Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high <u>electronegativity</u> as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an <u>inductive effect</u>. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be <u>weaker</u> and the compound will be more acid (because is easier to produce the hydronium ion
).
With this in mind, for A in the last compound, we have <u>2 Br atoms</u> near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have <u>more acidity</u>. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the <u>position</u> of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a <u>higher inductive effect</u> and more <u>acidity</u>.
See figure 1
I hope it helps!
Answer:
Adding sodium or potassium hydroxide in amounts sufficient to convert all the H2SO4 into Na2SO4 would approximately neutralize the solution. The error would be the result of the imbalance between the basicity of the hydroxide and the acidity of the bisulfate (HSO4) anion. An adjustment in concentration would have to be made to achieve an accurate approximate pH of 7. But then you didn’t ask how much we would need to add.
Explanation:
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To convert take 1.8 * degrees C + 32
1.8 * 35 + 32
63 + 32 = 95 so C
Answer:
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Explanation:
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%