In formation of a Type II Binary Compound, the metal atom present is<span>
NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.
The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
<h3>
Position of 18o-labeled methanol in the products</h3>
The 18O label will appear at position b in the product as indicated in the image.
This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).
While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.
Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
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Answer:
Explanation:
From the correct question above:
The reaction can be represented as:
From the above reaction; the ICE table can be represented as:
I (mol/L) 0.086 0.28 0 0
C -4x -3x +2x +6x
E 0.086 - 4x 0.28 - 3x +2x +6x
At equilibrium;
The water vapor =
Replacing the value of x, we have:
Answer:
=7.89013× 10^47 moles of zinc
Explanation:
1 atom contains 6.023×10^23 moles.
1.31×10^24 atoms of zinc contain6.023×10^24×1.31×10^.24
The greater amount is the solvent and the lesser amount is the solute.
Hence ethanol(200g) which is the greater amount is the solvent here.
And water (145g) which is lesser is the solute here.