Chemistry, 29.07.2019 12:30 moneyyfletcher

Asolution of sodium acetate (ch3coona) in water is weakly basic. a. true b. false

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Answer from: stacysadousky

The claim that a solution of sodium acetate in water is weakly basic is True.

What is Sodium Acetate?

Sodium acetate (CH3COONa) is a translucent deliquescent salt that can be used in a variety of ways.

It is used for the neutralization of sulfuric acid waste streams.Serves as Pickling agent in Chrome-tanning

The reaction of Sodium acetate with water results in the hydrolysis of salt.

This is because the sodium acetate is weakly basic and therefore splits into sodium and acetate ions.

The chemical reaction of sodium acetate with water can be expressed as follows:

$\mathbf{CH_3COONa \to Na^{+}_{(aq)} + CH_3COO^-_{(aq)}}$

$\mathbf{CH_3COO^{-}_{(aq)} +H_2O _{(l)} \to CH_3COOH +OH^{-} _{(aq)}}$

Learn more about sodium acetate here:

link

Answer from: kruzyoungblood8

Hello!

The statement that a solution of sodium acetate (CH₃COONa) is weakly basic is true:

Sodium acetate is the conjugate base of Acetic Acid. When sodium acetate is dissolved in water, it follows the equation that is shown below:

CH₃COONa(s) → CH₃COO⁻(aq) + Na⁺(aq)

Now the Acetate (CH₃COO⁻) ion, has an equilibrium in water to produce hydroxyl (OH⁻) ions and (Acetic Acid CH₃COOH)

CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻

This is a weak equilibrium, and the hydroxyl ions cause the solution to be weakly basic.

Have a nice day!

Answer from: Quest

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

$^{222}_{86} rn+ ^{4}_{2} he$ →

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.