V is more reactive, hope this helps!
Answer:
844.4cm³
Explanation:
Using Boyle's law equation;
P1V1 = P2V2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (cm³)
V2 = final volume (cm³)
According to the information in this question,
V1 = 650cm³
P1 = 760mmHg
P2 = ?
V2 = 10% reduction of V1
10% of 650 = 10/100 × 650
65
V2 = 650 - 65 = 585cm³
Using P1V1 = P2V2
P2 = P1V1 ÷ V2
P2 = (760 × 650) ÷ 585
P2 = 494000 ÷ 585
P2 = 844.44
final pressure (P2) = 844.4cm³
Answer:
in metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions. ... Metals are shiny.
Explanation: here is your answer for your please rate me the most brainlest in let me know if right
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!
