Answer:
1) mass and type of material
2) type of material
3) temperature
Explanation:
Explanation:
Let the mass of the compound be 100g.
Mass of Sulfur = 96g.
=> Moles of Sulfur
= 96g / (32.07g/mol) ≈ 3
Mass of Hydrogen = 4g.
=> Moles of Hydrogen
= 4g / (1.008g/mol) ≈ 4
Mole ratio of Sulfur to Hydrogen = 3 : 4,
Empirical formula of compound = S3H4.
a) 1 mole of Ne
b) i/2 mole of Mg
c) 1570 moles of Pb.
d) 2.18125*10^-13 moles of oxygen.
Explanation:
The number of moles calculated by Avogadro's number in 6.23*10^23 of Neon.
6.23*10^23= 1/ 6.23*10^23
= 1 mole
The number of moles calculated by Avogadro's number in 3.01*10^23 of Mg
3.2*10^23=1/6.23*10^23
= 1/2 moles of Pb.
Number of moles in 3.25*10^5 gm of lead.
atomic weight of Pb=
n=weight/atomic weight
= 3.25*10^5/ 207
= 1570 moles of Pb.
Number of moles 4.50 x 10-12 g O
number of moles= 4.50*10^-12/16
= 2.18125*10^-13 moles of oxygen.
Answer:
No, no precipitate is formed.
Explanation:
Hello there!
In this case, since the reaction between silver nitrate and sodium acetate is:
In such a way, we can calculate the concentration of silver and acetate ions in the solution as shown below, and considering that the final total volume is 50.00 mL or 0.0500 L:
![[Ag^+]=\frac{20.00mL*0.077M}{50.00mL}=0.0308M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D%5Cfrac%7B20.00mL%2A0.077M%7D%7B50.00mL%7D%3D0.0308M)
![[C_2H_3O_2^-]=\frac{30.00mL*0.043M}{50.00mL}=0.0258M](https://tex.z-dn.net/?f=%5BC_2H_3O_2%5E-%5D%3D%5Cfrac%7B30.00mL%2A0.043M%7D%7B50.00mL%7D%3D0.0258M)
In such a way, we can calculate the precipitation quotient by:
![Q=[Ag^+][C_2H_3O_2^-]=0.0308*0.0258=7.95x10^{-4}](https://tex.z-dn.net/?f=Q%3D%5BAg%5E%2B%5D%5BC_2H_3O_2%5E-%5D%3D0.0308%2A0.0258%3D7.95x10%5E%7B-4%7D)
Which is smaller than Ksp and meaning that the precipitation does not occur.
Regards!
