The complete table is shown in figure
a) NH3 is polar as the bonds between N and H are polar. Due to asymmetry in the molecule the molecule is polar
The shape of molecule is trigonal pyramidal while its electronic geometry is tetrahedral.
b) CO2: it is a non polar molecule with polar bonds. The molecule becomes non polar as the dipole moment cancel each other. [Dipole moment is a vector quantity]
The shape is linear.
<span>just find the percent mass of oxygen in sucrose again. and then multiply that by 50.00.</span>
Molecular equation
Hg₂(NO₃)₂ (aq) + KI(aq) ⇒Hg₂I₂(s) + 2KNO₃(aq)
Total Ionic equation
Hg²⁺(aq) + 2NO³⁻(aq) + 2K⁺aq) ⇒Hg₂I₂(s) + 2K⁺(aq) + NO³⁻ (aq)
Net Ionic equation
Hg²⁺(aq) + 2I⁻(aq) ⇒ Hg₂I₂(s)
<h3>What is the molecular equation?</h3>
Sometimes, a balanced equation is all that is used to refer to a chemical equation. Any ionic substances or acids are represented using their chemical formulas as neutral compounds in a molecular equation. Each substance's state is described in parenthesis after the formula. A complete ionic equation also contains the spectator ions, whereas a net ionic equation just displays the chemical species that are involved in a reaction.
The steps listed below can be used to determine the net ionic equation for a specific reaction:
Include the states of each chemical in the balanced molecular equation for the reaction.
To know more about the molecular equation, visit:
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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.
