Fe3N2, also known as Iron (II) nitride, is an ionic compound.
Ionic compounds are compounds that consists of metals and non-metals bonded with ionic bonds. The metal ion gives up electron(s) to the non-metals.
Since iron is a metal and nitrogen is an non-metal, the bond they would form would be an ionic bond. Iron gives up 2 electrons to form iron(II) ion, while nitrogen gains 3 electrons to form nitride ion. Since one iron cannot let a nitrogen gain 3 electrons, so in the compound, there would be 3 iron (ii) ions that has given up 6 electrons in total while 2 nitride ions have gained 6 electrons in total.
It would be false sulfur has 6
I’m pretty sure the correct answer is D.
Answer:
Explanation:
use the equation
moles = mass/mr
=19.9/79.5
=0.250moles of CuO
then do the same for
H = 2.02/1
=2.02
so CuO is the limiting reagent because there is less amount of it.
Hope this helps :)
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Answer:</h3>
2.125 g
<h3>
Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
