Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
Answer:
2) Their kinetic energy increases.
Explanation:
The particles have more kinetic energy because they move around more when they are a liquid.
The mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
<h3>What is the mole ratio of a reaction?</h3>
The mole ratio of a reaction is the ratio in which the reactants and products of a given reaction occur for the reaction to proceed to completion.
The mole ratio of a reaction is also known as the stoichiometry of the reaction.
The equation of the two reactions are given below:


From the equation of the reaction reaction, an equal volume of hydrogen gas will be produced by the two reactions.
Therefore, the mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
Learn more about mole ratio at: brainly.com/question/19099163
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The liquid is blue is an example of qualitative data. Qualitative data is data that can not be represented by numbers. The volume and density are both quantitative data.
Answer:

Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:

So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!