Answer:
B) +2; 6; -2
Explanation:
In order to know what the coordination number is, all we need to do is to count the number of ligands present in the coordination sphere. Coordination number refers to the number of ligands in the coordination sphere of a given complex. The coordination number here is 6.
The counter ion here is 2Na+. It means that the overall charge on the complex must be -2 because the magnitude of charge on the complex ion is exactly balanced by the number of counter ions.
The charge on the metal (x) must now be;
x + 0 + 4(-1) = -2
x - 4 = -2
x = -2 + 4
x = +2
Note that NH3 is designated as zero because it is a neutral molecule. Each NCS^- ion has a charge magnitude of -1.
Though all the options aren't visible i think she tried to find out the force friction at different surfaces.
Answer:
Here's what I get
Explanation:
![\rm A \xrightarrow{k_{1}}I \xrightarrow{k_{2}} P](https://tex.z-dn.net/?f=%5Crm%20A%20%5Cxrightarrow%7Bk_%7B1%7D%7DI%20%5Cxrightarrow%7Bk_%7B2%7D%7D%20P)
(a) Plot I against t
I assume that Equation 22.40 is something like
![\text{[I]} = \dfrac{k_{1}\text{[A]}_{0}}{k_{2} - k_{1}} \left(e^{-k_{1}t} - e^{-k_{2}t} \right )](https://tex.z-dn.net/?f=%5Ctext%7B%5BI%5D%7D%20%3D%20%5Cdfrac%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%7Bk_%7B2%7D%20-%20k_%7B1%7D%7D%20%5Cleft%28e%5E%7B-k_%7B1%7Dt%7D%20-%20e%5E%7B-k_%7B2%7Dt%7D%20%5Cright%20%29)
If the initial conditions are
[A]₀ = 1.0 mol·dm⁻³; k₁ = 10 s⁻¹; k₂ = 1 s⁻¹
The equation becomes
![\text{[I]} = \dfrac{10}{-9} \left(e^{-10t} - e^{-t} \right )](https://tex.z-dn.net/?f=%5Ctext%7B%5BI%5D%7D%20%3D%20%5Cdfrac%7B10%7D%7B-9%7D%20%5Cleft%28e%5E%7B-10t%7D%20-%20e%5E%7B-t%7D%20%5Cright%20%29)
The graph of [I] vs t is shown in Fig 1.
(b) Increasing k₂/k₁ ratio
In Fig. 2, I added the same plots, but with k₂ = 3, 11, and 51 s⁻¹ (black, green and purple).
The graphs show that, as k₂ becomes increasingly greater than k₁, the maximum concentration of I becomes smaller and the graph becomes (except for the very beginning) a flat line.
Thus, the approximation that
![\mathbf{\dfrac{\text{d[I]}}{\text{d}t}=0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Ctext%7Bd%5BI%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%3D0%7D)
becomes increasingly valid.
Answer:THIS IS NOT THE ANSWER
is this for the chemistry A final 10th grade for connexus (unit7 lesson 2)?
Explanation:
Answer:
![\boxed{\text{6.937 u; X = Li}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B6.937%20u%3B%20X%20%3D%20Li%7D%7D)
Explanation:
1. Write the unbalanced equation
X + N₂ ⟶ X₃N
2. Balance the equation and gather all the data.
MM: 28.01
6X + N₂ ⟶ 2X₃N
m/g 1.486 1.000
3. Calculate the moles of N₂
![\text{Moles of N}_{2} = \text{1.000 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{0.035 70 mol N}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20N%7D_%7B2%7D%20%3D%20%5Ctext%7B1.000%20g%20N%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20N%7D_%7B2%7D%7D%7B%5Ctext%7B28.01%20g%20N%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B0.035%2070%20mol%20N%7D_%7B2%7D)
4. Calculate the moles of X
The molar ratio is 6 mol X: 1 mol N₂
![\text{Moles of X} = \text{0.035 70 mol N}_{2} \times \dfrac{\text{6 mol X}}{\text{1 mol N}_{2}} = \text{0.2142 mol X}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20X%7D%20%3D%20%5Ctext%7B0.035%2070%20mol%20N%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B6%20mol%20X%7D%7D%7B%5Ctext%7B1%20mol%20N%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B0.2142%20mol%20X%7D)
5. Calculate the molar mass of X
![\text{Molar mass} = \dfrac{\text{mass}}{\text{moles}} = \dfrac{\text{1.486 g}}{\text{0.2142 mol}} = \text{6.937 g/mol}\\\text{The molar mass of X is 6.937 g/mol, so the atomic mass $\boxed{\textbf{6.937 u}}$}](https://tex.z-dn.net/?f=%5Ctext%7BMolar%20mass%7D%20%3D%20%5Cdfrac%7B%5Ctext%7Bmass%7D%7D%7B%5Ctext%7Bmoles%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B1.486%20g%7D%7D%7B%5Ctext%7B0.2142%20mol%7D%7D%20%3D%20%5Ctext%7B6.937%20g%2Fmol%7D%5C%5C%5Ctext%7BThe%20molar%20mass%20of%20X%20is%206.937%20g%2Fmol%2C%20so%20the%20atomic%20mass%20%24%5Cboxed%7B%5Ctextbf%7B6.937%20u%7D%7D%24%7D)
6. Identify X.
![\text{X has an atomic mass of 6.937 u, so $\boxed{\textbf{X = Li}}$ (at. mass 6.94 u)}](https://tex.z-dn.net/?f=%5Ctext%7BX%20has%20an%20atomic%20mass%20of%206.937%20u%2C%20so%20%24%5Cboxed%7B%5Ctextbf%7BX%20%3D%20Li%7D%7D%24%20%28at.%20mass%206.94%20u%29%7D)