When a metal bonds with another non-metal an ionic bond is formed
Answer:
The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Explanation:
We are given that
Aqueous solution that contains 22.9% NaOH by mass means
22.9 g NaOH in 100 g solution.
Mass of NaOH(WB)=22.9 g
Mass of water =100-22.9=77.1
Na=23
O=16
H=1.01
Molar mass of NaOH(MB)=23+16+1.01=40.01
Number of moles =![\frac{Given\;mass}{Molar\;mass}](https://tex.z-dn.net/?f=%5Cfrac%7BGiven%5C%3Bmass%7D%7BMolar%5C%3Bmass%7D)
Using the formula
Number of moles of NaOH![(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}](https://tex.z-dn.net/?f=%28n_B%29%3D%5Cfrac%7BW_B%7D%7BM_B%7D%3D%5Cfrac%7B22.9%7D%7B40.01%7D)
![n_B=0.572moles](https://tex.z-dn.net/?f=n_B%3D0.572moles)
Molar mass of water=16+2(1.01)=18.02g
Number of moles of water![(n_A)=\frac{77.1}{18.02}](https://tex.z-dn.net/?f=%28n_A%29%3D%5Cfrac%7B77.1%7D%7B18.02%7D)
![n_A=4.279 moles](https://tex.z-dn.net/?f=n_A%3D4.279%20moles)
Now, mole fraction of NaOH
=![\frac{n_B}{n_B+n_A}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_B%7D%7Bn_B%2Bn_A%7D)
![=\frac{4.279}{0.572+4.279}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B4.279%7D%7B0.572%2B4.279%7D)
=0.882
Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882
Answer:
the empirical (lowest raios) is
C2H4Cl
Explanation:
A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.
What is the empirical formula of this compound?
the compound has ONLY C, H, and Cl
the % Cl = 100% - 24.27% -4.03% = 71.7%
in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H
the number of moles are Cl=71.7/70.91 =1.01= ~ 1
C = 24.27/12.0 = 2.02 =~ 2
H = 403/1.01 = 3.97 =~ 4
so the empirical (lowest raios) is
C2H4Cl
It is an alkaline earth metal.