Ionic bonds is formed between a metal and a nonmetal because the metal transfers it's valence electrons to the nonmetal. Electrons doesn't transfer between two nonmetal because both are negatively charged.
Answer: 1.2 M
Explanation:
Given that:
volume of NaCl = 392 mL
Convert volume in milliliters to liters
(since 1000 mL = 1L
392 mL = 392/10000 = 0.392 L)
Concentration of NaCl = ?
Amount of moles of NaCl = 0.47 moles
Recall that the concentration of a solution depends on the amount of solute dissolved in a particular volume of solvent.
I.e Concentration in mol/L =
Amount in moles / Volume in liters
= 0.47 moles / 0.392 L
= 1.199 mol/L (Round up to the nearest tenth as 1.2 M)
Note that molarity is the same as concentration in moles per litres.
Thus, the molarity of 392 mL of solution that contains 0.47 mol NaCl is 1.2 M
Answer:
No, no precipitate is formed.
Explanation:
Hello there!
In this case, since the reaction between ammonium sulfide and aluminum nitrate is:
In such a way, we can calculate the concentration of aluminum and sulfide ions in the solution as shown below, and considering that the final total volume is 140.00 mL:
![[Al^3^+]=\frac{120.00mL*0.0082M}{140.00mL}=0.00703M](https://tex.z-dn.net/?f=%5BAl%5E3%5E%2B%5D%3D%5Cfrac%7B120.00mL%2A0.0082M%7D%7B140.00mL%7D%3D0.00703M)
![[S^2^-]=\frac{20.00mL*0.0090M}{140.00mL}=0.00129M](https://tex.z-dn.net/?f=%5BS%5E2%5E-%5D%3D%5Cfrac%7B20.00mL%2A0.0090M%7D%7B140.00mL%7D%3D0.00129M)
In such a way, we can calculate the precipitation quotient by:
![Q=[Al^3^+]^2[S^2^-]^3=(0.00703)^2(0.00129)^3=1.05x10^{-13}](https://tex.z-dn.net/?f=Q%3D%5BAl%5E3%5E%2B%5D%5E2%5BS%5E2%5E-%5D%5E3%3D%280.00703%29%5E2%280.00129%29%5E3%3D1.05x10%5E%7B-13%7D)
Which is smaller than Ksp and meaning that the precipitation does not occur.
Regards!
Answer:6 significant figures
Explanation:
