Cobalt has an atomic number of 27, which means that its atoms have 27 protons in their nuclei. In a neutral cobalt atom, there is also 27 electrons. The ground state<span>electron </span>configuration<span> of cobalt is [Ar]3d74s2 .</span>
Answer:
Copper is element number 29
4th row
11th column
21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced.
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.