<h2>Hello!</h2>
The answer is: C2H6
<h2>Why?</h2>
First, we need to find the empirical formula of the compound:
Looking for the relative atomic mass for each element:
Finding the number of moles for each element:
Then, we need to determinate the mole ratio by dividing each result into the smallest number:
So, the empirical formula would be CH3
We are given the molar mass of the compound is equal to 30 g/mol
We need to compare the empirical formula molecular mass and the molecular formula molar mass in order to find the number of atoms of the compound:
Empirical formula molar mass:
Molecular formula molar mass = 30.0 g/mol
So, by dividing the molecular formula molar mass by the empirical formula molar mass we have the number of atoms of the compound:
Therefore,
The molecular formula will be: (CH3)2= C2H6
Have a nice day!
Answer:
A. iodine
C. fluorine
F. bromine
Explanation:
Ionic bonds occur mostly between metals and non-metals. Usually, a wide electronegativity difference is preferred between the two atoms. This makes one atom more desirous to gain electron and other more willing to donate electrons.
To have Zn forming a compound in the ratio of 1 to 2, the combining power must be similar to this.
The dominant oxidation state of Zn is the is the +2 state.
The other combining atoms must have the ability to recieve the two electrons.
The halogens fit perfectly into this picture. They need just an electron to attain nobility. They are also highly electronegative. If two halogens combines with the Zn, then the ionic bond will result.
The halogens are fluorine,chlorine, bromine, iodine and astatine.
They will form these compounds:
ZnF₂, ZnBr₂ and ZnI₂
Answer:
1533.6 kg NO
Explanation:
The reaction that takes place is:
First we <u>convert the masses of ammonia (NH₃) and oxygen gas (O₂) into moles</u>, using<em> their respective molar masses</em>:
- NH₃ ⇒ 869 kg ÷ 17 kg/kmol = 51.12 kmol NH₃
- O₂ ⇒ 2480 kg ÷ 32 kg/kmol = 77.5 kmol O₂
77.5 kmol of O₂ would react completely with (77.5 kmol O₂ * ) 62 kmol of NH₃. There are not as many kmol of NH₃, so NH₃ is the limiting reactant.
Now we <u>calculate how many kmol of NO are produced</u>, using the <em>limiting reactant moles</em>:
- 51.12 kmol NH₃ * = 51.12 kmol NO
Finally we <u>convert kmol of NO to mass</u>, using its<em> molar mass</em>:
- 51.12 kmol NO * 30 kg/kmol = 1533.6 kg NO