We'll look at what happens<span> when you </span>dissolve ionic<span> and covalent </span>compounds<span> in </span>water<span>. </span>Ionic compounds<span> break apart into the </span>ions<span> that make them up, a process called dissociation, while covalent </span>compounds only break into the molecules, not the individual atoms.<span>When you immerse an </span>ionic compound<span> in </span>water<span>, the ions are attracted to the </span>water <span>molecules, each of which carries a polar charge. If the attraction between the ions and the </span>water <span>molecules </span>is<span> great enough to break the bonds holding the ions together, the compound </span><span>dissolves</span>
The answer is A.) because each carbon atom has six electrons meaning that there are six protons.
Answer:
3.24 mol/L
Explanation:
Given that:
mass of Boron triiodide = 6664 grams
molar mass of BI_3 = 391.52 g/mol
Recall that:
number of moles = mass/molar mass
∴
number of moles = 6664 g /391.52 g/mol
number of moles = 17.02 mol
Also;
Molarity = moles for solute/liter for solution
= 17.02 mol/5.25 L
= 3.24 mol/L
Answer:
The concentration of COF₂ at equilibrium is 0.296 M.
Explanation:
To solve this equilibrium problem we use an ICE Table. In this table, we recognize 3 stages: Initial(I), Change(C) and Equilibrium(E). In each row we record the <em>concentrations</em> or <em>changes in concentration</em> in that stage. For this reaction:
2 COF₂(g) ⇌ CO₂(g) + CF₄(g)
I 2.00 0 0
C -2x +x +x
E 2.00 - 2x x x
Then, we replace these equilibrium concentrations in the Kc expression, and solve for "x".
![Kc=8.30=\frac{[CO_{2}] \times [CF_{4}] }{[COF_{2}]^{2} } =\frac{x^{2} }{(2.00-2x)^{2} } \\8.30=(\frac{x}{2.00-2x} )^{2} \\\sqrt{8.30} =\frac{x}{2.00-2x}\\5.76-5.76x=x\\x=0.852](https://tex.z-dn.net/?f=Kc%3D8.30%3D%5Cfrac%7B%5BCO_%7B2%7D%5D%20%5Ctimes%20%5BCF_%7B4%7D%5D%20%7D%7B%5BCOF_%7B2%7D%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%282.00-2x%29%5E%7B2%7D%20%7D%20%5C%5C8.30%3D%28%5Cfrac%7Bx%7D%7B2.00-2x%7D%20%29%5E%7B2%7D%20%5C%5C%5Csqrt%7B8.30%7D%20%3D%5Cfrac%7Bx%7D%7B2.00-2x%7D%5C%5C5.76-5.76x%3Dx%5C%5Cx%3D0.852)
The concentration of COF₂ at equilibrium is 2.00 -2x = 2.00 - 2 × 0.852 = 0.296 M
