<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
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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If you really keep an eye on the flow chart, the only ions you can consider as being "Definitely not present" are: Cr3+, Fe3+, and Zn2+. The rest of the ions should be considered under "Possibly present", as we cannot conclude if any of the ions are "Definitely present".
Answer:
Acceleration
Explanation:
Acceleration Is the rate change in velocity
Answer:
This metal could be the aluminium with a specific heat of 
Explanation:
A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

Finally, the specific heat of this metal is

The aluminium could be the metal, its specific heat is similar to that found in this problem.
Finally, we can conclude that this metal could be the aluminium with a specific heat of 