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2 years ago

A solution of sodium acetate (ch3coona) in water is weakly basic. a. True b. False

1 answer:

7 0

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!

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According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by ________ valence electron

Mazyrski [523]

According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by__8__ valence electrons.

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2 years ago

It takes 839./kJmol to break a carbon-carbon triple bond. Calculate the maximum wavelength of light for which a carbon-carbon tr

tresset_1 [31]

Answer:

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

Explanation:

It takes 839 kJ/mol to break a carbon-carbon triple bond.

Energy required to break 1 mole of carbon-carbon triple bond = E = 839 kJ

E = 839 kJ/mol = 839,000 J/mol

Energy required to break 1 carbon-carbon triple bond = E'

$E'=\frac{ 839,000 J/mol}{N_A}=\frac{839,000 J}{6.022\times 10^{23} mol^{-1}}=1.393\times 10^{-18} J$

The energy require to single carbon-carbon triple bond will corresponds to wavelength which is required to break the bond.

$E'=\frac{hc}{\lambda }$ (Using planks equation)

$\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}$

$\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm$

$(1 m = 10^9 nm)$

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

6 0

2 years ago

A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?

Xelga [282]

let the actual height of car be x

now, according to question,

$\dfrac{1}{15} = \dfrac{8}{x}$

$x = 15 \times 8$

$x = 120 \: \: cm$

$height \: \: of \: \: car = 120 \: cm \: \: or \: \: 1.2 \: \: m$

5 0

1 year ago

When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction, 2 ch3oh(

umka21 [38]

The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH

Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction

8 0

2 years ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5