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1 year ago

What is the equation of proton transfer between NaC6H5CO2 and H2O

Chemistry

1 answer:

Anna35 [415]1 year ago

4 0

The equation of the proton transfer is:

NaC₆H₅CO₂ + H₂O ⇄ C₆H₅CO₂H + NaOH

<h3>What are acids?</h3>

Acids according to the Bronsted-Lowry definition, are substances that donates one or more hydrogen ions to another substance in a chemical reaction.

The substances that receive the proton are called bases.

Hydrogen ions are considered as protons. The hydrogen ions produced by acids results in a pH of less than 7 for acid solutions.

Considering the given substances, the equation of the proton transfer is given below:

NaC₆H₅CO₂ + H₂O ⇄ C₆H₅CO₂H + NaOH

The compound NaC₆H₅CO₂, sodium benzoate, is a base and accepts protons from H₂O which serves as the acid.

H₂O serve as the acid by transferring protons to NaC₆H₅CO₂ and then, NaOH is formed.

Learn more about protons at: brainly.com/question/24084759

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Here, the nitrogen atom is the central atom and its outermost shell is surrounded by four electron pairs - one lone pair and three bond pairs. This means that it has a tetrahedral electron pair geometry.

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3 years ago

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What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium

alex41 [277]

Answer:

Approximately $3.06 \times 10^{2}\; \rm g$ (approximately $306\; \rm g$ .)

Explanation:

Calculate the quantity $n$ of lithium phosphate in $V = 2.5\; \rm L$ of this $c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ lithium phosphate solution.

$\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}$ .

Empirical formula of lithium phosphate: $\rm Li_3PO_4$ .

Look up the relative atomic mass of $\rm Li$ , $\rm P$ ,and $\rm O$ on a modern periodic table:

$\rm Li$ : $6.94$ .
$\rm P$ : $30.974$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm Li_3PO_4$ :

$M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}$ .

Calculate the mass of that $n = 2.65\; \rm mol$ of $\rm Li_3PO_4$ formula units:

$\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}$ .

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