Answer:
<em>The answer to your question is </em><em>c. H2O</em>
Explanation:
<em>a. CaCO3 ----> Calcium Carbonate is obviously not a binary compound because it contains more than two elements. </em>
<em>b. CuSO4 ----> CuSO4 stands for copper(II) sulfate. Sulfate is the polyatomic ion SO4 -2.</em>
<em>c. H2O ----> Binary molecular compounds are composed of only two elements. Examples are H2O, NO, SF6 etc. </em>
<em>d. Pb(NO3)2 ----> Lead(II) nitrate is an inorganic compound with the chemical formula Pb(NO3)2.</em>
<em>I hope this helps and have a great day!</em>
Answer : The formula of hexaaquamanganese(II)sulfate is ![[Mn(H_2O)_6]SO_4](https://tex.z-dn.net/?f=%5BMn%28H_2O%29_6%5DSO_4)
.
Explanation :
Rules for writing formulas of coordination complexes :
- Central metal is written first and then the ligands are written along with their prefixes.
- The anion ligands are written before the neutral ligands.
- If there are more than one anion ligands or neutral ligands then they are written in alphabetical order.
The given name of complex is, hexaaquamanganese(ii) sulfate.
In this, the central metal is manganese which is written as Mn and the ligand is water (which is a neutral ligand) is written as 
and an anion is sulfate which is written as 
.
Water is a neutral ligand and sulphate has a charge of -2. As the overall complex is neutral the metal bears a +2 charge which is represented in Roman numeral after the metal.
Therefore, the formula of hexaaquamanganese(II)sulfate is written as .
Answer: 61%
The reaction equation should be
CaF2 + H2SO4 → 2HF + CaSO4
For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g
The percentage yield of the reaction would be: 35g/57.43g= 60.94%
16.94/18=.9411111
sig figs: 0.9411 mole of water
