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3 years ago

G brønsted-lowry proton transfer reaction, hydroxide functions as a/an ______.

Chemistry

2 answers:

Katen [24]3 years ago

6 0

In a bronsted lowry proton transfer reaction, the hydroxide functions as a/an <u>proton acceptor.</u>

Bases are the opposite of acids. Bases are basic since they take or accept protons. For example, a Hydroxide ion can accept a proton to form water.

san4es73 [151]3 years ago

3 0

Answer:

Base

Explanation:

A Brønsted-Lowry acid is simply a compound that supplies a hydrogen ion in a reaction. A Brønsted-Lowry base on the other hand, is a compound that accepts a hydrogen ion in a reaction.

Thus, the Brønsted-Lowry definitions of an acid and a base focus on the movement of hydrogen ions in a reaction, rather than on the production of hydrogen ions and hydroxide ions in an aqueous solution.

The hydroxide ion in these compounds accepts a proton (Hydrogen ion) from acids to form water: This means shydroxides functions as a base.

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Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a

musickatia [10]

Answer:

Approximately $1.876 \times 10^{-3}\; \rm mol$ .

Explanation:

Convert both volumes to standard units (that is: liters.)

$10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L$ .
$27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L$ .

Number of moles of $\rm HCl$ initially present (in the $10.00\; \rm mL$ solution at $1.00\; \rm M$ .)

$n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol$ .

Number of moles of $\rm NaOH$ from the titration:

$n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol$ .

$\rm NaOH$ neutralizes $\rm HCl$ at a $1:1$ ratio:

$\rm HCl + NaOH \to NaCl + H_2O$ .

Hence, $n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol$ .

$\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}$ .

3 0

2 years ago

For the reaction COCl2(g)⇌CO(g)+Cl2(g), K= 2.19×10−10 at 373 K

vladimir1956 [14]

Answer:

D) The equilibrium lies far to the left

Explanation:

According to the law of mass action, the equilibrium constant K for the reaction at 373K can be calculated as follows:

K = $\frac{[CO][Cl_{2}]}{[COCl_{2}]}$ = 2.19×10^{-10}

([X] means = concentration of X)

This means that in the equilibrium the concentration of the reactant (that is in the denominator) will be much higher (around 10^{10} fold) than the concentrations of the products (that are in the numerator), and this means that the equilibrium lies far to the left (to the reactants side) as very small amount of product is being formed.

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3 years ago

A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:

Arlecino [84]

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

$\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100$