Answer:
Percentage Yield = 56.9% (to 3 significant figures)
Explanation:
Firstly you need to calculate the amount of moles of Na used. Moles = Mass/Mr, so moles of Na = 1.72x10^-3/23. This gets you 7.47826087x10^-5 moles.
The mole ratio of Na:Na2O is 2:1 so the moles of Na2O will be the moles of Na divided by 2. This gets you 3.739130435x10^-5 moles.
Mass = Moles x Mr
The Mr of Na2O = 62
Mass = 3.739130435x10^-5 x 62 = 2.31826087x10^-3g
Percentage Yield = (Actual Yield / Theoretical Yield) x 100
Percentage Yield = (1.32x10^-3 / 2.31826087x10^-3) x 100
Percentage Yield = 56.93923481%
Percentage Yield = 56.9% (to 3 significant figures)
That means that one (s) type orbital and three (p) type orbitals (all of the p orbitals in the valence set) have been mixed together to form four NEW atomic orbitals, all of equal energy, that point in specific directions (to the four vertices of a tetrahedron) the directions that the orbitals point is THE point of doing this because that enables you to explain why certain compounds like CH4 have a tetrahedral structure. It's kind of circular logic that you have to basically accept, but it makes sense if you delve deeper into the theory
This is answer open this image I am not sure is this wright
True because freshwater is regular water
Answer:

Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant

3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.

From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)