The answer is Hydrogyn bonding. It keeps the water molocules bonded together and in a liquid state, without it it'd be in a gashious state.
Answer:
I believe that there are 70.2
Explanation:
There are 2 atoms in sodium chloride so I x 35.1 by 2 and got 70.2
The volume of H₂ : = 15.2208 L
<h3>Further explanation</h3>
Given
Reaction
2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)
34.0g of As
Required
The volume of H₂ at STP
Solution
mol As (Ar = 75 g/mol) :
= mass : Ar
= 34 g : 75 g/mol
= 0.453 mol
From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :
=3/2 x mol As
=3/2 x 0.453
= 0.6795
At STP, 1 mol = 22.4 L, so :
= 0.6795 x 22.4 L
= 15.2208 L
Octahedral is the geometry of the complex ion which has d2sp3 hybridization.
The octahedral has eight faces hence the prefix octa. Octahedral molecular geometry describe the shape of compound with six atom or groups of atoms or ligands symmetrically arranged around a central atom.The octahedron is one of the platonic solids.
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:

Moles of ethane = 
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g


The percent yield of chloro-ethane in the reaction is 82.98%.