All categories

2 years ago

In this equilibrium, hydroxide ion acts as a A) Lewis base B) Bronsted base C) both A & B D) neither A or B

Chemistry

1 answer:

Nata [24]2 years ago

5 0

Answer:

(b) Bronsted base

Explanation:

The hydroxide ion is formed when $H_2O$ losses $H^+$ ion

We know hat the compound which losses $H^+$ is known as the bronsted acid and the ion which is formed after losing $H^+$ is known as bronsted base

As the hydroxide ion is formed after losing the hydrogen ion from $H_2O$ so it is a bronsted base.

You might be interested in

Hey ignore ths i didn mean to post

Sauron [17]

Answer:

WHYYYYYYYYYYYYYYYYYYYYYYY

Explanation:

you said you didnt mean to post it but still typed it and posted it LOLLLLLLLLLLLLLL

7 0

1 year ago

Nvmmmmmmmmmmmmmmmmmmmmmmmmm lol

Vlad1618 [11]

Answer:

coooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooool lol

Explanation:

7 0

2 years ago

Read 2 more answers

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

Calculate the temperature in k of 3.05 moles of gas occupying 3.70 L at 4.12 atm

Gnoma [55]

Answer:

${ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times {10}^{4} \: kelvin }}$

6 0

2 years ago

At some temperature for the equilibrium PX3(g) + X2(g) # PX5(g) the equilibrium constant is 0.74. At the same temperature the eq

mariarad [96]

Answer: The equilibrium constant for $PX_5(g)\rightarrow PX_3(g)+X_2(g)$ is 1.35

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$