Cu (s) + 2AgNO3 ---> 2Ag + Cu(NO3)2
=13.5 moles Silver Nitrate
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.
Answer:
The C4+ ion has the smallest ionic radius
Explanation:
In the periodic table, the size of an atom increases on going downwards in a group.
On the other hand, in a period, size decreases on going from left to right
The size of cation is smaller than its parent atom.
Lithium-ion (Li+) is the most left so it has the biggest radius.
More to the right ( this means a smaller ionic radius) , we have Berylium-ion (Be2+).
More to the right, so smaller than Be2+, we have B3+
The ion the most to the right, we have C4+
The C4+ ion has the smallest ionic radius
Li+ > Be2+ > Be3+ > C4+
Answer:
c. 2 OH⁻(aq) + 2 H⁺(aq) ⇒ 2 H₂O(l)
Explanation:
Step 1: Write the molecular equation
The molecular equation includes all the molecular species.
H₂A(aq) + 2 NaOH(aq) ⇒ Na₂A(aq) + 2 H₂O(l)
Step 2: Write the complete ionic equation
The complete ionic equation includes all the ions and the molecular species.
2 H⁺(aq) + A²⁻(aq) + 2 Na⁺(aq) + 2 OH⁻(aq) ⇒ 2 Na⁺(aq) + A²⁻(aq) + 2 H₂O(l)
Step 3: Write the net ionic equation
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
2 OH⁻(aq) + 2 H⁺(aq) ⇒ 2 H₂O(l)
When 2 moles burn 6 moles of O2 is produced !
so for 16 moles , (6/2)×16= 48 moles will be produced !
so answer is A , 48 moles !
