Explanation:
Magnesium reacts with dilute hydrochloric acid in a conical flask which is connected to an inverted measuring cylinder in a trough of water. The volume of hydrogen gas produced is measured over a few minutes, and the results are used to plot a graph
This is intended as a class practical. It is best if the students work in pairs because setting up and starting the experiment requires more than one pair of hands. One student can add the magnesium ribbon to the acid and stopper the flask, while the other starts the stopclock. During the experiment, one student can take the readings while the other records them. The experiment itself takes only a few minutes. But allow at least 30 minutes to give students time to set up, take readings and draw graph.
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A gas because if you freeze water, it turns into a shape as where gas can’t be frozen
Answer:
These are the chemical names and molecular formula of the compounds formed from these elements:
- Rubidium + Bromine = <em>Rubidium bromide, RbBr</em>
- Strontium + Sulphur = <em>Strontium sulfide, SrS</em>
- Barium and Chlorine = <em>Barium chloride, </em>
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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