The correct answer is option 2. Chlorine gas molecules are held together by strong covalent bonds. This type of bond requires an ample amount of energy in order to break it and turn gas into liquid. That is why, high pressure and low temperature is needed.
M(H₂O) = 97,2 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 97,2 g ÷ 18 g/mol.
n(H₂O) = 5,4 mol.
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 5,4 mol · 6,023·10²³ 1/mol.
N(H₂O) = 3,25·10²⁴ molecules of water.
n - amount of substance.
Na - Avogadro number.
<span>Read the Fig 4 description. From there you can conclude that it does scatter light, because its ot a solution. So it must be a mixture, the substances are not dissolved so therfore they aren't a solution.</span>
Answer:
Keq=0.866
Explanation:
Hello,
In this case, the undergone chemical reaction is:
In such a way, since 0.0055 mol of N₂O₄ remains in the flask, one infers that the reacted amount (
) was:
In addition, the produced amount of NO₂ is:
Finally, considering the flask's volume, the equilibrium constant is then computed as follows:
Best regards.
<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it.
Crude appearance.
18 18 0
F --> O + e
9 8 1
Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts.
The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons.
Followed by a right arrow to indicate the direction the reaction is going.
Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons.
Followed by a plus sign to indicate more.
Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>
