ammonia
then Y moles of nitrogen produces 10 moles of ammonia
Y = (1 × 10) / 2
Y = 5 moles of nitrogen
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of nitrogen (N₂) = 5 × 28 = 140 g
Learn more about:
Avogadro's number
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The statement that best explains why magnesium and chlorine combine in a 1:2 ratio is; Magnesium has two valence electrons, and chlorine can accept one electron in its outer shell.
The number of electrons that an atom of an element has in its outermost shell determines the chemical formula of the compounds formed by atoms such elements.
Magnesium is in group 2, as such it has two electrons in its outermost shell while chlorine in group 17 only accepts one electron in its outermost shell. This one electron will give chlorine an inert gas configuration while the loss of two electrons give magnesium an inert gas configuration.
Therefore; The compound MgCl2 is formed in the ratio of 1:2 because Magnesium has two valence electrons, and chlorine can accept one electron in its outer shell.
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The answer is b) the highest occupied orbital is a “d”orbital.
Transition metals are metals where the highest energy electrons partially fill the d subshells. There are some elements with complete d subshells but on forming cations they have incomplete d subshells.
These transition metals have some properties that are different from the other metals .