Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
Answer:
The final balanced equation is
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
Explanation:
It is given that sodium hydroxide is added to collect the solid nickel(II) hydroxide product
The empirical equation for this statement is
Ni2+ + NaOH --> Ni (OH)2 + Na+
We will first balance the hydroxide molecule. On the right side there are two OH molecules.
Thus, on the left side we will take 2 sodium hydroxide
Ni2+ + 2NaOH --> Ni (OH)2 + Na+
Now we will balance the sodium ion which are 2 in numbers on the left side and 1 on the right side
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
So, the final balanced equation is
Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+
1. True
2. False
3. False
4. True
5. True
6. True
7. All of the above
8. According to their reactivity with other substances
9. Standard temp and pressure
10. Flush your eyes in the eye flush for 15-20 minutes
The answer is: when the aim is to show electron distributions in shells
An orbital notation is more appropriate if you want to show how the electrons of an atom are distributed in each subshell. This is because there are some atoms that have special electronic configurations that aren't obvious in just written configurations.
They all are correct , so with that being said anyone of them can be right
