They held together by metallic bonds.
Hopefully this has helped! :)
We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
M1V1 = M2V2
0.266 M x V1 = 0.075 M x 150 mL
V1 = 42.29 mL
Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.
Answer:
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
Explanation:
Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.
Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
Now, we have to balance Li by multiplying LiNO₂ by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
The equation is now balanced.
Answer: the answer should and most definitely be D.
Explanation: I mean think about it after a while only a few radioactive nuclei are left which means it will dye down after a while which also makes it very boring hope this helps :)
Example:
Mass = ?
Density = 25 g/mL
Volume = 5 mL
therefore:
d = m / V
25 = m / 5
m = 25 x 5
m = 125 g
hope this helps!
