Answer:
If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Answer:
covalent bond is the answer
9L!!! You take the moles of N2 in the reaction and then divide by one mole of nitrogen. Then you multiply that by the number of hydrogen moles in the reaction which in this case is 3. This gives you an answer of 3. You then take the number of hydrogen moles and multiply it by the 3L giving you 9L of hydrogen.
Answer:
The electronengativity values of given elements is as follows.
Fluorine - 4
Chlorine -3
Bromine - 2.9
Iodine- 2.5
Explanation:
Electronegativity =consant (I.E-E.A)
The electron affinity and ionization energy values of the given elements is as follows.
(In attachment)
First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".
<u>Fluorine:</u>
![4=constant[1678-(-327.8)]](https://tex.z-dn.net/?f=4%3Dconstant%5B1678-%28-327.8%29%5D)

By using this constant values we can find electronegatvity values of remaining elements.
<u>Chlorine:</u>
![Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1255%2B348.7%5D%3D3.1980%5Csim%203)
Therefore, electronegativity of chlorine is 3.
<u>Bromine:</u>
![Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1138%2B324.5%5D%3D2.91%5Csim%202.9)
Therefore, electronegativity of bromine is 2.9.
<u>Iodine:</u>
![Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1007%2B295.7%5D%3D2.59%5Csim%202.5)
Therefore, electronegativity of iodine is 2.5.